text:
How to find the second order perturbation to wave function?
Today, I'm looking for how to find the 2nd perturbation to the base in Rayleigh Schrödinger Perturbation Theory (RSPT).
SETUP
Starting from the 2nd order perturbation in Dirac's notation:
\begin{equation}
\hat{H}^0 |n^2\rangle + \hat{V}  |n^1\rangle  = E_n^0  |n^2\rangle + E_n^1 |n^1\rangle  +  E_n^2  |n^0\rangle
\end{equation}
Multiply to the left with $\langle k^0 |$
\begin{equation}
\langle k^0| \hat{H}^0 |n^2\rangle + \langle k^0| \hat{V}  |n^1\rangle  = \langle k^0|  E_n^0  |n^2\rangle + \langle k^0|  E_n^1 |n^1\rangle  + \langle k^0|  E_n^2  |n^0\rangle
\end{equation}
The hermicity of the Hamilton operator:
\begin{equation}
E_k^0 \langle k^0 |n^2\rangle + \langle k^0| \hat{V}  |n^1\rangle  = E_n^0 \langle k^0  |n^2\rangle + E_n^1 \langle k^0 |n^1\rangle  + E_n^2 \langle k^0  |n^0\rangle
\end{equation}
Cancel the third term on the right
\begin{equation}
E_k^0 \langle k^0 |n^2\rangle + \langle k^0| \hat{V}  |n^1\rangle  = E_n^0 \langle k^0  |n^2\rangle + E_n^1 \langle k^0 |n^1\rangle
\end{equation}
Gathering same inner products:
\begin{equation}
\langle k^0| \hat{V}  |n^1\rangle - E_n^1 \langle k^0 |n^1\rangle = ( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle
\end{equation}
\begin{equation}
( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle  = \langle k^0| \hat{V}  |n^1\rangle - E_n^1 \langle k^0 |n^1\rangle
\end{equation}
Replacing $| N^1 \rangle$ and $E_N^1$ with the first order perturbation to the base and to the energy:

\begin{equation}
|n^1\rangle =\sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}
\end{equation}
    \begin{equation}
E_N^{1} = \langle n^0 | \hat{V} | n^0 \rangle
\end{equation}

\begin{equation}
( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle   = \langle k^0| \hat{V}  \sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}  - E_n^1 \langle k^0 \sum_{m \neq n} |m^0 \rangle \frac{\langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}
\end{equation}
\begin{equation}
( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle   = \sum_{m \neq n} \frac{ \langle k^0| \hat{V} |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}  - \langle n^0 | \hat{V} | n^0 \rangle \sum_{m \neq n}  \frac{\langle k^0 |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}
\end{equation}
At this point, I'm not sure if the bracket algebra was done properly not even forward.
Then ... are another some algebra steps that I cannot figure out the properly path to find the respective coefficient:
$$\langle k^0 | n^2 \rangle$$
In order to replace it in:
\begin{equation}
|n^2\rangle =\sum_{k \neq n} |k^0 \rangle \frac{\langle k^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_k^0 ???}
\end{equation}
QUESTION
Please, Would someone help me out to show me how to get the properly 2nd perturbation to the wave function?

I'm not sure what you mean by "base". Do you want an expression for the second order contribution to the energy or the second order correction to the wave function?

to the wave function and base is a synonym in physics jargon

Disclaimer This post is some kind of a legacy post. Find the notation used in the question in the other answer. I added this proof as I was not entirely certain I understood the notation correctly. As it turned out, I did not. As a result I posted the complete derivation of RSPT up to second order, trying to guide anyone through it using a different notation.$\require{cancel}$
Obviously, I will come back to the initial problem only at the end, if you are impatient, just skip ahead.
Please also note, that I removed the parts that became redundant when I posted the second answer. If you are interested in legacy versions, just look them up in the history.
Also, this page is heavily loaded with $\mathcal{M}^{ath}\mathrm{J_{a}X}$, give it time to load completely.

In Rayleigh Schrödinger perturbation theory, we set up the Hamiltonian as $$\mathbf{H}=\mathbf{H_0}+\lambda\mathbf{H'}.\tag1$$
The perturbed Schrödinger equation is therefore $$\mathbf{H}\Psi=W\Psi_\mathrm{ST}.\tag2$$
You can develop this into Taylor series:
\begin{align}
W                &=
     \lambda^0W_0    + \lambda^1W_1    + \lambda^2W_2    + \cdots
     &&= \sum_{i=0}\lambda^i W_i\tag3\\
\Psi_\mathrm{ST} &=
     \lambda^0\Psi_0 + \lambda^1\Psi_1 + \lambda^2\Psi_2 + \cdots
     &&= \sum_{i=0}\lambda^i \Psi_i\tag4\\
\end{align}
Your unperturbed reference system is a complete solution denoted as
$$\mathbf{H}_0\Phi_i = E_0\Phi_i,\tag5$$
where $i\in\mathbb{N}$.
You chose your system to be intermediate normalised.
$$\langle\Psi_\mathrm{ST}|\Phi_0\rangle = 1\tag6$$
Since all your solutions of the unperturbed system are orthogonal,
$\langle\Phi_i|\Phi_j\rangle=\delta_{ij},$
the same holds for your perturbed system, hence
$$\langle\Psi_{i\neq0}|\Phi_0\rangle = 0.\tag7$$
Now we can combine $(1)$ through $(4)$ and gather them according to their order.
\begin{align}
\lambda^0&:&
  \mathbf{H}_0\Psi_0
    &= W_0\Psi_0\tag{8a}\\
\lambda^1&:&
  \mathbf{H}_0\Psi_1 + \mathbf{H'}\Psi_0
    &= W_0\Psi_1 + W_1\Psi_0\tag{8b}\\
\lambda^2&:&
  \mathbf{H}_0\Psi_2 + \mathbf{H'}\Psi_1
    &= W_0\Psi_2 + W_1\Psi_1 + W_2\Psi_0\tag{8c}\\
\lambda^n&:&
  \mathbf{H}_0\Psi_n + \mathbf{H'}\Psi_{n-1}
    &= \sum_{i=0}^n W_i\Psi_{n-i}\tag{8d}\\
\end{align}
We do have a complete set of solutions, hence we can express perturbations as linear combinations of these.
\begin{aligned}
\Psi_0 &= \Phi_0 &
\Psi_1 &= \sum_i c_i \Phi_i &
\Psi_2 &=\sum_i d_i \Phi_i &
\dots
\end{aligned}
Now we project equations $8$ onto $\Psi_0$. For $\mathrm{8a}$ this is trivial and it leads back to the expectation value of the energy for the unperturbed system. (In the last transformation we use the linear combinations.)
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_0\rangle
    &= \langle\Phi_0|W_0|\Psi_0\rangle\\
\langle\Phi_0|\mathbf{H}_0|\Psi_0\rangle
    &= W_0\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\tag{9a}\\
W_0 &= \langle\Phi_0|\mathbf{H}|\Phi_0\rangle = E_0\tag{9a'}
\end{align}
The next step is similar, and everything else will follow from there analogously. For $\mathrm{8b}$ we obtain:
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_1\rangle +
\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
    &=   \langle\Phi_0|W_0|\Psi_1\rangle
       + \langle\Phi_0|W_1|\Psi_0\rangle\\
\langle\Phi_0|\mathbf{H}_0|\Psi_1\rangle +
\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
    &= W_0\cancelto{c_0}{\langle\Phi_0|\Psi_1\rangle} +
       W_1\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\\
\cancelto{c_o\cdot E_0}{\sum_i c_i \langle\Phi_0|\mathbf{H}_0|\Phi_i\rangle}
       + \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
    &= W_0\cdot c_0 + W_1\tag{9b}\\
W_1 &= \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle\tag{9a'}
\end{align}
Similarly we can obtain the coefficients by projecting on some other function than $\Phi_0$. I will cut this a little bit shorter.
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_1\rangle
     + \langle\Phi_j|\mathbf{H'}|\Psi_0\rangle
    &= \langle\Phi_j|W_0|\Psi_1\rangle +
       \langle\Phi_j|W_1|\Psi_0\rangle\\
\cancelto{c_j\cdot E_j}{\langle\Phi_j|\mathbf{H}_0|\Psi_1\rangle}
     + \langle\Phi_j|\mathbf{H'}|\Psi_0\rangle
    &= \cancelto{c_j\cdot E_0}{\langle\Phi_j|W_0|\Psi_1\rangle}
     + \cancelto{0}{\langle\Phi_j|W_1|\Psi_0\rangle}\tag{10b}\\
c_j &= \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{E_0-E_j}\tag{10b'}
\end{align}
Now this is impossible to solve for $c_0$, but we still have the intermediate normalisation.
\begin{align}
0 &= \langle\Phi_0|\Psi_1\rangle
   = \sum_i c_i \langle\Phi_0|\Phi_i\rangle
   = c_0\cancelto{1}{\langle\Phi_0|\Phi_i\rangle} +
     \sum_{i\neq0} c_i \cancelto{0}{\langle\Phi_0|\Phi_i\rangle} = c_0\\
\end{align}
Similarly this applies to all zeroth coefficients, $c_0 = d_0 = \dots = 0$.
Let's get back to business and continue with $\mathrm{8c}$.
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_2\rangle
  + \langle\Phi_0|\mathbf{H'}|\Psi_1\rangle
  &= \langle\Phi_0|W_0|\Psi_2\rangle
   + \langle\Phi_0|W_1|\Psi_1\rangle
   + \langle\Phi_0|W_2|\Psi_0\rangle\\
\scriptsize\cancelto{d_0\cdot E_0}{\langle\Phi_0|\mathbf{H}_0|\Psi_2\rangle}
   + \sum_i c_i\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle
  &= \scriptsize W_0\cancelto{d_0=0}{\langle\Phi_0|\Psi_2\rangle}
   + W_1\cancelto{c_0=0}{\langle\Phi_0|\Psi_1\rangle}
   + W_2\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\tag{9c}\\
W_2 &= \sum_i c_i\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle\tag{9c'}
\end{align}
We do now insert $\mathrm{10b'}$ and get the second order correction:
\begin{align}
W_2 &= \sum_{i\neq0}\frac{
             \langle\Phi_0|\mathbf{H'}|\Phi_i\rangle
             \langle\Phi_i|\mathbf{H'}|\Psi_0\rangle
            }{E_0-E_i}\tag{9c''}
\end{align}
Now we have the expression for the second order energy, let's do the messy stuff for the second order contribution to the wave function.
Skip here for the short version, which is still very long, but that's how we roll for the second order correction to the wave function
We start by multiplying $\mathrm{8c}$ with $\Phi_j$ from the left and integrating (projection). I will include a few more steps for this problem, hopefully that will help transferring it to your notation.
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle +
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle &=
\langle\Phi_j|W_0|\Psi_2\rangle +
\langle\Phi_j|W_1|\Psi_1\rangle +
\langle\Phi_j|W_2|\Psi_0\rangle\\
\end{align}
I will go through this term by term this time, as it is quite a long equation.
Let's start with the most simplest term, the third one on the right hand side. As you noted, it cancels.
\begin{align}
\langle\Phi_j|W_2|\Psi_0\rangle
   &= W_2 \cancelto{0}{\langle\Phi_j|\Psi_0\rangle }\\
   &= 0
\end{align}
Let's continue on the right hand side:
\begin{align}
\langle\Phi_j|W_1|\Psi_1\rangle
   &= W_1 \langle\Phi_j|\Psi_1\rangle\\
   &= W_1 \sum_i c_i \langle\Phi_j|\Phi_i\rangle\\
   &= W_1 \left(c_j\cancelto{1}{\langle\Phi_j|\Phi_j\rangle} +
         \sum_{i\neq j} c_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\right)\\
   &= c_j W_1
\end{align}
And the same for the last contribution:
\begin{align}
\langle\Phi_j|W_0|\Psi_2\rangle
   &= W_0 \langle\Phi_j|\Psi_2\rangle\\
   &= W_0 \sum_i d_i \langle\Phi_j|\Phi_i\rangle\\
   &= W_0 \left(d_j\cancelto{1}{\langle\Phi_j|\Phi_j\rangle}
        + \sum_{i\neq j} d_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\right)\\
   &= d_j E_0 &\mathrm{9a':}&(W_0 = E_0)
\end{align}
Let's go to the left side:
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle
   &= \sum_i d_i \langle\Phi_j|\mathbf{H}_0|\Phi_i\rangle \\
   &= \sum_i d_i E_i \langle\Phi_j|\Phi_i\rangle\\
   &= d_j E_j \cancelto{1}{\langle\Phi_j|\Phi_j\rangle}
      + \sum_{i\neq j} d_i E_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\\
   &= d_j E_j
\end{align}
And the last one can barely be simplified:
\begin{align}
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle
  &= \sum_i c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
  &= \cancelto{0}{c_0 \langle\Phi_0|\mathbf{H'}|\Phi_0\rangle}
   + \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
\end{align}
Now, Avengers reassemble, rearrange, and insert the expression for $W_1$:
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle +
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle &=
\langle\Phi_j|W_0|\Psi_2\rangle +
\langle\Phi_j|W_1|\Psi_1\rangle +
\langle\Phi_j|W_2|\Psi_0\rangle\\
d_j E_j + \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle
 &= c_j W_1 + d_j E_0\\
d_j (E_j - E_0)
 &= c_j \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
   - \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
d_j &=
   c_j \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_j - E_0)}
   - \sum_{i\neq 0} c_i \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(E_j - E_0)}
\end{align}
Now introduce the first order base, $c_i$ and $c_j$ from $\mathrm{10b'}$, there will be some minus shuffling as well.
\begin{align}
d_j
  &=
   \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)}
   \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_j - E_0)}
   - \sum_{i\neq 0} \frac{\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_i)}
     \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(E_j - E_0)}\\
  &=
   \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)} \cdot
   \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(-1)(E_0 - E_j)}
   - \sum_{i\neq 0} \frac{\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_i)}
     \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(-1)(E_0 - E_j)}\\
 d_j &=
   \sum_{i\neq 0}
   \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle%
         \langle\Phi_i|\mathbf{H'}|\Psi_0\rangle%
        }{(E_0 - E_j)(E_0 - E_i)}
   -\frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle%
          \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)^2} \tag{10c'}
\end{align}
Well, I am officially exhausted right now.

And lastly, a small remark on your notation (I keep this paragraph for nostalgic reasons). It took me quite some time understanding it, and as we saw (in some earlier version of this answer), I was mistaken.
While I do not mind the hat notation for operators, indicating order on top of the quantities is problematic in my opinion. To me this place is reserved for powers. I suggest you change it to one of the following: $E_{n,0},\ {}^{(0)}\!E_n,\ E^{(0)}_n$. When you have a look at the other answer, you will notice, I have decided to stick to the last option. But as I do not know which book your are using it is also not easy to recommend. I believe there are as many different notations as there are books out there. I personally preferred the book "Introduction to Computational Chemistry, 2nd Edition"by Frank Jensen.
As we settled in the comments, I appreciate that you stick to Dirac's original notation. And also, notation is like Marvel and DC, you pick one and stick with it, you can still appreciate the other, but just not as much. Or a football (Americans pronounce this as "soccer") club, or ...
As long as you understand it and can explain it to someone else, you are fine.
And now we all deserve some chocolate and some coffee and cake. Thank you for reading this post until the end.

thank you so much and the way you do it is more than astonishing (+1). The only but that I found is that you found the 2nd order perturbation to the energy and I would like to find the 2nd perturbation to the wave function

btw Martin, the notation that I tried to use is the Dirac's one. In some way is easier than the traditional, I prefer it because is slightly more direct and I don't get lost with mountains of indices

@beginner I realised that, just before I left my office, but I did not have the time to include it. It becomes a really messy and complicated term, so I might need a little more time, but I am going to insert it. The choice of notation is of course a very subjective one, use whatever you are comfortable with, I was just adding the remark, because of the confusion order - power. Once you are stuck with a notation, you are hardly ever able to change your mind ;) I becomes too routine.

@beginner You are welcome. Please see my recent addition. Let me gather my strength and I will (at least try to) transfer it to your notation.

@beginner I am very happy to help, and it still bothers me, that I cannot get my head around the notation. It will bother me surely for quite some time, but I printed the question and I will come back to it. This has to be solved. I need that epiphany. And it was really fun to think about that again, after all, I do not want to get my brains rusty on the tough mathematics part.

@Nicolau Thanks. It was indeed a little painful. Especially considering that you cannot turn off the preview when you answer the first time. And all that MathJax will be processed again when you type a single letter. That made it very slow. I have therfore a feature request on meta: http://meta.stackexchange.com/q/253011/260760 As for the TeX part, that was actually a lot of fun. I cannot use it at work as often as I want to, so this was a very welcome exercise. Don't be demotivated about it, a lot was just copy/paste ;) Nards4eva!!1

@beginner I finally understood the notation completely and was able to derive the whole shenanigans again. I hope that halps you even moar. :D

Since I had troubles with the notation the first time around, I did some real quantum chemistry (with a pencil and some paper) and finally was able to derive the whole RSPT ansatz. Since this is equivalent to the first approach, but yet also standalone, I decided to add it as a separate answer. In a few key steps, this approach is slightly different, so a one to one transformation will not be possible.
$$%Introducing some shortcuts
\require{cancel}
\newcommand{\op}[1]{\hat{\mathrm{#1}}}  %\op{H}
\newcommand{\order}[1]{^{(#1)}}         %E_n\order{1}
\newcommand{\overlap}[3]{\mathcal{#1}_{#2}\order{#3}} %\overlap{S}{m}{1}
\newcommand{\integral}[3]{\mathcal{#1}_{#2,#3}} %\integral{V}{i}{j}
\newcommand{\tagref}[1]{\mathrm{#1}}
$$
We start again by setting up the Hamiltonian, as a reference operator with the perturbation operator.
$$\op{H} = \op{H}\order{0} +\lambda\op{V}  \tag1$$
The perturbed Schrödinger equation therefore becomes
$$(\op{H}\order{0} +\lambda\op{V})|n\rangle = E_n|n\rangle, \tag2$$
and we again develop the terms of the energy and wave function into Taylor power series.
\begin{align}
E_n &= E_n\order{0} + \lambda E_n\order{1} + \lambda^2 E_n\order{2} + \dots &
    &= \sum_i \lambda^i E_n\order{i}  \tag3\\
|n\rangle &= |n\order{0}\rangle + \lambda |n\order{1}\rangle
             + \lambda^2 |n\order{2}\rangle + \dots &
          &= \sum_i \lambda^i |n\order{i}\rangle  \tag4\\
\end{align}
We can write the unperturbed system, that provides us with a complete solution and therefore with a complete set of energies and wave functions, i.e. $n\in\mathbb{N}$.
$$\op{H}\order{0} |n\order{0}\rangle = E_n\order{0} |n\order{0}\rangle \tag5$$
We also know from this, that all solutions are orthonormal, i.e.
\begin{align}
\langle n\order{0} |n\order{0}\rangle &= 1,
\langle n\order{0} |m\order{0}\rangle &= 0. \tag6
\end{align}
Further we choose our perturbed wave function to be (intermediate) normalised, i.e.
\begin{align}
\langle n\order0|n\rangle &= 1, &
\langle n|n\rangle &= 1, &
\langle n\order{i}|n\order{j}\rangle &= \delta_{ij}
   \begin{cases}
      1, & i=j\\
      0, & i\neq j\\
   \end{cases}.\tag7
\end{align}
We continue with gathering the equations for all our orders.
\begin{align}
  \lambda^0 &:&
    \op{H}\order0 |n\order0\rangle
       &= E_n\order0 |n\order0\rangle\tag{8a}\\
  \lambda^1 &:&
    \op{H}\order0 |n\order1\rangle + \op{V} |n\order0\rangle
       &= E_n\order0 |n\order1\rangle + E_n\order1 |n\order0\rangle\tag{8b}\\
  \lambda^2 &:&
    \op{H}\order0 |n\order2\rangle + \op{V} |n\order1\rangle
       &= E_n\order0 |n\order2\rangle + E_n\order1 |n\order1\rangle
          E_n\order1 |n\order0\rangle\tag{8c}\\
  \vdots\\
  \lambda^n &:&
    \op{H}\order0 |n\order{n}\rangle + \op{V} |n\order{n-1}\rangle
       &= \sum_{i=0}^n E_n\order{i} |n\order{n-i}\rangle\tag{8d}\\
\end{align}
The two fundamental tricks of algebra are adding zero and multiplying by one.
We use this to set up our bases, we basically expand our bases into an auxiliary base (also known as 'resolution-of-the-identity'), which is possible, since the unperturbed system provides the complete set. This is what makes this approach somewhat different. While in my other answer we are just expanding the wafer function into an ordinary boring linear combination, we are using here the complete Hilbert space to do the same thing, but a little more elegant. However, this is also the key step: Multiplying by one.
At the same time we introduce an overlap integral $\integral{S}{m}{i}$, which makes writing a little easier further on. For any order $i\neq0$ we can therefore write:
\begin{align}
|n\order{i}\rangle &=
   \sum_{m} |m\order0\rangle\langle m\order0|n\order{i}\rangle\\
                   &=
   \sum_{m\neq n} |m\order0\rangle
    \underbrace{\langle m\order0|n\order{i}\rangle}_{\overlap{S}{m}{i}}
     + |n\order0\rangle\cancelto{0}{\langle n\order0|n\order{i}\rangle}\\
|n\order{i}\rangle &= \sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{i}\\
\end{align}
The bases we need are for first and second order, therefore we write
\begin{align}
|n\order{1}\rangle &= \sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{1}, &
|n\order{2}\rangle &= \sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{2},
\end{align}
which seems rather trivial.
The next step is getting the expressions for the energies and corrections to the wave function.
For $\tagref{8a}$ this is trivial and it just yields the expectation value of the unperturbed Hamiltonian. We project on (multiply from the left with) $\langle n\order0|$, hence
\begin{align}
  \langle n\order0|E_n\order0 |n\order0\rangle &=
  \langle n\order0|\op{H}\order0 |n\order0\rangle \\
  E_n\order0 \cancelto{1}{\langle n\order0|n\order0\rangle} &=
  \langle n\order0|\op{H}\order0 |n\order0\rangle \\
  E_n\order0 &= \langle n\order0|\op{H}\order0 |n\order0\rangle. \tag{9a}\\
\end{align}
For the first order correction to the energy we use the hermiticity of the Hamiltonian, i.e. we rewrite $\tagref{8a}$ in bra instead of the ket notation we used all way through.
$$\langle n\order0| \op{H}\order0 = E_n\order0 \langle n\order0|$$
We then further go about and project $\tagref{8b}$ on $\langle n\order0|$, hence
\begin{align}
\langle n\order0|\left(\op{H}\order0|n\order1\rangle
                       + \op{V}|n\order0\rangle \right) &=
   \langle n\order0|\left(E_n\order0|n\order1\rangle
                       + E_n\order1|n\order0\rangle \right)\\
\langle n\order0|\op{H}\order0|n\order1\rangle
     + \langle n\order0| \op{V}|n\order0\rangle  &=
   \langle n\order0|E_n\order0|n\order1\rangle
     + \langle n\order0| E_n\order1|n\order0\rangle\\
E_n\order0 \cancelto{0}{\langle n\order0|n\order1\rangle}
     + \langle n\order0| \op{V}|n\order0\rangle  &=
   E_n\order0\cancelto{0}{\langle n\order0|n\order1\rangle}
     + E_n\order1 \cancelto{1}{\langle n\order0| n\order0\rangle}\\
E_n\order1 &= \langle n\order0| \op{V}|n\order0\rangle  \tag{9b}
\end{align}
Now for the first order correction to the waffle function. I have to admit, this was the part where I struggled most. We will project on $\langle k\order0|\neq\langle n\order0|$, therfore obtaining:
\begin{align}
\langle k\order0|\left(\op{H}\order0|n\order1\rangle
                       + \op{V}|n\order0\rangle \right) &=
   \langle k\order0|\left(E_n\order0|n\order1\rangle
                       + E_n\order1|n\order0\rangle \right)\\
\langle k\order0|\op{H}\order0|n\order1\rangle
     + \langle k\order0| \op{V}|n\order0\rangle  &=
   \langle k\order0|E_n\order0|n\order1\rangle
     + \langle k\order0| E_n\order1|n\order0\rangle\\
E_k\order0 \langle k\order0|n\order1\rangle
     + \langle k\order0| \op{V}|n\order0\rangle  &=
   E_n\order0\langle k\order0|n\order1\rangle
     + E_n\order1 \cancelto{0}{\langle k\order0| n\order0\rangle}\\
E_k\order0 \langle k\order0|\sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{1}
     + \langle k\order0| \op{V}|n\order0\rangle  &=
   E_n\order0\langle k\order0|\sum_{m\neq n}
                    |m\order0\rangle \overlap{S}{m}{1}\\
\end{align}
Before it gets to messy, we are going to reduce each of the terms individually and then reinsert, to obtain our expression.
\begin{align}
E_k\order0 \langle k\order0|\sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{1}
  &= \sum_{m\neq n,k}E_k\order0
       \cancelto{0}{\langle k\order0|m\order0\rangle} \overlap{S}{m}{1}
     + E_k\order0
       \cancelto{1}{\langle k\order0|k\order0\rangle} \overlap{S}{k}{1}\\
  &= E_k\order0 \overlap{S}{k}{1}\\
\end{align}
The second term stays the same, and the last one simplifies analogously.
\begin{align}
E_n\order0 \langle k\order0|\sum_{m\neq n} |m\order0\rangle \overlap{S}{m}{1}
  &= \sum_{m\neq n,k}E_n\order0
       \cancelto{0}{\langle k\order0|m\order0\rangle} \overlap{S}{m}{1}
     + E_n\order0
       \cancelto{1}{\langle k\order0|k\order0\rangle} \overlap{S}{k}{1}\\
  &= E_n\order0 \integral{S}{k}{1}\\
\end{align}
And reinsterting we can write and resubstitute for the overlap integral $\overlap{S}{k}{1}$, with $m=k$:
\begin{align}
E_k\order0 \overlap{S}{k}{1} + \langle k\order0| \op{V}|n\order0\rangle
   &= E_n\order0 \overlap{S}{k}{1}\\
\integral{S}{k}{1}
   &= \frac{\langle k\order0| \op{V}|n\order0\rangle}{E_n\order0 -E_k\order0}\\
|n\order1\rangle
   &= \sum_{k\neq n} |k\order0\rangle
      \frac{\langle k\order0| \op{V}
              |n\order0\rangle}{E_n\order0 -E_k\order0} \tag{10b}\\
\end{align}
To make matters a little bit easier further on, we will introduce another short notation for the perturbed integrals:
$$\integral{V}{i}{j} := \langle i\order0|\op{V}|j\order0\rangle$$
And therefore we can write
\begin{align}
|n\order1\rangle
   &= \sum_{k\neq n} |k\order0\rangle
      \frac{\integral{V}{k}{n}}{E_n\order0 -E_k\order0} \tag{10b}\\
\end{align}
For the second order correction to the energy we project $\tagref{8c}$ on $\langle n\order0|$. This is again fairly straight forward, the key is to insert $\tagref{10b}$ at the end to obtain the expression.
\begin{align}
\langle n\order0|\left(\op{H}\order0|n\order2\rangle
  + \op{V}|n\order1\rangle\right)
  &= \langle n\order0|\left( E_n\order0|n\order2\rangle
       + E_n\order1|n\order1\rangle + E_n\order2|n\order0\rangle\right)\\
\langle n\order0|\op{H}\order0|n\order2\rangle
  + \langle n\order0|\op{V}|n\order1\rangle
    &=   E_n\order0 \langle n\order0|n\order2\rangle
       + E_n\order1 \langle n\order0|n\order1\rangle
       + E_n\order2 \langle n\order0|n\order0\rangle\\
\small
 E_n\order0 \cancelto{0}{\langle n\order0|n\order2\rangle}
  + \langle n\order0|\op{V}|n\order1\rangle
    &= \small
         E_n\order0 \cancelto{0}{\langle n\order0|n\order2\rangle}
       + E_n\order1 \cancelto{0}{\langle n\order0|n\order1\rangle}
       + E_n\order2 \cancelto{1}{\langle n\order0|n\order0\rangle}\\
E_n\order2 &= \langle n\order0|\op{V}|n\order1\rangle\\
%           &= \langle n\order0|\op{V} \sum_{k\neq n} |k\order0\rangle
%      \frac{\langle k\order0| \op{V}
%              |n\order0\rangle}{E_n\order0 -E_k\order0}\\
           &= \sum_{k\neq n} \frac{\langle n\order0|\op{V} |k\order0\rangle
      \langle k\order0| \op{V} |n\order0\rangle}{E_n\order0
                     -E_k\order0}\tag{9c}\\
           &= \sum_{k\neq n} \frac{\integral{V}{n}{k}
                \integral{V}{k}{n}}{E_n\order0 -E_k\order0}
\end{align}
Now let's get to the real fancy part: The second order correction to the basil. We project once again $\tagref{8c}$ on $\langle k\order| \neq \langle n\order 0|$. We do of course need our expression for the first order base and our expansion for the second order. But let us take one step at the time.
\begin{align}
\langle k\order0|\left( \op{H}\order0|n\order2\rangle
   + \op{V}|n\order1\rangle\right)
  &= \langle k\order0 \left(
      E_n\order0|n\order2\rangle
    + E_n\order1|n\order1\rangle
    + E_n\order2|n\order0\rangle\right)\\
\small
\langle k\order0| \op{H}\order0|n\order2\rangle
   + \langle k\order0| \op{V}|n\order1\rangle
  &= \small
      \langle k\order0| E_n\order0|n\order2\rangle
    + \langle k\order0| E_n\order1|n\order1\rangle
    + \langle k\order0| E_n\order2|n\order0\rangle\\
\end{align}
And we proceed again term by term, because it is pretty messy. We are starting with the right hand side, and there with the third term, as this one cancels, because of our normalisation.
\begin{align}
\langle k\order0| E_n\order2|n\order0\rangle
  &= E_n\order2 \cancelto{0}{\langle k\order0| n\order0\rangle} =0\\
\end{align}
For the second term on the right we need our definition of the base. Since we are projecting on $k$ we choose the base to be represented by $m$. We also need the expression for the first order correction to the energy $\tagref{9b}$. Also note, that we are using the contracted notation.
\begin{align}
\langle k\order0| E_n\order1|n\order1\rangle
  &= E_n\order1 \langle k\order0|n\order1\rangle\\
  &= E_n\order1 \langle k\order0| \sum_{m\neq n} |m\order0\rangle
       \frac{\integral{V}{m}{n}}{E_n\order0 - E_m\order0}\\
  &= \sum_{m\neq n,k} E_n\order1 \cancelto{0}{\langle k\order0| m\order0\rangle}
       \frac{\integral{V}{m}{n}}{E_n\order0 - E_m\order0}
     + E_n\order1 \cancelto{1}{\langle k\order0| k\order0\rangle}
       \frac{\integral{V}{k}{n}}{E_n\order0 - E_k\order0}\\
  &= \frac{\integral{V}{n}{n}\integral{V}{k}{n}}{E_n\order0 - E_k\order0}\\
\end{align}
For the first term on the right hand side we just need our expansion of the second order correction of the base.
\begin{align}
\langle k\order0| E_n\order0|n\order2\rangle
  &= E_n\order0 \langle k\order0|n\order2\rangle\\
  &= E_n\order0 \langle k\order0| \sum_{m\neq n} |m\order0\rangle
     \overlap{S}{m}{2}\\
  &= \sum_{m\neq n,k} E_n\order0 \cancelto{0}{\langle k\order0|m\order0\rangle}
     \overlap{S}{m}{2}
     + E_n\order0 \cancelto{1}{\langle k\order0|k\order0\rangle}
     \overlap{S}{k}{2}\\
  &= E_n\order0 \overlap{S}{k}{2}\\
\end{align}
We have to use the same for the first term on the left hand side. In addition we are using the bra notation for the energy to simplify.
\begin{align}
\langle k\order0| \op{H}\order0|n\order2\rangle
  &= E_k\order0 \langle k\order0|n\order2\rangle\\
  &= E_k\order0 \langle k\order0|\sum_{m\neq n} |m\order0\rangle
     \overlap{S}{m}{2} \\
  &= \sum_{m\neq n,k} E_k\order0 \cancelto{0}{\langle k\order0|m\order0\rangle}
     \overlap{S}{m}{2}
     + E_k\order0 \cancelto{1}{\langle k\order0|k\order0\rangle}
     \overlap{S}{k}{2}\\
  &= E_k\order0 \overlap{S}{k}{2}\\
\end{align}
For the second term on the left side we once again need the expression for the first order correction to bass, choosing $m$, as we are already projecting on $k$.
\begin{align}
\langle k\order0| \op{V}|n\order1\rangle
  &= \langle k\order0| \op{V} \sum_{m\neq n}|m\order0\rangle
     \frac{\integral{V}{m}{n}}{E_n\order0 -E_m\order0}\\
  &= \sum_{m\neq n}\frac{\integral{V}{k}{m}
     \integral{V}{m}{n}}{E_n\order0 -E_m\order0}\\
\end{align}
Now, let's put all the stuffing back into the turkey. Then we rearrange to complete bird and solve for our desired expression for the base.
\begin{align}
E_k\order0 \overlap{S}{k}{2}
  + \sum_{m\neq n}\frac{\integral{V}{k}{m}
     \integral{V}{m}{n}}{E_n\order0 -E_m\order0}
  &= E_n\order0 \overlap{S}{k}{2}
     + \frac{\integral{V}{n}{n}\integral{V}{k}{n}}{E_n\order0 - E_k\order0}\\
(E_n\order0 - E_k\order0)\overlap{S}{k}{2}
  &= \sum_{m\neq n}\frac{\integral{V}{k}{m}
     \integral{V}{m}{n}}{E_n\order0 -E_m\order0}
    - \frac{\integral{V}{n}{n}\integral{V}{k}{n}}{E_n\order0 - E_k\order0}\\
\overlap{S}{k}{2}
  &= \sum_{m\neq n}\frac{\integral{V}{k}{m}
     \integral{V}{m}{n}}{(E_n\order0 -E_m\order0)(E_n\order0 - E_k\order0)}
    - \frac{\integral{V}{n}{n}\integral{V}{k}{n}}{(E_n\order0 - E_k\order0)^2}\\
|n\order2\rangle
  &= \sum_{k\neq n} \sum_{m\neq n} |k\order0\rangle
     \frac{\integral{V}{k}{m}
     \integral{V}{m}{n}}{(E_n\order0 -E_m\order0)(E_n\order0 - E_k\order0)}\\
%split into two lines
  &\phantom{=~}
    - \sum_{k\neq n} |k\order0\rangle
      \frac{\integral{V}{k}{n}\integral{V}{n}{n}}{(E_n\order0 - E_k\order0)^2}\\
\end{align}
And last, but not least, let's remove that nasty shorthand and make everything pretty again.
\begin{align}
|n\order2\rangle
  &= \sum_{k\neq n} \sum_{m\neq n} |k\order0\rangle
     \frac{
       \langle k\order0|\op{V}|m\order0\rangle
       \langle m\order0|\op{V}|n\order0\rangle
          }{(E_n\order0 -E_m\order0)(E_n\order0 - E_k\order0)}
    - \sum_{k\neq n} |k\order0\rangle
      \frac{
       \langle k\order0|\op{V}|n\order0\rangle
       \langle n\order0|\op{V}|n\order0\rangle
           }{(E_n\order0 - E_k\order0)^2}\\
\end{align}
And there we are. Pretty.

Now you will probably have noticed, that you were one step away from the ultron goal. In your last equation you simply had to reduce to the non-zero parts.

\begin{equation}
( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle   = \sum_{m \neq n} \frac{ \langle k^0| \hat{V} |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}  - \langle n^0 | \hat{V} | n^0 \rangle \sum_{m \neq n}  \frac{\langle k^0 |m^0 \rangle \langle m^0 | \hat{V} | n^0 \rangle}{E_n^0 - E_m^0}
\end{equation}

\begin{align}
( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle
 &= \sum_{m \neq n}
    \frac{
          \langle k^0| \hat{V} |m^0 \rangle
          \langle m^0| \hat{V} |n^0 \rangle
         }{E_n^0 - E_m^0}\\
 &\phantom{=~}
  - \sum_{m \neq n,k}
      \frac{ \cancelto{0}{\langle k^0 |m^0 \rangle}
            \langle n^0 | \hat{V} | n^0 \rangle
            \langle m^0 | \hat{V} | n^0 \rangle
           }{E_n^0 - E_m^0}\\
 &\phantom{=~}
  -   \frac{ \cancelto{1}{\langle k^0 |k^0 \rangle}
            \langle k^0 | \hat{V} | n^0 \rangle
            \langle n^0 | \hat{V} | n^0 \rangle
           }{E_n^0 - E_k^0}\\
( E_n^0 - E_k^0  ) \langle k^0  |n^2\rangle
 &= \sum_{m \neq n}
    \frac{
          \langle k^0| \hat{V} |m^0 \rangle
          \langle m^0| \hat{V} |n^0 \rangle
         }{E_n^0 - E_m^0}
  -   \frac{
            \langle k^0 | \hat{V} | n^0 \rangle
            \langle n^0 | \hat{V} | n^0 \rangle
           }{E_n^0 - E_k^0}\\
\langle k^0  |n^2\rangle
 &= \sum_{m \neq n}
    \frac{
          \langle k^0| \hat{V} |m^0 \rangle
          \langle m^0| \hat{V} |n^0 \rangle
         }{(E_n^0 - E_m^0)(E_n^0 - E_k^0)}
  -   \frac{
            \langle k^0 | \hat{V} | n^0 \rangle
            \langle n^0 | \hat{V} | n^0 \rangle
           }{\left(E_n^0 - E_k^0\right)^2}\\
\end{align}
Now you just have to expand into the complete base by multiplying with $\sum_{k\neq n}|k^0\rangle$ and you get to the same expression as above.
And as a theoretician a proof is only complete with a coffee stain, and trust me, there have been many.

Image is courtesy of Roger Karlsson (http://www.free-photo-gallery.org/photos/coffee-stain/) obtained from flicker.

The final result is slightly different from the Wikipedia one: https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)

@jack I'm sorry, but I won't have the time to compare notes and check where what deviates and for what reason. If you are more specific I might check this and come back to it. No promises though. Alternatively you can share your detailed approach as another answer.

@Jack I've had a glance at the Wikipedia article. I believe the difference is due to normalisation. Wikipedia uses only $\langle n| n \rangle =1$. This post assumes intermediate normalisation $\langle n| n \rangle =1$ and $\langle n^{(0)} | n \rangle = 1$. This should give you a good starting point to do the maths for your answer.


text:
Whats the Balanced chemical equation for heating sodium chloride and what type of a reaction is it?
For my chemistry lab we had to heat sodium chloride and there was a colour change and a sizzling sound. Is it a decomposition or synthesis reaction, and what is the balanced chemical equation for it?

Doesn't sound like sodium chloride to me; salt is pretty thermally stable, although you could conceivably melt it with a blow torch (mp. is 801 degrees C).

My guess would be that this was rather the dissociation of Sodium Carbonate or Sodium Hydrogen Carbonate.
The equations for the dissociation of Sodium Carbonate and Sodium Hydrogen Carbonate, respectively, are as follows:

$$\ce{Na2CO3 -> Na2O + CO2}$$

$$\ce{and}$$

$$\ce{2 NaHCO3 -> Na2O + H2O + 2 CO2}$$

a rather wild guess, i think

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Can covalently bonded hydrocarbons be influenced by a magnetic field?
I am doing a University project on Magnetically Assisted Combustion and am curious to know more about how magnetic fields may or may not affect gases such as butane and propane.
I think I am right in saying that due to the gas molecules being covalently bonded, there are no free electrons to be influenced in anyway by an induced magnetic field.
Someone briefly mentioned something about that depending on whether they are saturated or unsaturated but the conversation didn't go any further as he then promptly left.. so I am still unsure.
If anyone could shed any light on this or has any thoughts at all I'd be very grateful for your input.
The method of some experiments I have done was attach a neodynium magnetic arrangement around the fuel line of a gas burner and measuring heat transfer to water.  So far the results have been fairly inconclusive as I am using a camping stove which has problems sustaining a constant gas flow as well as a number of inconsistonsies with heat loss and human error etc .. but that's another issue.

You question is not precise, a bit of clarification could be useful. Flame test is not 'pure' as flames typically contain charged species, definitely interacting with magnetic field. Anyway, hydrocarbons DO interact with magnetic fields, exhibiting diamagnetic behavior (see wikipedia://diamagnetism). Also in most atoms the nucleus has non-zero magnetic moment, also (typically very weakly) interacting with magnetic field (used in NMR spectroscopy).

Just because there are no unpaired electrons doesn't mean magnetic fields can't influence something. Unpaired electrons tend to lead to strong interactions with magnetic fields (the dioxygen in air has two and is paramagnetic, an effect best seen in the liquid phase where a magnet will attract the cold liquid).
But compounds with paired electrons also interact with magnetic fields, though the effects are usually weak. But the repulsive effect is sometimes strong enough to levitate a diamagnetic substance (see this youtube video showing how strong magnets can levitate plates of graphite which is strongly paramagnetic).
It is unlikely that such effects would be easy to see in gases. But you are talking about combustion and the most likely effect will not be on the neutral gas but on the flame where the gas is not neutral but a plasma of charged particles. Charged particles or molecules will have a much stronger interaction with a magnetic field though it might still require sensitive apparatus to detect how the field makes a difference.

Effects of magnetic fields on combustion and gas-flow were studied. Methane, propane and hydrogen gases were burned, and flames of these gases were exposed to gradient magnetic fields up to 1.6 T and 220 T/m. Flames bent so as to escape from magnetic fields of higher intensities. Apart from the combustion experiments, flows of gases such as carbon dioxide and oxygen were exposed to magnetic fields up to 2.2 T and 300 T/m. The flows of these gases with a flow velocity 20-140 ml/min were blocked or modified by the magnetic fields. The changes of flame-shape and gas-flow by magnetic fields are understood to be the result of the role of oxygen. Under the intensities of magnetic fields concerned, oxygen gases as paramagnetic molecules are not concentrated but are aligned so as to make a "wall of oxygen". The wall of oxygen presses back flames and other gases.

Published in:
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text:
Why does rock salt change colour when it's heated
We heated rock salt for a chemistry lab and it changed from white crystally rock to brown rock and ash-like materiak on the bottom of the test tube. Also there was a sizzling\popping noise. Why did it change colour and why was there a noise? What caused this?

"Rock salt" contains many impurities, in addition to its main ingredient, sodium chloride. Some trace chemicals in rock salt, such as lithium chloride or sodium carbonate, hold onto water, which might boil out, sizzling and popping, and if your specimen got hot enough, carbonates could decompose.
Even if you had used "table salt", rather than rock salt, chemicals are intentionally added, such as sodium ferrocyanide (to prevent caking), which has a yellowish color and potassium iodide (as a food supplement).
Had you heated pure sodium chloride, there should not have been a color change.

I'm not too sure what you are asking. If you are refering to rock salt as in sodium chloride (NaCl), I too am at a loss at how you would have created such an effect.
I will assume that you have heated some random other compound since NaCl does not decompose/ react to heat so readily.
It is possible that the "rock salt" reacted to the heat and perhaps decomposed.
For example, CaCO3 (calcium carbonate) decomposes to CaO and Oxygen, perhaps forming the sizzling noise.


text:
Ways to change the spontaneity of a reaction
I can think of any other way to influence the value of $\Delta G$ than (1) change the temperature (2) providing an external source to overcome the negative cell potential
The relationship between $\Delta G$ and $E_{\mathrm{cell}}$ is $\Delta G =-nFE_{\mathrm{cell}}$
Between $\Delta H$, $\Delta S$ and $\Delta G$ is $\Delta G=\Delta H - T\Delta S$
Sometimes, temperature won't do anything to change the spontaneity. But if we change $\Delta G$ through $\Delta E_{\mathrm{cell}}$ by providing external voltage in excess of the negative voltage, and though $\Delta G$ we are also able to influence enthalpy and entropy(???).
This is the part I am not sure about. Besides the title question, a sub-question is: how in this way can we change the enthalpy and entropy?

Natural variables of $\Delta G$ are $p$, $T$, and $N_i$.
$p$ = pressure
$T$ = Temperature
$N_i$ = number of particles  (or number of moles)
The derivation showing this is listed on the Wikipedia.

The Nernst Equation relates the reduction potential of a cell to the chemical reaction that is occurring in the cell, and is written:
$$E_{\mathrm{cell}}=E^\mathrm{o}-\frac{RT}{nF}\ln(Q)$$
$E_{\mathrm{cell}}$, and thereby $\Delta G$, can be affected by changes in the concentration of the ions in either half cell.

But how will both enthalpy and entropy be affected in turn?

Okay, so given: $E_{cell}=E^o−\frac{RT}{nF}ln(Q)$, $\Delta G=−nFE_{cell}$, and $\Delta G=\Delta H-\Delta S·T$, we can figure a few things out. We know that $\Delta H$ is a state function, so no matter the pathway we take for a given reaction, $\Delta H$ remains constant. We must, therefore, be affecting $\Delta S$ by changes in $Q$. An increase in the concentration of the products will increase $\Delta G$, meaning $\Delta S$ is decreasing. The opposite will be true for an increase in the concentration of the reactants. If you're still having trouble, try messing around with the equations.

You can't simply call only enthalpy a state function. A quick search shows a wiki list of state function. Both free energy and entropy are state functions. I think you need to adjust your terminology.

I'll admit my initial reasoning may be faulty in saying that $\Delta H$ will not change because it is a state function, but I stand by the idea that $\Delta H$ won't be affected by a change in concentration. Since $\Delta H=\Delta U_{system}+P·\Delta V$ and $\Delta U_{system}=q-P·\Delta V$ by combining the two equations, we find that $\Delta H=q$, where $q$ is the heat flow into or out of the system. Since we are not adding or removing heat to the system by changing the concentration of solute, we can finally conclude that $\Delta H$ is not affected by changes in solute concentration.

$\Delta H^\circ$ is not the same as $\Delta H$.  Both are state functions.  But changing the concentration is changing the state.  So $\Delta H$ will definitely change with solute concentration.


text:
Assigning R/S stereochemistry of a cyclic molecule
I've been dealing with assigning stereochemistry and I seem to be doing okay, but I've come across a molecule that I've found a little tricky to deal with.

I've identified three stereocentres, highlighted below. My first question would be to confirm that this is correct.

No. 1 is standard fare and I got that it was an R enantiomer.
But for 2 & 3, does the fact that both carbons are bonded to the oxygen have an effect on the rotation of the molecule? As in, when you go through the process of rotating the lowest priority group to the back and then tracing from first priority to third, does this affect the process? It may seem a silly question, but I don't have ball & stick models to easily visualise it so I'm struggling a bit. I treated them as normal and got that they were both R enantiomers, but I'm wondering if I've missed a trick.

Yes, 3 centers and they are all "R".  You got it right.  If you plan to take other chemistry courses get some models now.  BTW, don't worry about rotating the lowest priority group to the back - if it's in the front just reverse the R/S assignment.


text:
How does one determine the relative basicity of a nitrogen in a molecule?
I'm trying to solve this problem:

Specifically, how does one know which nitrogen is more basic? I'm assuming that for the first molecule, that it's the nitrogen without the proton, since it seems like the other nitrogen has already been protonated (and already has acted as a base.)
Is my reasoning correct? How does one solve the other molecules?

Use a combination of

the exclusion principle: sort out the unlikely centres with low electron density and
paper work to figure out those nitrogen atoms for which the additional positive charge can be distributed over a larger part of the molecule without moving the proton.

In the case of caffeine (1) the amide and the imide are rather unlikely centres for protonation! You have probably already ruled out the amide nitrogen in the benzamide case (your second molecule) in favour of the aniline site.
From an initial number of four possible centres in 1, only two remain.
Now, let's see whether the positive charge resulting from protonation can be stabilized.

This looks fair, but creates a quarternary ammonium ion for which no obvious charge distribution seems to be available.
Let's examine the last possibility:

Would you agree that this looks much more promising?

My pleasure! I reckon that you figured out the other ones easily!

For amine basicity:
Basicity decreases with increasing s character
Basicity decreases with increasing stability, due to resonance/ delocalization
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/amine1.htm (taken from this webpage)
For the first molecule for example, the double bonded nitrogen is less basic due to increasing s character (closer to nucleus, less likely to bind to H).
I think most of your examples can be reasonably thought out with these two key points.

double bonded nitrogen is more basic, because the other one has delocalized lone pair  - your second point...

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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SF3Cl Vsepr Most stable structure
I was asked to " Use VSEPR theory to predict the most stable structure for $\ce{SF3Cl}$ and justify your answer. "
I managed to interpret the structure as seesaw and with a lone pair on the equatorial position. However, I had chosen that the most stable structure was that with Chlorine on the equatorial position because I thought that the Chlorine-Sulfur bond length would be greater than that of Chlorine-Fluorine.
However, the answer key stated the Fluorine was more electronegative, and therefore would be more stable on the axial (with chlorine as equatorial).
I was wondering if the answer key was accurate and if so, how would I compare the effects of electronegativity and bond length? I had considered electronegativity but I had thought it would simply make the bond even shorter, increasing repulsion.
Thanks!

You can visit this page, this page and this one  on how to make your future posts better.

The bond lengths have little to do with the stability of the atom positioning around the central $\ce{S}$ atom. Rather, the relative sizes of the electron clouds are what should be considered when positioning these atoms. An atom of $\ce{Cl}$ has a 8 more electrons than does an atom of $\ce{F}$, so naturally its electron cloud will be larger and therefore more repulsive. The $\ce{Cl}$ should therefore be positioned in the equatorial position for the same reason you predicted that the lone pair would be present in the equatorial position--because these positions offer the least repulsion between the electron clouds of the atoms.
In the axial position, the 3 equatorial positions, present at 90 degree angles to the axial position, repel the electron cloud.
In an equatorial position, however, only the 2 axial positions are present at 90 degree angles to repel the electron cloud. The other 2 equatorial positions are positioned away from the equatorial bond in question, and repel its electron cloud only slightly. For this reason unbonded electron pairs and atoms with large electron clouds occupy these spaces first-- they offer the least repulsion and the most stable positioning.


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Why are alpha-carbonyl relatively less stable?
So if we have an alpha halogen ketone, why is an Sn1 type mechanism so unfavorable for that species?
I don't understand. It seems like the C+ would be resonance stabilized by the carbonyl's double bond.

Welcome to Chemistry.SE. Would you mind to explain how a carbocation next to an electrophilic centre is stabilized by resonance?

Could you push the pi electrons in the carbonyl to form an alkene from the alpha carbon to the carbonyl carbon and a + charged oxygen?

A plus charge on oxygen is highly unfavourable situation even more worse than carbocation

The carbon atom of the carbonyl group has a partial positive charge due to electron withdrawal by the electronegative oxygen atom. An $S_{N}1$ mechanism would result in an intermediate with a positive charge next to a partial positive charge, which would be high in energy.


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What is the difference between the units 'torr' and 'mm of Hg'?
According to present definition torr and mm of Hg differ though slightly.
I am still confused over it. Can someone write an explicit answer differentiating between them and the reason for differentiation as earlier they were considered same. The explanation is probably related to the changes is pressure at different heights so I want the exact definitions of these units that was used earlier and what modifications have been made?

In addition to the confusion of the two different units, the unit names and unit symbols are sometimes mixed up. The name of one unit is “torr” and its symbol is “Torr”. The name of the other unit is “conventional millimetre of mercury” and its symbol is “mmHg”.

The two units torr, and mm of Hg were the same until they were redefined.
The torr was named after the Italian Evangelista Torricelli. 1 atmosphere is $101325\ \mathrm{Pa}$. The torr is defined as $1/760$ of an atmosphere. This is equal to $133.322\overline{368421052631578947}~\mathrm{Pa}$, which periodically infinitely repeats.
The mm of Hg was defined as the pressure generated by a column of mercury one millimeter high. Since the pressure of mercury depends on temperature and gravity, it is now redefined as
$$\mathrm{0.001\ mL \cdot 13595.1~mg \cdot 9.80665\ m/s^2 = 133.322387415 ~Pa}$$
This has caused a difference between the units of $0.000015~\%$, so unless you require extremely high precision, you can assume that $1\ \mathrm{torr} = 1\ \mathrm{mmHg}$

that means 760 mm Hg is not exactly equal to 1 atm

Also is 1 mm of Hg an exactly defined quantity

@DSinghvi: Yes to both questions.

As I said before, the difference between the values is so negligible, you can just assume that 1 torr = 1 mm Hg

Very very few chemistry experiments are ever done to 0.000015% precision. Something like trying to measure Avogadro's Constant would be an exception.

Note that the error in assuming 1 torr = 1 mm Hg is probably spectacularly smaller than the local variation of g depending on altitude, latitude and longitude, or the error introduced by local atmospheric densities.

$0.001 mL$?  Maybe $0.001 m$ for the first factor in the definition of mm Hg?  Also the factors in both definitions are rendered to only six significant figures therefore please report the SI values of to that accuracy.


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What is the reaction in a child's 'add water and watch the egg hatch' toy?
My son and daughter got given gifts where you place a plastic egg in water, and the (rubber?) toy inside expands (over three days) and breaks the shell of the egg so it 'hatches'.

I also observed that three weeks later the toys shrank again, but grew once placed in water.
I found the chemical reaction fascinating.
My question is: What is the reaction in a child's 'add water and watch the egg hatch' toy?

The egg hatch toys are composed of a superabsorbent polymer. These polymers are able to absorb up to 30 to 60 time their volume in water by forming hydrogen bonds with the water molecules.
Sodium polyacrylate ($\ce{[-CH2-CH(CO2Na){}-]}_n%edit$) is the most common superabsorbent polymer and according to its wikipedia page, is the one used in the egg hatch toys.


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Is it possible to synthesize amino acids from hydrocarbons?
I recently heard that during some period of time Purina pet food was being produced from crude oil via a set of complex chemical transformations. Which, I think, is mostly bluff and marketing. So is it possible, and have somebody had any real positive practical experience of getting amino acids from hydrocarbons?

Out of memory, I don't recall a method for the direct synthesis of $\alpha$-amino acids from plain hydrocarbons (including alkenes and alkynes).
The situation is different for $\beta$-amino acids. The $\ce{Pd(II)}$-catalyzed reaction of alkenes with amines in the presence of carbon monoxide, known as aminocarbonylation can furnish these amino acids or their derivates.

[…] from crude oil via a set of complex chemical transformations

however, is pretty vague and may involve anything, including the formation of aldehydes and their subsequent use in a Strecker reaction.

Well (almost) everything is possible. A better question might be whether or not it is profitable, but remember that A LOT of organic compounds used in synthesis originate from the petrochemical industry (these include other than simple alkanes as well).
A possible industrial pathway from an alkane to an amino acid could be: 1) chlorination of the alkane using a radical chain reaction 2) reaction of the chlorinated functionalized compound with ammonia to yield the alkylamine 3) chlorinate again and react with $\ce{NaCN}$ to obtain the nitrile amine 4) hydrolyze in aqueous acid which yields the desired amino acid.
Of course the above procedure is highly simplified as it does neither consider stereo chemistry nor the question of how to chlorinate at the right carbon (regioselectivity), but you do get the idea that even completely unfunctionalized alkanes can be turned into complex molecules with enough manipulations. Notice that for each functional group (at least for the first) we introduce, we must do some radical chain reaction to functionalize the otherwise unreactive alkane. Chlorination of alkanes is done widely on industrial scale as it is one of the few ways to functionalize an alkane.


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What determines whether an aldol reaction will occur again or eliminate?
The product of an aldol reaction is a $\beta$-hydroxy ketone. This can go on to eliminate to form an alkene thus:

However, the species on the far left of the above diagram (the $\beta$-hydroxy ketone product of the first aldol reaction) can react through an aldol reaction again given that it has two more available protons at the $\alpha$ position.
How is it possible to tell which reaction pathway will occur? Is this the issue that is referred to when the Aldol reaction is referred to as "hard to control"?

Partially yes. The big problem with aldol reactions in general is that there are a lot of species that can react together to form stable compounds resulting in low yield. There are numerous ways to counter this:

Specific enolate equivalants, like lithium enolates, silyl enol ethers, aza enolates, enamines, etc. These are among the most important.
Lack of proton in $\alpha$-position.
Strong thermodynamic favoring of certain products in combination with reversibility of the reaction.
etc.

As for your example the $\beta$-hydroxy ketone is more sterically hindered than acetone and is therefore not very likely to be the major contributor (although polymerization can still happen of course).
In general you cannot predict whether or not elimination of $\ce{H2O}$ occurs after the aldol reaction, but the likeliness can be increased by forcing certain reaction conditions.


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Why do some element symbols contain a single letter while others contain two letters?
As the title says, why do some element symbols contain a single letter while others contain two letters?

Would it be as simple as # Elements - # Letters in Alphabet > 0? So you need more letters per element if you want to abbreviate them all.

@Eljee I think that's the answer.

@Eljee - at the time people recognized there were elements, they already had names for a number of substances that were elements.  Even without knowing how many elements there were they already had namespace collisions if they went with single letters.  A lucky few elements got the prestigious single letters, the rest were left with two (and lets not mention those poor unnamed elements with 3!). An interesting clash of language, science, and nature...

There is a bit of history behind this.
In the website Development of the chemical symbols and the Periodic Table, it is stated that the Swedish chemist Berzelius updated Dalton's previous method.  He established a protocol for the symbols, still in use today:

In the class which I call metalloids, I shall employ the initial letter only, even when this letter is common to the metalloid and some metal.
In the class of metals, I shall distinguish those that have the same initials with another metal, or a metalloid, by writing the first two letters of the word.
If the first two letters be common to two metals, I shall, in that case, add to the initial letter the first consonant which they have not in common:

for example, S = sulphur, Si = silicium, Sb = stibium (antimony), Sn =
  stannum (tin), C = carbonicum, Co = cobaltum (cobalt), Cu = cuprum
  (copper), O = oxygen, Os = osmium, &c

The basis of the symbols are the Latin word for the elements.
A notes as what Berzellius was refering to when he used the term 'metalloid', from the Wikipedia page Origin and use of the term metalloid:

In 1811, Berzelius referred to nonmetallic elements as metalloids, in reference to their ability to form oxyanions.

.

Interesting, but by the time Berzelius made this rule the metals potassium (K), uranium (U), tungsten (W) and yttrium (Y) were all known.  What happened, is there something I am missing?


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NMR population difference from a pulse of known flip angle
I want to know how a radiofrequency pulse of known flip angle $\beta$ changes the population difference of a set of isolated spin 1/2 nuclei.
I know that the equilibrium magnetization is related to the population difference by:

$M_0= \frac{1}{2}\gamma\hbar\Delta{n_{eq}}$

I would like to know how to calculate the x, y, z components of the magnetization after the pulse, and then calculate the population difference from these values.

immediately after the pulse:
$M_z = M_0cos(\beta )$
$M_x = M_0sin(\beta )$
$M_y = 0$
so for example after a 90 degree pulse, all the magnetization has been rotated into the x-y plane, and after a 180 pulse, the magnetization is in the -z direction.

Thanks. How can I use this to calculate population differences ?

@J.LS  $M_z= \frac{1}{2}\gamma\hbar\Delta{n_{eq}}$, so no population difference after 90 degree pulse, and inverted population after 180 degree pulse

If there is no population difference after a 90 degree pulse, why does spin-spin relaxation occur ?

@J.LS the applied magnetic field is in the z direction.  The axes about which the protons are spinning are tilted, either up or down relative to the z direction.  The axes are precessing about the z axis, and are free to have any x and y value (not quantized with respect to these directions).   After the pulse, the x and y components of the spins are not randomly oriented in the x-y plane, but instead are preferentially oriented in the +x direction.  (continues below)

(continued from above) Due to local magnetic field inhomgeneity, the spins do not all precess at the same rate, and become out of phase with respect to each other, eventually becoming randomly oriented with respect to x and y.


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Mechanism for the formation of butanoic acid from ethylacetoacetate
My problem sheet asks me: "Draw a mechanism for the formation of $\ce{EtCH2COOH}$ from the reaction of $\ce{MeCOCH(Et)COOEt}$ with $\ce{NaOH}$ in $\ce{EtOH}$"
This has got me stumped. The mechanisms that I have attempted use the hydroxide ion as the base/nucleophile, is this right? My initial thoughts were to make the enolate by deprotonating between the caronyl groups then doing and aldol into the ketone part of another molecule of ethylacetoacetate but this just gives me a complicated molecule that I can't see reacting further.
I simply cannot find a way to make the product using only the reactants in the question.

Formally, it looks like reversing a mixed Claisen ester condensation, followed by saponification.

Thank you very much. Should I use the hydroxide ion as the nucleophile or should I use the ethoxide ion? Given that sodium hydroxide is dissolved in ethanol.

I thought about adding solid $\ce{NaOH}$ to ethanol and therefore used $\ce{EtO-}$ ;) :

Final arrow pointing towards me? Weird :D

Why not the hydroxide ion though? The hyroxide ion is used in the saponification.

Actually you're right: ethanolate is the stronger base! Hydroxide is correct and would explain the saponification step!

@RobChem It doesn't matter which (hydroxide or ethoxide) you use to initiate the retro-Claisen. If the ethoxide participates, then hydroxide will saponify the resulting ester.

I have one concern about this approach. How much of the substrate is protonated. The enolate will not cleave. This may only be a rate problem. Another route is to do an acid hydrolysis/decarboxylation followed by a haloform reaction.


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Corrosion of iron
First question regarding corrosion of iron:

Why is anodic area always located at nicks and scratches, where energy state is high? (does"high [potential] energy" mean unstable structure?)

Second questions(refer to the image):

Why and how does $\ce{Fe^{2+}}$ move to the top of the iron piling(part near the surface)?

Based on what I learn about conductor, there cannot charges piled up at a particular region of conductor, as opposed to insulator? Also the image is sorta misleading, the ions move through the iron piling or the electrolytes in sea water?

Can you show the full picture? Is this meant to be an iron pole in an ocean? The ions are probably meant to be moving up in the water and not through the pole.

A full image similar to your diagram with chemical reactions is found at a blog Chemistry Tutoring:

According to the UC Davis ChemWiki page Electrochemistry 7: Electrochemical Corrosion, the oxidation and reduction reactions occur in separate places in the metal - in the case of your diagram:

Oxidation occurs in the anode part of the iron pole submerged in water:

$$\ce{Fe_{(s)}->Fe^{2+}_{(aq)} +2e-}$$
As Burak Ulgat correctly noted in his comment, the $\ce{Fe^{2+}_{(aq)}}$ ions are dissolved in the water and are outside of the pole and are transported by the water. The electrons move through the metal in the pole (as iron is a metal, these electrons conduct easily) - this part is not shown in your diagram.
At the anodic region, for whatever reason, the iron is exposed to the seawater, may already have damage etc, so comes under 'attack'.  As $\ce{Fe^{2+}_{(aq)}}$ ions are 'liberated', pitting occurs as material is removed (dissolved) from this area of the pole. Often, these pits are deprived of oxygen, increasing the rate of pitting. The presence of seawater speeds up this process.

Reduction occurs at the cathode region, oxygen from the air is reduced to form hydroxide:

$$\ce{O2_{(g)} + H2O_{(l)} + 4e^{-}->4OH^{-}}$$
Then rust formation occurs as per the diagram above (summarised below):
$$\ce{Fe^{2+}_{(aq)} + 2OH^{-}_{(aq)}->Fe(OH)2_{(s)}}$$
$$\ce{4Fe(OH)2_{(s)} + O2_{(g)} + H2O_{(l)} ->Fe(OH)3_{(s)}}$$
According to the UC Davis ChemWiki, the process of corrosion continues if the presence of a depolarizer, which removes electrons from the metal, depolarizers include more noble metals, acids and in the case relevant to the diagram, oxygen. So, the oxygen depolarizer is removing electrons out of the iron, resulting in more iron ions being dissolved into the water at the anode.
The electrochemical process is summarised by the HyperPhysics page about Corrosion as an Electrochemical Process, in the diagram below:


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Cis/Trans-Decalin in ChemBio3D
According to my organic chemistry textbook, Cis-Decalin has the two rings nearly perpendicular to one another, with the middle hydrogens cis to one another. Trans-Decalin has the ring nearly flat (planar), with the middle hydrogens trans to one another.
However, typing Cis-Decalin and Trans-Decalin into ChemBioDraw (where they appear correctly in 2D form) and copying them into ChemBio3D shows the exact opposite conformations.
Is this a bug with ChemBio3D, or a mistake in the textbook?

I couldn't reproduce your problem. I my chem3d structures look as they should.


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When disqualifying trans fat, are we qualifying cis-fat as healthy?
I have a question about fats. This may sound cliché! Because we are in the era of "trans-fat" free, "unsaturated" food. Not to mention it's the period where people get sicker and die faster (oh may be, the statistics weren't obvious in the past...).
Fats are important part of our diet and major source of energy (the mighty omega-3 fatty acids). But, The trans-fat, it's a linear molecule, it tends to stick and stacks up, builds up bad cholesterol, so that's not good for our cardiovascular system...
Does it mean, the opposite configuration, cis-fat is healthy for our body?
If cis is what's naturally found, by default it's unsaturated fat isn't it?
PS:
It's that time of the year again, for a community project we (the people who are new to biology, organic, biochemistry education) need to explain that to kids of age 7 and 10. They are going to go for shopping with us and we have to pop our eye balls to make sure they don't buy the wrong stuff. Because everything they adore seems to contain trans-fat from breakfast cereals, syrups, chips to doughnuts...list goes on. Saying something isn't bad not going to convince them an atom! So we gotta explain why they should go for trans fat free, unsaturated food or in the worst case just choose a saturated food instead of a trans-fatty thing and prevent any "kids-melo-drama".
I am struggling with this explaination and I doubt it's going to be any easier for the kids.
The breakfast cereals reference.

If cis is what's naturally found, by default it's unsaturated fat isn't it?

Both cis and trans are unsaturated.  "Unsaturated" means having at least one C-C bond that is not a single bond (but instead a double or triple bond).  The terms cis and trans refer to the geometry of the molecule about the double bond.
Trans fat also occurs naturally in milk, for example fats of vaccenic acid
Also, since a fatty acid can have more than one double bond, it can be both cis and trans, for example fats of Rumenic acid found in milk.

When disqualifying trans fat, are we qualifying cis-fat as healthy?

Cis-fat can also have negative health consequences.  For example, according to the review article Dietary fat, insulin sensitivity and the metabolic syndrome Clinical Nutrition vol. 23 pages 447-456.

monounsaturated and ω-6 polyunsaturated fatty acids have less negative effect on insulin sensitivity than saturated fat

In other words, even consumption of too much unsaturated, naturally occuring, cis fat can have negative health consequences, such as insulin resistance.
Your question says:

Because everything they adore seems to contain trans-fat from breakfast cereals, syrups, chips...

but is this really true?

From what I've seen, cereals have little or no fat of any kind, let alone trans-fat.
Similarly, maple syrup has no fat.

Even potato chips don't necessarily contain any trans fat:

So we gotta explain why they should go for trans fat free, unsaturated food or in the worst case just choose a saturated food instead of a trans-fatty thing

The amounts of partially hydrogenated trans fat in foods has been greatly reduced in the last 10 years.  Milk and meats naturally contain trans fat.
I wouldn't want you telling my kids they can't drink milk or eat a hamburger.  Sure it's bad to eat too many hamburgers.  The focus should be a balanced diet.

@DavidPhD putting this into perspective and jargon for kids is quite a daunting task. Here's what I grasp so far, 1. Nothing is too good e.g. insulin resistance, milk issue- so eat the fats (unsat, cis, naturally found) in moderation. Select a list of food with those. 2. tran is anyway bad, so avoid it. Select a list of food with those and a list of substitutes for those. I still have no answer to a question like, "if these aren't good, why is it on sale?"...

@bonCodigo I added to my answer

As for the current regulation: label may not list trans fat since manufacturers are not required to list trans fat on the label if the food contains 0.5 g of trans fat or less. Well it's kinda like "smoke in moderation"... :) I did update my question a bit with the list of cereals I had found earlier, which was not added into the question. Because out there in the market it's not just one type of cereal or just one chip or one syrup.

@bonCodigo Corn Pops has 0.04g of trans fat http://www.bodbot.com/Nutrition/Foods/8068/cereals-ready-to-eat-kellogg-kelloggs-corn-pops compare to 0.2 g from a cup of milk.  Even human milk contains trans-fat, so basically you will be ripping babies from their mothers' breasts with your zero tolerance approach.


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Why does water evaporate spontaneously at room temperature despite ΔG > 0?
Standard Gibbs free energy of formation of liquid water at $\pu{298 K}$ is $\pu{−237.17 kJ mol-1}$ and that of water vapour is $\pu{−228.57 kJ mol-1}$ therefore,
$$\ce{H2O (l) -> H2O (g)}\qquad\Delta G =\pu{8.43{kJ mol-1}}$$
Since $\Delta G>0$, it should not be a spontaneous process but from common observation, water does turn into vapour from liquid over time without any apparent interference. Why does this happen?

possible duplicate of Why does water evaporate at room temperature?

@MARamezani The marked question does not discuss the process in terms of the second/third law of thermodynamics. (i wish to know about holes in my understanding of free energy)

Also, take a look at http://physics.stackexchange.com/questions/106754/how-can-water-evaporate-at-room-temperature also. AFAIK, this thing has nothing to do with $\Delta G$

thanks for the link, there are many quality answers but i don't feel satisfied with them as i still can't find justification for the above mentioned scenario.

I retracted my close vote. The other answers didn't mention how the Delta G issue can be explained. This question does require a different approach, anyway.

Maybe an appropriate example

Standard Gibbs free energy of formation of liquid water at 298 K is −237.17 kJ/mol and that of water vapour is −228.57 kJ/mol. Therefore, $$\ce{H2O(l)->H2O(g)}~~\Delta G=8.43~\mathrm{kJ/mol}$$
Since $\Delta G>0$, it should not be a spontaneous process but from common observation, water does turn into vapour from liquid over time without any apparent interference.

Your math is correct but you left out a very important symbol from your equations.  There is a big difference between $\Delta G$ and $\Delta G^\circ$.  Only $\Delta G^\circ$ means the Gibbs energy change under standard conditions, and as you noted in the question, the free energy values you quoted are the standard gibbs free energy of water and water vapor.
Whether or not something is spontaneous under standard conditions is determined by $\Delta G^\circ$.  Whether something is spontaneous under other conditions is determined by $\Delta G$.  To find $\Delta G$ for real conditions, we need to know how they differ from standard conditions.
Usually "standard" conditions for gases correspond to one bar of partial pressure for that gas.  But the partial pressure of water in our atmosphere is usually much lower than this.  Assuming water vapor is an ideal gas, then the free energy change as a function of partial pressure is given by $G = G^\circ + RT \ln{\frac{p}{p^\circ}}$.  If the atmosphere were perfectly 100% dry, then the water vapor partial pressure would be 0, so $\ln{\frac{p}{p^\circ}}$ would be negative infinity.  That would translate to an infinitely negative -- i.e. highly spontaneous -- $\Delta G$ for the water evaporation reaction.
Small but not-quite zero dryness in the atmosphere would still lead to the $\Delta G$ of water vapor that is more negative than liquid water.  So water evaporation is still spontaneous.
Extra credit: given the standard formation energies you found, and assuming water is an ideal gas, you could calculate the partial pressure of water vapor at which $\Delta G = 0$ for water evaporation.  And the answer had better be the vapor pressure of water, or else there is a thermodynamic inconsistency in your data set!

There are two considerations about why water evaporates but only one of them has to do with the Gibbs free energy of vaporisation.
As the first answer correctly states, there is an equilibrium (heavily biased in favour of the liquid not the vapour) at room temperature. In equilibrium (which is what the $\Delta G$ measures) there is enough thermal energy to put a small amount of water into the vapour phase but most water will remain liquid. This is the correct analysis in a closed vessel. The water doesn't continue to evaporate once the equilibrium is reached.
But, mostly, you don't observe water in closed vessels. This explains why the water continues to evaporate. The point is that, when the vapour can be removed from the vessel containing the liquid, the normal equilibrium is not established and the water can slowly evaporate (at least as long as the surrounding environment has a lower humidity than the equilibrium vapour pressure).
In open systems the simple thermodynamic numbers don't correctly describe the long term behaviour because equilibrium isn't what happens.

I think this example might be appropriate: Water in an open vessel evaporates (open system) while the one in ones bottle doesn't (closed system). Very good answer!

$\Delta G = -RT \ln K$
Which means there for any given process an equilibrium constant can be deduced.  Given the products are 'gaseous' they will escape the system, and further drive the equilibrium.
In this case,
$K = 0.05$
Applied pragmatically, we define a system as 'near the puddle'.  When the water evaporates, it can either condense again, or move 'far from the puddle' and hence leaves the system.  Over time the 'moving away' portion of this process leads to what we observe humanisticly as the puddle disappearing.

So basically, the vapour leaves the system and is hence out of consideration?

Roughly yes.  Even though a small amount is vapor, that vapor is constantly leaving at some rate, and the equilibrium shifts to more vapor to try to balance.

The Gibbs energy is defined for systems at a specified temperature and pressure. Think of a container closed to any transfer of material, immersed in a heat bath at a specified temperature, but with adjustable volume, like a cylinder with a movable piston. Put nothing but water in it, set the heat bath to 25 C, and adjust the volume so the pressure is 1 atmosphere. Since the vapor pressure of water at 25 C is 0.0313 atmospheres, there will be no vapor, the piston will be in contact with the liquid water, pressing down with one atmosphere pressure. Your equation and conclusions are correct.
If you lower the pressure to 0.0313 atm by increasing the volume , then you can set the volume to anything between the volume of the water and the volume of the vapor when the last bit of water just evaporated and still maintain 0.0313 atm. If you increase the volume more, you are setting the pressure below the vapor pressure, and all the water will evaporate.
When you note that water will evaporate at 25 C, you are specifying a huge volume with air at 1 atm, and a little water, so you are effectively setting the partial pressure of the vapor to something very small, and of course, it will evaporate.

I think the starting assumption is not correct.
Water does not always evaporate.
For instance, at night, the temperature decreases due to radiative cooling. If the temperature drops to the dew point temperature, then dew will form upon further cooling.
This is an example of an open system, where water does not evaporate.
Also near a large body of water, water will not always evaporate. That would mean we would not have oceans. They would all have evaporated.
The air over the sea will be at equilibrium, and it is sort of a'closed space' whose boundaries are determined by the fact that water molecules need to diffuse away from the water air interface, and they do so at finite speed.
Also there is energy required for evaporation, which has to come from somewhere. In general, this means that the temperature will drop, which will in turn bring equilibrium conditions closer.
The whole notion of Gibbs Free Energy was introduced to explain that systems tend to equilibrium and a minimum in the Gibbs free energy.
Now an air/water system will also tend to equilibrium. And evaporation will stop.
In general processes don't go on forever spontaneously. Things can basically evaporate because the heat the sun supplies and condense back at night, when the earth cools down due to radiative cooling. And then you have to correct for winter and summer.

Water is evaporating at frost temperatures

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Decarboxylation product of bicyclic compound
What is the decarboxylation product of the following species:

It may seem straightforward but the product enolate involves a double bond with one end at the bridgehead carbon which is impossible/very unstable. What is the correct product and how is the product reached mechanistically.

Could the resulting enol tautomerize?

Yes it could. But the enolate must surely form initially?

Enolization is a blindingly fast reaction.  Even though the intermediate is unstable, the easy pathway forward might allow formation of the resulting ketone, despite the bridgehead.

${\beta}$-keto acids typically decarboxylate very smoothly at temperatures between 100-200°C. The mechanism involves a cyclic, 6-membered transition state that first produces an enol intermediate (see figure).

However in the case of your bicyclo[2.2.1] ${\beta}$-keto acid, a bridgehead double bond would result.  As you correctly point out, such a structure would be very high energy.  Therefore this compound does not decarboxylate at normal temperatures (reference, see p. 218, scheme 6.53).
I'm sure if you heated it hot enough something would happen.  It may decarboxylate via a free radical mechanism. The resultant bridgehead free radical could fragment or dimerize.  Alternately, and my preferred guess, would have the starting compound, in equilibrium with its non-bridgehead enol tautomer, undergo a reverse Diels-Alder reaction as pictured above.

My term hasn't started again yet. I'll let you know

she said that it does not decarboxylate

Thanks for getting back on this.  So is my answer acceptable?

Well, she just said it wouldn't decarboxylate. I didn't offer her your response as a possibility I'm afraid. Although, she seemed to suggest heating it wouldn't decarboxylate it. She didn't suggest anything else so I guess nothing happens until it combusts.

You're a hard grader, but that's fine :)  You asked if the compound decarboxylated, I correctly answered "no" and correctly explained why.  I then added an aside for the interested reader who might want to know something beyond the original question.


text:
Alkaline glycine buffer molarity
Having never prepared a buffer solution before, I am hoping for a little help in understanding the process.
This method I plan to use calls for a $\pu{0.1M}$ glycine buffer at pH $10.6$. Some internet searching yielded this recipe (very bottom of the page) for making a glycine-$\ce{NaOH}$ buffer at $10.6$ pH, which I assume is what I'm looking for.
Most of it seems to make sense to me. $\pu{0.2M}$ stock solutions of both glycine and $\ce{NaOH}$ would be prepared as:
0.2M NaOH solution = 0.2 * 39.9971 = ~8g NaOH/l
0.2M glycine solution = 0.2 * 75.07 = ~15g glycine/l
(Molar mass: NaOH = 39.9971; glycine = 75.07 g/mol)

And, to yield a buffer at $10.6$ pH, mix:
25ml glycine stock + 22.75ml NaOH stock + deionized water to 100ml

Where I get lost is figuring out the molar concentration of the resulting solution, since there are two different solutes. Would it be just the total number of moles present of both, something like:
(25ml + 22.75ml) * 0.2M / 100ml = 0.0955M ?

It seems pretty close to the $\pu{0.1M}$ concentration called for, but I can't seem to find any confirmation of this.

As per comments and revisiting the question and my first answer, I am amending my answer to reflect the correct one.
Because you are converting glycine to its sodium salt, the final solution concentration will be 2 times the concentration of glycine.
How many moles of glycine are in $25\,\mathrm{ml}$ of stock solution?
$${0.2\,\mathrm{mol}\,\ce{glycine}\over 1000\,\mathrm{ml}} = {x\,\mathrm{mol}\,\ce{glycine}\over 25\,\mathrm{ml}}\implies x = 0.005\,\mathrm{mol}\,\ce{glycine}$$
You need $0.005\,\mathrm{mol}$ of $\ce{NaOH}$ to convert the glycine to the sodium salt, so:
Total amount of substance in buffer solution:
$$0.005\,\mathrm{mol} + 0.005\,\mathrm{mol} = 0.01\,\mathrm{mol}$$
Total volume of buffer solution:
$$100\,\mathrm{ml} = 0.1\,\mathrm{L}$$
Molarity of buffer solution:
$${0.01\,\mathrm{mol}\over 0.1\,\mathrm{L}} = 0.1\,\mathrm{M}$$

The first answer might be wrong. Please verify, because when you use NaOH to convert glycine to its sodium salt you should actually substract but not add NaOH moles in the total mole calculation and then add them together. As such the molarity will not change. Just think about it the pKa for glycine-COOH is 9.6 and your desired pH is 10.6. Therefore the number of salt moles must be greater than the number of acid moles. In this case most of the glycine is converted to its sodium salt.

@bob I also don't think this answer can be correct. The only substance that buffers in the scenario is glycine and the converted sodium salt. Therefore the initial amount of substance of glycine is also the total amount of substance of the buffering components.

It is much more simple. Just make 0.1 M solution of glycine by dissolving 0.75g glycine per 100ml volume (0.1 * 0.1 * 75). In the beginning dissolve it only in about 70 ml water. Adjust the pH of this glycine solution to 10.6 with 1N NaoH. Adjust the volume to 100ml. By my HH calculation it will consume about 900 micro liters of 1N NaoH. You can use more dilute NaOH to have a better control while adjusting the pH. The molarity of the final solution will be 0.1N not any fraction.
You can use this approach for making virtually any buffer. The approach is that you will start with the free acid weighed to the desired molarity and adjust the pH upwards.
Starting with a sodium salt of the free acid and adjusting the pH downwards with HCl is not desirable because that would introduce chloride ion into the system in the form of sodium chloride which can interfere with electropheresis.


text:
Figure Headings in Laboratory Reports
How do I write a figure legend for a thin layer chromatography plate drawing?

Provide all the data that are necessary to reproduce the experiment:

TLC material (silica gel or aluminium oxide)
composition of the mobile phase
special tricks, such as running the TLC in methanol first for 5 mm to squeeze the initial spot to a very narrow band
detection of the separated bands/spots:

$\mathrm{R_f}$ values
do the spots have a colour
do the show fluorescence
which reagents did you use for staining

UPDATE
If you report on a series of measurements, it's usually good practise describe the conditions that didn't change (type of TLC plate, staining solutions, etc.) in the general section of the experimental part of the report/article/thesis.

At the bottom of each spot mark your legends by alpha/numerals. Outside put the meaning of your legends. Mention solvent system used for elution.


text:
Reaction of ethylacetoacetate with Iodine
A "reaction map" in my lecture handout suggests that the product of this reaction is:
So, basically overall it just merges two ethylacetoacetate molecules together at the central carbon. What is the mechanism for this? My thoughts are to swap the acidic proton for an Iodine atom by making the enol (in acidic conditions to avoid iodoform reaction) and react with $I_2$. Then, the enol of another molecule of EAA substitutes the Iodine to make the product. However, I don't know whether this is $S_N1$ or $S_N2$. Typically alkylation is $S_N2$ (I don't know why, please explain if you can) but in this case its a bulky secondary substituent and the solvent is protic (acidic conditions). Surely it's $S_N1$, right?

Are you certain that the coupling itself isn't a radical reaction?
The bond energy of an ordinary $\ce{C-I}$ bond is in the range of $210\,\mathrm{kJ\cdot\,mol^{-1}}$, which isn't a hell of a lot. Moreover, the resulting radical of an 1,3-dicarbonyl compound is nicely stabilized.

Iodine is definitely involved.

Sorry if I didn't make myself clear enough :)  I meant: homolytical cleavage of the iodinated $\beta$-ketoester is the first step,  yielding an iodine atom and a stabilized radical.

On a side note, these radicals of 1,3-dicarbonyl compounds can also be generated (this time without the iodine) using $\ce{Mn(III)}$acetate. Barry Snider publishes a lot on that.

@RobChem My pleasure!


text:
What causes spray cans to get cooler when shaken?
I've been searching the web for this phenomenon and surprisingly I found only one post about it. And the reply to this question about canned air begins with words:

I don't know about the shaking part (it is hard for me to see how that would make a difference)

This beginning has strongly shaken my faith, so for sure I picked up a can with deodorant and shaken it. And it does become cooler. Significantly! After standing still, the can doesn't seem to become warmer, so this reaction doesn't revert itself and maybe is irreversible.
As I imagine it, when being shaken the average pressure in the can as a whole can't possibly change. Also it's interesting how you add energy by shaking only to have more energy drained by cooling.
So please, can you explain this strange behavior? Could you also cover the energy flow? When does the energy return back out of the can?

OK! I admit it! You got me interested! Hmm, as the deodorant isn't very pressurized, adiabatic expansion upon releasing isn't playing an effective role here. But, if you have the bottle really close to your skin, you might be feeling the alkanes that instantly evaporate when sprayed. Shivering

I just tried this myself with a can in each hand simultaneously and only shook one, but I could not reproduce the effect to my satisfaction, even after some alternation. To me, if there is a difference, it is not large. Also, I can see some confounding factors, the main one being that we don't detect temperature, but rather the rate of heat exchange; the shaking hand will grip tighter, and also some heat is being generated by the hand while it is gripping and shaking, both of which favour removal of energy from the hand at a faster rate.

@NicolauSakerNeto You make good points. I tried to grip the can as hard as I can and I stil observed the effect. But I indeed need some measured evidence only I don't know how to get it. Also it doesn't seem to work with all cans. Definitelly does not work with bottles of water or glasses. According to your theory, I should observe the same effect even with empty glass. I'll also try to ask somebody unaware of the phenomenon without telling him what I expect them to observe. That's the only alternative to precise laboratory equipment which I don't have.

Here's another important factor; metal cans are quite thin and have small thermal capacity, which means most of the heat is transferred quickly and predominantly into the liquid, and from there on heat transfer is limited in part by how quickly warmer liquid molecules can diffuse away from the surface in contact with the hand. Shaking will allow cooler molecules to more quickly reach the contact region, once again speeding up transfer. If the container is made of a thermal insulator or has a high thermal capacity (thick walls), the liquid inside may make no difference in a short experiment.

@TomášZato Adiabatic expansion.

TL;DR: Spray cans don't actually get colder when shaken.  However, shaking a can does increase heat conduction from your hand to the can, making it feel colder.

Humans don't actually sense external temperature directly; our thermoreceptors are located under the skin, and thus effectively measure the rate at which body heat is lost through the skin.  This is why, for example, touching a metal surface will feel noticeably colder than a wooden or plastic surface at the same ambient temperature.
When you pick up a spray can, it is typically at room temperature, i.e. significantly below body temperature.  The can itself is a thin metal cylinder, and thus conducts heat well, but has very little heat capacity in itself.  Thus, the rate of heat loss from your hands (and thus the sensation of coolness) mainly depends on how quickly whatever's on the inside of the can will absorb heat from the surface.
Now, a typical aerosol spray can contains a propellant (mixed with the actual payload fluid) which would be slightly above its boiling point (and thus in gaseous form) at room temperature and pressure, but is kept liquid by the elevated pressure inside the can.  As the spray valve is opened and the pressure reduced, some of the propellant fluid will boil, bringing the pressure back up until the liquid and gas phases are again at equilibrium.
What this means is that, in a half-empty spray can, the top half of the can will contain gaseous propellant (which, being a gas, will not absorb heat very effectively), while the bottom half will contain liquid propellant (mixed with the payload fluid) at just under its boiling point.  This liquid will absorb heat much better than the gas at the top, both because it's denser (and thus has higher heat conductance and capacity), and also because heating it will cause some of it to boil, converting part of the absorbed heat into its latent heat of vaporization.  (The boiling effect actually causes a half-empty spray can to act, to some extent, as a diode heat pipe.)
Thus, even without shaking the can, it's possible to observe that the bottom end of a half-empty spray can will feel cooler to your hand than the top end.
Now, when you hold the can and shake it, you cause the propellant liquid to slosh around inside, coating all of the inner surface of the can instead of just the bottom.  The sloshing also significantly increases the heat transfer from the can to the liquid, by causing the warmer liquid near the surface of the can to mix with the colder liquid in the interior.  (This is essentially the same reason as why moving air feels cooler on your skin than still air.)  Thus, a shaken can will generally feel even cooler to touch than either end of an unshaken can.

agitating the can causes the liquid / gas equilibrium to shift toward the gas phase. since the can is not a closed thermodynamic system, the required energy (latent heat of evaporation) is "borrowed" from the environment.

Why does agitation shift the equilibrium towards the gas phase?

Surface area is affected: dG = σ dA

this was merely a hypothesis, but seemed plausible enough


text:
ordering the hydrophilicity of some polymers
I need to compare the hydrophilicity of those materials to compare contact angles of water with those materials' surfaces for surface tension measurement. Materials are:
PTFE (Teflon), PVC, PS, PVDF, Si-Wafer-Si-SiO$_x$, Parafilm and PMMA
Ordering the polarity can also work. Where can I found such a comparison? Can someone help me with the ordering. Thanks.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.


text:
Which form of chromium oxide is formed naturally in the atmosphere?
I'm trying to figure out which form of chromium oxide forms naturally in our atmosphere, i.e., if you take a piece of elemental $\ce{Cr}$ from vacuum to a normal, room temperature environment with air, does it form $\ce{CrO}$ on the surface? Or $\ce{Cr2O3}$? Etc.
This site says:

Chromium (III) oxide, $\ce{Cr2O3}$ is the main oxide of chromium.

So it seems like it might be that one, but "main" is pretty vague. Does anyone have a more definitive answer, with a source?
thank you!

An investigation of the phase diagram reveals that the only stable chrome oxide near room temperature is $\ce{Cr2O3}$.  $\ce{Cr3O4}$ appears, but only near 2000K.  There is no stable chrome analogue to $\ce{FeO}$ on the phase diagram.
Thus, the one stable oxide that will form under normal atmospheric conditions is $\ce{Cr2O3}$. However, note that it is easy to find mixed metal-oxygen compounds with high mutual solubility.
If you are interested in the thermodynamics of the $\ce{Cr-O}$ system, you could start with M. Kowalski and P.J. Spencer, Calphad 19(3) 229-243 (1995), which covers the $\ce{Cr-O, Fe-O}$ and $\ce{Ni-O}$ systems

Thanks! That paper was actually perfect because I had the same question about Nickel Oxide!

Quite welcome! If your institution has access, the ASM Alloy Phase Diagram Database at asminternational.org is a great place to start looking for info. Thermodynamic information can be found in various journals including Calphad, J. Alloys and Compounds, Intermetallics, amd Met. Trans A.

Drawing a parallel to the formation of rust (oxides of iron), in which the following reactions:
$$\ce{Fe<=>Fe^2+ +2e-}$$
$$\ce{4Fe^2+ +O2<=>4Fe^3+ +2O^2-}$$
form both of the oxidation states of iron, reacting to eventually form both $\ce{Fe2O3}$ and $\ce{FeO}$, the two most common forms of rust. The more common of the two forms is $\ce{Fe2O3}$, which is likely because $\ce{Fe^3+}$ has an electron configuration of $[\ce{Ar}] 3d^5$, which is more stable than the electron configuration of $\ce{Fe^2+}$, which is $[\ce{Ar}] 3d^6$.
It is likely then that both $\ce{CrO}$ and $\ce{Cr2O3}$ form when $\ce{Cr}$ is exposed to the atmosphere, and also that $\ce{Cr^3+}$ has a more stable electron configuration than $\ce{Cr^2+}$, though this is not intuitive since neither half an empty or half-filled 3d orbital.

I am not sure about CrO forming under normal atmospheric conditions. If it did, it would likely be the predominant oxidation product (similar to FeO predominating when iron is oxidized at high temperature). Passivation (therefore "stainless" in stainless steel) would be much less effective with Cr. In reality Cr(II) seems to exist primarily as a solution species at high temperature.

As explained by Jon Custer, only chromium(III) oxide is thermodynamically stable as a pure chromium oxide and only that oxide is normally formed on chromium in our atmosphere. However, chomium(II) oxide may exist as a component in solid solution with other metal(II) oxides. Gasik[1] gives a phase diagram for chromium oxide in magnesium oxide at high temperature under an oxygen fugacity equal to that of air (0.2 atmosphere). The $\ce{CrO}$ component is soluble in magnesia up to about 16 weight percent $\ce{CrO}$ at 1500°C, with the solubility increasing at higher temperature. Adding chromium above the solubility limit, or cooling a given composition so that the chromium(II) solubility drops, results in formation of the spinel phase $\ce{MgCr^{III}_2O4}$. Since magnesia is plentiful in Earth's mantle, chromium delivered there may take the form of this chromium(II) solution species, releasing oxygen or oxidizing other elements (such as iron) as it is reduced from chromium(III).

[Eechout et al2](https://doi.org/10.2138/am.2007.2318) studied the oxidation state of chromium in minerals synthesizedunder physical n chemical conditions simulating the lower mantle; they also report that chromjum(III) used in the starting material reduces to chromium(II) and thus, in oxide minerals to the $\ce{CrO}$ component:

$\ce{Cr}$ K-edge X-ray absorption near-edge structure (XANES) spectra were recorded on $\ce{Cr:MgO}$ periclase and $\ce{Cr:(Mg,Fe)O}$ ferropericlase synthesized at different pressures (4 and 12 GPa) and temperatures (1200 to 1400 °C) at reducing oxygen fugacity conditions (~iron-wüstite buffer IW to IW - 2), and on $\ce{Cr:MgSiO3}$ perovskite with 0.5 wt% $\ce{Cr2O3}$... In $\ce{Cr:(Mg,Fe)O}$ the relative $\ce{Cr^{2+}}$ content reached at most 12.5% but results from Mössbauer spectroscopy combined with chemical composition data suggest that some $\ce{Cr^{2+}}$ oxidized during cooling through the reaction $\ce{Cr^{2+} + Fe^{3+} -> Cr^{3+} + Fe^{2+}}$. In iron-free $\ce{Cr:MgO}$, the $\ce{Cr^{2+}}$ content is much higher and reaches ~40%. In $\ce{Cr:MgSiO3}$ perovskite with 0.006 $\ce{Cr}$ pfu (similar to estimated lower mantle abundance), chromium is mainly divalent.

References

Mihail I. Gasik, "Chapter 8 - Technology of Chromium and Its Ferroalloys",
Editor(s): Michael Gasik,
Handbook of Ferroalloys,
Butterworth-Heinemann, 2013,
Pages 267-316,
ISBN 9780080977539,
https://doi.org/10.1016/B978-0-08-097753-9.00008-3.

Eeckhout, Sigrid Griet; Bolfan-Casanova, Nathalie; Mccammon, Catherine; Klemme, Stephan; and Amiguet, Elodie. "XANES study of the oxidation state of Cr in lower mantle phases: Periclase and magnesium silicate perovskite" American Mineralogist, 92(5-6), 2007, pp. 966-972. https://doi.org/10.2138/am.2007.2318