text:
Will the total equilibrium pressure increase in gas phase reaction PCl5(g) ⇌ PCl3(g) + Cl2(g) after chlorine is removed?

Will the total pressure be greater than the initial equilibrium pressure if all $\ce{Cl2}$ is removed and the following reaction is allowed to re-equilibrate?
$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$
Assume constant temperature and volume.

I know that when $\ce{Cl2}$ is removed, the reaction will go forward and I also know expressions of $K_p$ and $K_c$. I also know Le Chatelier's principle. I have thought about it for a long time but have not made any progress.

The final pressure will simply not change, if the temperature is maintained constant, whatever the value of the equilibrium constant.. If the system contains $n$ moles $\ce{PCl5}$ at the beginning, and if $x$ moles $\ce{PCl5}$ are transformed into $\ce{PCl3}$ and if $\ce{Cl2}$ is removed by reaction with a metal for example, the final amount of $\ce{PCl5}$ is $n-x$ and the amount of $\ce{PCl3}$ is $x$. As a consequence, the total amount of gas is $n-x + x = n$. So it does not change.

@Maurice note the question has been edited to make it more clear.

This answer may help https://chemistry.stackexchange.com/questions/150785/the-effect-of-dilution-on-the-rate-of-a-system-in-equilibrium/150799#150799

@Buck Thorn. Whatever the changes in the text, the number of gaseous moles does not change as the reaction proceeds, if the $\ce{Cl2}$ is eliminated from the gaseous phase (for example by reaction with iron producing solid $\ce{FeCl3}$).

First it is necessary to understand that due to the stoichiometry of the reaction, if you move the equilibrium to the products side, pressure will be increased. This is based on the fact that 1 mole of gas produces 2 moles of gas. In other words the number of molecules in the system is increased. By removing all of the $\ce{Cl2}$ you force the reaction to move to the products side (Le Chatelier's Principle). Thus more $\ce{PCl3}$ will be produced along with $\ce{Cl2}$. Eventually you will end up with more atoms inside the system and that will lead to increase of the pressure.

@Kelesedes Adonios I agree that the total number of molecules generated increases, but we are removing the chlorine fraction entirely, right? So, a product of partial pressures equal to the earlier one will develop gradually, right? But, will this increase the number of molecules present in the reaction mixture? Would you please support your answer with maths?

No.
For this problem I'm going to assume ideal gas conditions, mostly because looking into nonideal behavior of PCl3 and PCl5 at unspecified pressure and temperature sounds painful.  There are some intermolecular forces with phosphorus compounds to consider, but I didn't see much about PCl3/PCl5 dimerization in a quick search and this table gives an entry for PF3 that doesn't seem extraordinary, though it doesn't have PCl3 or PCl5.  My guess is PCl3 ought to be smaller than PCl5 if anything.
For ideal conditions, you can just say every particle smacks up against the walls of the container with a certain average energy that depends only on temperature, not what the particle is.  1 kPa = 1 J/L, representing the work needed to push back all the particles in one degree of freedom for a distance x area.  (The energy in one degree of freedom is half that, but thermodynamics won't let you stop them in their tracks at absolute zero, so you have to pay double).
In other words, the amount of pressure is proportional to the number of particles ("Avogadro's law")  If you remove the chlorine, the number of particles won't change when you turn PCl5 into PCl3.  Assuming the question means first you remove the chlorine then you see the equilibrium reestablish itself, with more chlorine being removed, you ought to see no change in pressure.  Otherwise you see a reduction in proportion to the original amount of chlorine.

The statements in the last paragraph are incorrect. If you continuously remove chlorine, the pressure will decrease a lot. If you remove chlorine once, it will also decrease.

@Karsten - under an ideal gas approximation, a mole of PCl3, PCl5, Cl2, or burrito gas exerts the same pressure when at the same volume or temperature.  Removing the Cl2 therefore cannot reduce the pressure below what it started at, unless the PCl3 would show non-ideal characteristics (such as dimerization) that reduce its pressure to a greater degree than PCl5.

I know that when $\ce{Cl2}$ is removed, the reaction will go forward and I also know expressions of $K_p$ and $K_c$. I also know Le Chatelier's principle. I have thought about it for a long time but have not made any progress.

Seven years have passed for thinking about the answer. Knowing the direction of the reaction is a good first step. We can also summarize what happens to the pressure when the reaction goes forward or reverse:
$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$
For every molecule of $\ce{PCl5}$ that reacts, we gain one molecule. Said differently, the increase in gas molecules total is equal to the increase in chlorine molecules.
So if you remove all the chlorine gas, the reaction would have to replace all of it by going forward to reach the same pressure (assuming ideal gas). Now we are back at the original pressure. Are we at equilibrium?
The chlorine concentration is the same, but the $\ce{PCl5}$ is lower than the starting point (we had to use up some to replenish the chlorine gas), and the $\ce{PCl3}$ concentration is higher. If you look at the equilibrium constant expression, the reaction is not at equilibrium but has to reverse to reach it, decreasing the pressure.
Here is a different way to think about this. You start with the initial state $S_0$. You take away the chlorine and let the system reach equilibrium. This is state $S_1$, which has a certain pressure and a certain concentration of chlorine $c_1$. Now you ask what happens when you add the chlorine again. To remain at the same pressure, all the added chlorine would have to react, using up $\ce{PCl3}$ and making $\ce{PCl5}$. The chlorine concentration would still be $c_1$, but there would be more $\ce{PCl5}$ and less $\ce{PCl3}$ now than in state $S_1$. To reach equilibrium, the reaction has to go forward, increasing the pressure when it goes back to the equilibrium state $S_0$.
Both ways of thinking about it give the same conclusion: Removing chlorine will reduce the equilibrium pressure. This is reflected in one way of explaining Le Chatelier. "The reaction will go in the direction that partially undoes the change imposed from the outside." The principle does not always give you the right answer, so it is good to check with a specific example using the equilibrium constant expression and the criterion $Q$ vs $K$.

Let:
$A$ represent $\ce{PCl5}$
$C$ represent $\ce{PCl3}$
$D$ represent $\ce{Cl2}$
Then the reaction becomes:
$$\ce{A <=> C +D}$$
Let:
(1) represent "before removing $\ce{Cl2}$"
(2) represent "after removing $\ce{Cl2}$"
Assuming that, initially, we have $P_{A0}$ amount of $A$ and no $C$ or $D$ present:
$$P{A_1= P_{A0}-x}$$
$$P{C_1=0+x}$$
$$P{D_1=0+x}$$
$$---------$$
$$P_1=P_{A0}+x$$
Then, we remove all the $\ce{Cl2}$, so more products are formed:
$$P{A_2=P_{A0}-x-y}$$
$$P{C_2=x+y}$$
$$P{D_2=y}$$
$$---------$$
$$P_2=P_{A0}+y$$
Then, we define the difference in total equilibrium pressure as:
$$\Delta P=P_2-P_1=y-x$$
Temperature is constant, so $K_P$ won't change after removing $\ce{Cl2}$:
$$K_P=\frac{P_{C2}\;P_{D2}}{P_{A2}}=\frac{P_{C1}\;P_{D1}}{P_{A1}}$$
Rearranging:
$$\left(\frac{P_{C2}}{P_{C1}}\right)\left(\frac{P_{D2}}{P_{D1}}\right)=\left(\frac{P_{A2}}{P_{A1}}\right)$$
Substituting in terms of $x$, $y$, and $P_{A0}$:
$$\left(\frac{x+y}{x}\right)\left(\frac{y}{x}\right)=\left(\frac{P_{A0}-x-y}{P_{A0}-x}\right)$$
Now, we define the dissociation fractions of $A$ for equilibrium system (1) and (2), respectively as:
$$\alpha=\frac{x}{P_{A0}}$$
$$\beta=\frac{y}{P_{A1}}=\frac{y}{P_{A0}-x}$$
Solving for $x$ and $y$:
$$x=P_{A0}\;\alpha$$
$$y=P_{A0}\;\beta-x\;\beta=P_{A0}\;\beta-P_{A0}\;\alpha\;\beta$$
Substituting $x$ and $y$ in the expression we derived earlier:
$$\left(\frac{P_{A0}\;\alpha+P_{A0}\;\beta-P_{A0}\;\alpha\;\beta}{P_{A0}\;\alpha}\right)\left(\frac{P_{A0}\;\beta-P_{A0}\;\alpha\;\beta}{P_{A0}\;\alpha}\right)=\left(\frac{P_{A0}-P_{A0}\;\alpha-P_{A0}\;\beta-P_{A0}\;\alpha\;\beta}{P_{A0}-P_{A0}\;\alpha}\right)$$
All the $P_{A0}$ terms cancel out:
$$\left(\frac{\alpha+\beta-\alpha\;\beta}{\alpha}\right)\left(\frac{\beta-\alpha\;\beta}{\alpha}\right)=\left(\frac{1-\alpha-\beta-\alpha\;\beta}{1-\alpha}\right)$$
After expanding, we get:
$$\alpha^3-\alpha^3\beta^2+3\alpha^2\beta^2-\alpha^2\beta-\alpha^2-3\alpha\beta^2+\alpha\beta+\beta^2=0$$
The positive solution for $\beta$ in terms of $\alpha$ is:
$$\beta=\frac{\alpha\left(\sqrt{4\alpha^2-8\alpha+5}-1\right)}{2(1-\alpha)^2}=f(\alpha)$$
Recalling our derived expression for $\Delta P$:
$$\Delta P=y-x=P_{A0}\;\beta-P_{A0}\;\alpha\;\beta-P_{A0}\;\alpha=P_{A0}(\beta-\alpha\beta-\alpha)$$
$$\Delta P=P_{A0}[(\beta(1-\alpha)-\alpha)]$$
Since $P_{A0}>0$, the term in between "[ ]" controls the sign of $\Delta P$, so we can evaluate 3 different outcomes.
Assuming just for a moment that $\Delta P=0$, we would get:
$$\beta(1-\alpha)-\alpha=0$$
Solving for $\beta$:
$$\beta=\frac{\alpha}{1-\alpha}=g(\alpha)$$
Then the three possible outcomes are:
(i) If $\beta>g(\alpha)$, then $\Delta P >0$
(ii) If $\beta=g(\alpha)$, then $\Delta P =0$
(iii) If $\beta<g(\alpha)$, then $\Delta P <0$
Now, we can compare $f(\alpha)$ vs $g(\alpha)$, noting that $f$ represents the actual value of $\beta$, and $g$ represents the exact boundary at which $\Delta P$ =0.
Considering $\alpha$ is a dissociation fraction, and the fact that for equilibrium to make sense neither 0% nor 100% of dissociation is possible, the domain of $\alpha$ is:
$$\alpha ∈ (0,1)$$
We can now tabulate $\alpha$ vs $f(\alpha)$ vs $g(\alpha)$:
\begin{array} {|r|r|}\hline \alpha & f(\alpha) & g(\alpha) \\ \hline \to 0 & \to 0 & \to 0 \\ \hline 0.1 & 0.06538 & 0.1111 \\ \hline 0.2 & 0.1386 & 0.25 \\ \hline 0.3 & 0.2206 & 0.4286 \\ \hline 0.4 & 0.3122 & 0.6667 \\ \hline 0.5 & 0.4142 & 1 \\ \hline 0.6 & 0.5262 & 1.5 \\ \hline 0.7 & 0.6463 & 2.333 \\ \hline 0.8 & 0.7703 & 4 \\ \hline 0.9 & 0.8912 & 9 \\ \hline \to 1 & \to 1 & \to \infty \\ \hline  \end{array}
From the table, we can observe that for every value within the domain of $\alpha$, $f$ is smaller than $g$:
$$\alpha ∈ (0,1) \to f(\alpha)<g(\alpha)$$
In other words, condition (iii) is met, and $\Delta P$ is negative, so total equilibrium pressure decreases after removing all the $\ce{Cl2}$:
$$\beta<g(\alpha)\to \Delta P<0\to P_2<P_1$$
Although it's not very important, as a side note it also follows that:
$$\alpha ∈ (0,1) \to \beta<\alpha$$
I included a graph below to illustrate the solution of the problem, where I also added a $\beta=\alpha$ function as a reference line:


text:
Alternative Program for Orbital Population Analysis
Gaussian has Orbital-by-Orbital Population Analysis.
From the manual:
# UHF/6-311+G(d) Pop=Orbitals=3
Here is the resulting output from a calculation on FeO+ quartet:

Atomic contributions to Alpha molecular orbitals:
Alpha occ 16 OE=-0.923 is Fe1-d=1.00
Alpha occ 17 OE=-0.699 is O2-p=0.88
Alpha occ 18 OE=-0.690 is O2-p=0.68 Fe1-s=0.21
Alpha vir 19 OE=-0.253 is Fe1-s=0.70 Fe1-p=0.27
Alpha vir 20 OE=-0.188 is Fe1-p=0.71 O2-p=0.29
Alpha vir 21 OE=-0.133 is Fe1-p=1.04

Atomic contributions to Beta  molecular orbitals:
Beta  occ 13 OE=-0.801 is O2-p=0.79
Beta  occ 14 OE=-0.783 is Fe1-d=1.00
Beta  occ 15 OE=-0.758 is O2-p=0.89
Beta  vir 16 OE=-0.241 is Fe1-s=0.81 Fe1-p=0.17
Beta  vir 17 OE=-0.139 is Fe1-p=0.91 Fe1-d=0.14

This can be used to provide a more quantitative characterization of the nature of canonical molecular orbitals.
Due to certain restrictions on the use of Gaussian (http://www.bannedbygaussian.org/), I am unable to use Gaussian. Are there other programs that can perform such a calculation? NBO only seems to provide such a breakdown in terms of NBOs, and I can't find the feature available in any other quantum chemistry program. I realize that I can do this myself by extracting the C matrix, figuring which coefficients go with which orbital, and summing up myself. However, this would probably take me a day to program.
Note: My calculations use B3LYP, ECPs, and the COSMO solvent model. So the program would need to be able to at least produce single points with such a method.

I'm sure, almost any program can do that. Since you mentioned some restrictions on the use of Gaussian, I suggest you try free programs, such as, GAMESS (US), NWChem, ORCA. I quickly checked their manuals, so I'm sure all three support everything you need (Population Analysis, B3LYP, ECPs, COSMO). And all these programs are gratis (freeware) for academics, NWChem being libre as well (free software).

Maybe it is not in the manual, but most programs put this info in the detailed output file.

ORCA can perform population analysis for each individual molecular orbital. This is an excerpt from the output file:
-----------------------------------
MULLIKEN ORBITAL POPULATIONS PER MO
-----------------------------------
THRESHOLD FOR PRINTING IS 0.1%
SPIN UP
                      0         1         2         3         4         5
                 -177.49372 -100.54186 -100.54186 -100.54184 -100.54179 -100.54174
                   1.00000   1.00000   1.00000   1.00000   1.00000   1.00000
                  --------  --------  --------  --------  --------  --------
  0Ti  1s             99.6       0.0       0.0       0.0       0.0       0.0
  0Ti  2s              0.4      -0.0      -0.0      -0.0      -0.0      -0.0
  1Cl  1s             -0.0       4.5       3.8       0.0      48.9       0.4
  1Cl  2s              0.0       0.0       0.0       0.0       0.3       0.0
  2Cl  1s             -0.0       4.9       4.1       0.0      50.5       0.4
  2Cl  2s              0.0       0.0       0.0       0.0       0.3       0.0
  3Cl  1s             -0.0      69.2       3.8       0.0       0.0      23.3
  3Cl  2s              0.0       0.5       0.0       0.0       0.0       0.2
  4Cl  1s             -0.0       1.4      71.5       0.0       0.0      23.4
  4Cl  2s              0.0       0.0       0.5       0.0       0.0       0.2
  5Cl  1s             -0.0       8.9       7.5      50.7       0.0      26.2
  5Cl  2s              0.0       0.1       0.1       0.3       0.0       0.2
  6Cl  1s             -0.0      10.4       8.7      48.6       0.0      25.6
  6Cl  2s              0.0       0.1       0.1       0.3       0.0       0.2

The program supports all options that you listed, and this input file header should get you started:
! B3LYP def2-SVP ECP{def2-SVP} COSMO(water) LargePrint
*xyz charge multiplicity
coordinates go here
*

You can find more detailed information in the manual.

That's what I wanted, I looked through the ORCA manual and somehow could not find it, thanks!

@Jonathon If you just want to turn on this reporting, without all of the other extra output afforded by LargePrint, you can add %output Print[ OrbPopMO_M ] 1 end on its own line for the Mulliken populations. Replace _M with _L for Loewdin populations.

I found those flags, but thanks. Also, as a note to other users, the program segfaults if you use ROKS, but only after printing all the necessary information (I think it has to do with the fact that there are no beta orbitals for high spin complexes).

NBO's natural population analysis (NPA) may in fact be a better option since it lacks many deficiencies Mulliken population analysis has (e.g. non-orthogonality of the basis set, sensitivity to the basis set choice).
You may run a NPA analysis in the standalone executable GENNBO after obtaining a suitable input for it:
$ gennbo < water.gen > water.nboout

A proper input may be generated by NWChem by the use of the nbofile keyword:
property
 nbofile
end

NWChem has all the features you've cited.

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text:
What exactly is an orbital?
What exactly is an orbital? Atomic or molecular.
Is it the function that describes the behaviour of the electron?
Is it the Schroedinger's equation solution, e.g., for Hydrogen atom?
Is it the probability of finding any electron of an atom in any specific region around the atom's nucleus?
Maybe all these describe the same thing. What is it?

I think you're not gonna get any satisfying answer for this question. Your question basically boils down to "what is the wavefunction?" for which there are hundreds of discussions. That said the wavefunction is interpreted as a probability density function. That is, you integrate it to find the probability of the electron being at some interval of space.

Terminological questions in chemistry usually are better answered using IUPAC recommended definitions and the IUPAC Gold Book defines orbital (atomic or molecular) as

Wavefunction depending explicitly on the spatial coordinates of only
  one electron.

But to be honest, I don't like this particular definition. The problem is that a one-electron wave function is, strictly speaking, a function of not only the spatial coordinates of an electron ($\vec{r}$) but also the spin one ($m_s$). In quantum chemistry we distinguish the so-called spin orbitals, which are one-electron wave functions, from the so-called spatial orbitals, which can be though of wave functions that describes a hypothetical spin-less electron.
The connection between spin and spatial orbitals is not so trivial. In the most general case a spin orbital $\psi$ can be represented as follows,
$$
  \psi(\vec{x}) = \phi_{\alpha}(\vec{r}) \alpha(m_s) + \phi_{\beta}(\vec{r}) \beta(m_s) \, ,
$$
where $\vec{x} = \{ \vec{r}, m_s \}$ is the joint spin-spatial coordinate of an electron, and $\phi_{\alpha}$ and $\phi_{\beta}$ are two different spatial orbitals. What we have here is essentially the linear combination of two possible spin states of an electron ("spin-up" and "spin-down") each with its own spatial probability.
And since OP mentioned in comments the fully relativistic Dirac's treatment of a single electron, it is a good idea to note an alternative way of expressing the idea of a spin orbital: in a form of the  so-called two-component wave function, or Pauli spinor, that represents the state of a single electron in a non-relativistic theory,
$$
    \left| \psi \right>
    \longleftrightarrow
    \begin{pmatrix}
        \phi_{\alpha}(\vec{r})  \\
        \phi_{\beta}(\vec{r})
    \end{pmatrix} \, .
$$
This is to be contrasted with a four-component Dirac spinor that describes a state of an individual electron in a relativistic theory,
$$
    \left| \psi \right>
    \longleftrightarrow
    \begin{pmatrix}
        \phi_{\alpha}(\vec{r})  \\
        \phi_{\beta}(\vec{r})   \\
        \chi_{\alpha}(\vec{r})  \\
        \chi_{\beta}(\vec{r})
    \end{pmatrix} \, ,
$$
in which the last two components describe a "spin-up" and a "spin-down" states of the associated positron.
Now, back to chemistry, we note that in general chemistry we always think of each and every electron as being in a pure spin sate: either "spin-up" or "spin-down", rather than in a superposition of these two. Besides, we assume that spin orbitals come in pairs: for each and every spatial orbital $\phi(\vec{r})$ we have two spin orbitals given as follows,
$$
  \psi(\vec{x}) = \phi(\vec{r}) \alpha(m_s), \quad \psi(\vec{x}) = \phi(\vec{r}) \beta(m_s) \, .
$$

So the spin orbitals are one-electron wave functions which...describe what? And one-electron for the Hydrogen atom, right? Thank you!

@NickyR, well, surprisingly, a one-electron wave functions describes... drumroll please... a one electron! :) A spin orbital describes the quantum state of a single electron in a chemical system.

A chemical system might be a hydrogen atom in which case spin orbitals describe possible states of a single electron in it. It might be a molecule as well in which case spin orbitals describe possible states of a single electron in this molecule.

I sincerely believe that dealing with Dirac's equation (just a feeling) is much easier than dealing with these spin orbitals and Slater Determinants.

@NickyR, hmmm... I wouldn't say so. Dirac's fully relativistic one-electron wave function is a four-component one, while in non-relativistic picture a one-electron wave function has only two components (the most general case in my answer). Besides, in chemical applications the two-component description is always reduced even further to one-component one-electron wave functions by restricting electrons to be in pure spin states. Life is much more easier in non-relativistic theory. :)

I like your sense of humour! As you seem quite into quantum mechanics and possibly further theories and models, would you have a suggestion of a book or something for someone who wants to find the deep meaning of such "terms"? You know, get deeper in the mystery of the chemical bond and atomic properties!

@NickyR, there are few good books on the subject. To give an advice I need to know a bit more about your level of knowledge of quantum theory.

Applications of Schroedinger's equation using approximations such as perturbation theory for more complicated systems than Hydrogen. And the other essential things you need to know in order to do that.

Let us continue this discussion in chat.

If I can add something, sometimes they can see S. Equation as $H\psi=E\psi$ as you can see its the same equation you need to solve when you are seeking eigenvectors :) Here the eigenvectors are the orbitals and they are associate to an eigenvalue which is here the energy. You can solve this equation for all eigenvalues you want to find :)

I decided add this answer because I have a different viewpoint, although I agree with Wildcat on the IUPAC definition.
What exactly is an orbital? Atomic or molecular.
The term orbital is used with more than one meaning. Normally they are classified into different kinds. In a technical sense, they are just functions of the variables on only one electron. If the variables are spatial coordinates they are called spatial orbital, if they also include the spin coordinate they are called spin orbitals.
Generally speaking (for more than one electron particles) they are just mathematical constructions useful get the wavefunction or an approximation to it (from which we can get estimation of measurable values).
Is it the Schroedinger's equation solution, e.g., for Hydrogen atom?
I assume that you are speaking about the Time Independent Schrödinger Equation (TISE). In such case, generally not. In the particular case of H atom they are solutions of TISE, as well as linear combinations of them. In such case they represent stationary states of the system, in other words, are functions that represent (somehow) time independent states in which the system can be in (but only in systems with one electron!).
In polyelectronic species, they represent nothing, although it can be given (in some context) some approximate interpretations. Going further, if you decompose the system into one particle systems, and get their orbitals, it is mathematically true (because this: see last example) that they can be combined (how to is the hard part) to represent the wavefunction of the original system.
Is it the probability of finding any electron of an atom in any specific region around the atom's nucleus?
No. They are rather arbitrary. It is a common practice to transform them according to one needs. In any case, normally they represent something like the stationary state the one electron embed in a space region where it feels the mean interaction with the remaining electrons. But do not dig to much in those interpretations.
Maybe all these describe the same thing. What is it?
The introductory textbooks call orbital to an space region. It is not compatible with the usage of the word above. In a polyelectronic system, sometimes, for some kind of orbitals, in very well defined contexts, they can have a sort of interpretation, in essence they are just helpful functions.
You should check the answers to this question: What does an orbital mean in atoms with multiple electrons?... They give some details that I omitted.

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text:
Is my textbook using the law of partial pressures correctly?
My textbook shows that pressure fraction is equal to mole fraction because the same volume cancels:
But it violates (if I am not mistaken) the law in this example it gives:

the main components of dry air, by volume, are N2, 78.08%, O2, 20.95%, Ar, 0.93%, and CO2, 0.04%. What is the partial pressure of each gas in a sample of air at 1atm.

The strategy it gives for solving this problem is by treating the volume percent as mole percent, and in turn as pressure percent. But obviously, each gas occupies a different volume, the law of partial pressure doesn't apply. does it?
Also, is it possible for different components of air to occupy different amounts of space in atmosphere? If atmosphere is a container, doesn't all components just occupy the same amount of space-the size of container?

I'd guess that you have something confused. Textbook? page? Need to see whole problem to understand the context. In general though PV=nRT, so P of a particular gas component would be  nRT/V and the pressures of the individual components would sum to the total pressure. Note that for some container at some temperature R, T and V will all be constants.

"If atmosphere is a container, doesn't all components just occupy the same amount of space-the size of container?" Yes, but consider having a liter container with 200 ml of oxygen and 800 ml of nitrogen. Now there is also a "force field" between the nitrogen and oxygen. If the temperature was exactly the same and the pressure was exactly 760 mm in both, then the ideal gas laws tell us that releasing the force field would yield 1000 ml mixture of oxygen and nitrogen at 760 mm, and the same temperature.

Whenever they use the term "by volume," they always mean mole fraction (particularly in atmospheric science).  So, when I see the words "by volume," I mentally translate it into the words "mole fraction."  It has never made sense to me why they use this terminology because all the components of the gas occupy the same volume.


text:
Uncertainties after Averaging Trials
If I conducted 6 trials for a lab, all with the uncertainty of (+/- 0.1 mL)... After averaging the 6 trials to plot the results, what is the new uncertainty?

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

In any case, you should calculate the mean $\overline X$, the standard deviation $\sigma$, and the standard error of the mean $\sigma_{\overline X}$ for the six trials.
However, the estimated uncertainty $u(y)$ of the final result $y$ depends on what is meant by “all with the uncertainty of $\pm0.1\ \mathrm{ml}$”, i.e. $u(x)=0.1\ \mathrm{ml}$.
For example, if this uncertainty $u(x)=0.1\ \mathrm{ml}$ is caused by the unknown error of a pipette and the same pipette was used for all six trials, each of the six individual results includes the same systematic error. You can neither eliminate this error nor reduce the caused uncertainty by making additional trials. In this case, you should combine (by using the rules for error propagation) the calculated standard error $\sigma_{\overline X}$ and the assumed uncertainty $u(x)=0.1\ \mathrm{ml}$. Therefore, the estimated uncertainty $u(y)$ of the final result cannot be smaller than $u(x)=0.1\ \mathrm{ml}$.
If the uncertainty $u(x)=0.1\ \mathrm{ml}$ is caused by a random error of each individual trial, it is already included in the observed standard deviation $\sigma$  and thus already taken into account for the calculated standard error $\sigma_{\overline X}$. Therefore, the observed standard deviation $\sigma$ should be greater than (or at least equal to) the assumed uncertainty $u(x)=0.1\ \mathrm{ml}$. In this case, you may assume that the estimated uncertainty of the final result $u(y)$ is approximately equal to the calculated standard error $\sigma_{\overline X}$.


text:
Does H2CO3 exist in solution?
When $\ce{CO2}$ reacts with water, it supposedly forms $\ce{H2CO3}$, which is the aqueous form of $\ce{CO2}$. But it is hard to isolate, but its $K_\mathrm{a}$ value is pretty low (on the order of $10^{-7}$) indicating it should be found in a non dissociated form? So does $\ce{H2CO3}$ exist in solution?

The equilibrium of $\ce{CO2(aq) + H2O(l) <=> H2CO3 (aq)}$ lies extremely far to the left (you might find data on the Internet... I remember reading it in Skoog et al) so if you were to have $\ce{H2CO3}$ in solution, it won't dissociate to $\ce{H+ + HCO3-}$ but instead to $\ce{CO2 + H2O}$ (I believe, not entirely sure).

@orthocresol so how does CO2 exist in solution?

The given low $K_\mathrm a$ value applies to $\ce{[CO2(aq)] + [H2CO3]}$. True $\ce{H2CO3}$ is a much stronger acid with $\mathrm pK_{\mathrm a}\approx3.5$, which is stronger than acetic acid and formic acid.

Aqueous carbonate solutions contain four different solute species:

$\ce{CO2(aq)}$
$\ce{H2CO3}$
$\ce{HCO3-}$
$\ce{CO3^2-}$

The equilibrium for the hydration reaction
$$\ce{H2O + CO2(aq) <=> H2CO3}\tag1$$
lies to the left. The corresponding equilibrium constant $K$ at $25\ \mathrm{^\circ C}$ is about
$$K=\frac{\left[\ce{CO2(aq)}\right]}{\left[\ce{H2CO3}\right]}\approx650$$
(literature values of $K$ range from $350$ to $990$.)
The commonly used first acid dissociation constant of carbonic acid $\mathrm pK_{\mathrm a1}=6.35$ (at $25\ \mathrm{^\circ C}$) actually is a composite constant that includes both the hydration reaction $\text{(1)}$ and the protolysis of true $\ce{H2CO3}$ $\text{(2)}$.
$$\ce{H2CO3 <=> H+ + HCO3-}\tag2$$
The constants $K$, $K_{\mathrm a1}$, and the actual acid dissociation constant $K_{\ce{H2CO3}}$ of true $\ce{H2CO3}$ are interrelated according to
$$K_{\mathrm a1}=\frac{K_{\ce{H2CO3}}}{1+K}\approx\frac{K_{\ce{H2CO3}}}{K}$$
Calculated values for $\mathrm pK_{\ce{H2CO3}}$ range from $3.8$ to $3.4$. Thus, true $\ce{H2CO3}$ is stronger than acetic acid and formic acid.
The second acid dissociation constant of carbonic acid is $\mathrm pK_{\mathrm a2}=10.33$ (at $25\ \mathrm{^\circ C}$).
Therefore, the predominant species in aqueous carbonate solutions (at $25\ \mathrm{^\circ C}$) are

$\ce{CO3^2-}$ at $10.33\lt\mathrm{pH}$
$\ce{HCO3-}$ at $6.35\lt\mathrm{pH}\lt10.33$
$\ce{CO2(aq)}$ and $\ce{H2CO3}$ at $\mathrm{pH}\lt6.35$

Thus, true $\ce{H2CO3}$ primarily exists in acidic solutions. However, because of the hydration equilibrium $\text{(1)}$ and the corresponding equilibrium constant $K\approx650$, most of the true $\ce{H2CO3}$ is converted to $\ce{CO2(aq)}$.
(All values taken from: Stumm, W.; Morgan, J. J. Aquatic Chemistry, Third Edition; John Wiley & Sons: New York, NY, 1996; pp 150–152.)

No doubt that you've written the "traditional" model correctly in equations 1 and 2. The math and chemical equations work. The real question is how long does a particular molecule of H2CO2 on average last? What is the mean time between collisions for molecules in water? So if a molecule on average lasts for trillions of collisions then it is stable. Certainly if the "molecule" last for only a trillionth of the average time between collisions then it is unstable. But where is the absolute cutoff between stable and unstable? // In other words the kinetics of your equation 1 are very very fast.

I think that the answer is rather orthogonal to your question. If you think of any sort of acid $\ce{H+}\ce{A-}$ then the "molecule" of HA exists so that a Ka can be tabulated. But let's tag the particular ions $\ce{H+}$ and $\ce{A-}$) of one molecule and drop it into the mixture. It won't stay as as one molecule but rather due to the transfer of protons (hydronium ions really) the two ions will become separated. Due to mathematical combinatorial  possibilities it is unlikely that our two tagged ions would get together again.
From wikipedia

"The long-held belief that carbonic acid could not exist as a pure compound has reportedly been recently disproved by the preparation of the pure substance in both solid and gas form by University of Innsbruck researchers."

So the question is really about how long does a particular "molecule" of HA last in solution? For $\ce{H2}\ce{C}\ce{O2}$ the answer would be "not long." So in solution I think you really have to define "exists" in terms of the average time (T) between molecular collisions. So how many multiples of T are necessary for the molecule to "exist" in solution?

It hasn't been isolated in an aqueous solution, and that's why they say it doesn't "exist" in there. Wikipedia tries to be skeptical of the widely held belief by linking to a news article that doesn't link to the journal article? I'm a bit skeptical of Wikipedia's skeptical skeptic skepticism.

Wikipedia links to this article http://phys.org/news/2011-01-international-gas-phase-carbonic-acid-isolated.html which has journal information.  // I'm certainly not an expert on this particular controversy, but I think my explanation is reasonable. The "stability" of a molecule which is in equilibrium is fuzzy since the "equilibrium" doesn't  account for the kinetics.


text:
Calculating binding energies for excitated levels in potassium?
Suppose a laser is used to excite electronic transitions in potassium from the $3p$ orbitals to ones higher than $4s$. When I attempt to use the Bohr model to calculate binding energies, what I obtain for $n\le4$ disagrees significantly with this source, calling into doubt my results for $n>4$. I think it might be possible to use the Rydberg formula, but I'm having trouble finding a working example that computes the right energies; in addition, there is another form, and the two seem inconsistent.  How does one calculate the binding and transition energies in potassium for levels $n>4$? Are the quantum defects for arbitrary elements available online?
What changes here if we're talking not about pure potassium but potassium chloride?

Perhaps relevant: http://chemistry.stackexchange.com/questions/32350/can-the-bohr-atomic-model-be-applied-to-sodium-and-other-single-valence-electron

What do you expect by using the Victorian technology (Bohr model)? :) More than a century ago Bohr himself attempted to use his model to calculate energy levels in many-electron atoms and quickly found that it gives unacceptable numbers already for the lightest of then, $\ce{He}$ atom. For the next couple of decades after this discovery a scientific thought in that area had developed revolutionary and a good physical theory that gives way better predictions for atomic energy levels (amongst many other things) was formulated.

Fair enough. :)  But my problem is a practical one, and it would be nice to have realistic transition energies, so that I will know what photon wavelengths are needed to excite the transitions.  Any help is appreciated.


text:
What's the pKa (range) of a carbonyl functional group?
Why is a $\mathrm pK_\mathrm a$ value for carbonyls never mentioned anywhere in textbooks I've use while other functional groups have these values?

Are you talking about aldehydes or enolisable ketones?

I guess ketones, although information about both would be nice.

I can't believe they're not mentioned, but look for enols instead ;)

Just to clarify, are you talking about the $\mathrm{pK_a}$ of the $\alpha$ hydrogens or the $\mathrm{pK_a}$ of hydrogens attached directly to the carbonyl carbon?

Directly attached to the carbonyl carbon

As per comments you are asking for the $\mathrm{p}K_\mathrm{a}$ value of a hydrogen directly attached to a carbonyl carbon. This means, you are asking only for non-enolisable aldehydes as ketones have no hydrogens on the carbonyl carbon. Aldehydes that have a hydrogen in α-position will be deprotonated there generating an enolate ($\mathrm{p}K_\mathrm{a} \approx 20$) — more acidic than a carbonyl hydrogen. So all we have left is 2,2-dimethylpropanal (or pivaloylaldehyde). Its $\mathrm{p}K_\mathrm{a}$ is not given on SciFinder, but should be somewhere between $45$ (an alkane) and $25$ (an alkyne), probably just slightly higher than an alkene ($40$).


text:
How does the solvent determine whether mono- or tribromination of phenol occurs?
In phenol, the $-\ce{OH}$ group activates the benzene ring towards electrophilic substitution. But why is it that the electrophilic bromination of phenol gives trisubstitution (forming 2,4,6-tribromophenol) when aqueous bromine is used, but monosubstitution (either 2- or 4-bromophenol) when a $\ce{CCl4}$ solution of bromine is used? How does the solvent control the extent of bromination?
Related but does not explain completely: How do we know whether we will get a mono substituted or tri-substituted product during electrophilic aromatic substitution?

In aqueous medium, phenol is deprotonated to a certain extent, forming the phenoxide ion in which the ortho and para positions are even more activated than in phenol itself. Hence, trisubstitution occurs here.
$$\ce{PhOH + H2O <<=> PhO- + H3O+}$$
In an organic solvent, phenol is not deprotonated and monosubstitution tends to occur.
ron also pointed out in a comment that:

The concentration of $\ce{Br^+}$ is higher in water than in $\ce{CCl4}$ making the reaction faster in water.
The dielectric constant of water is much greater than that of $\ce{CCl4}$, so the former solvent will better stabilize the charged intermediate, again accelerating the reaction.

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


text:
Effect of Temperature on pH of Water

The $K_\mathrm w$ is a function of temperature. It is $10^{-14}$ at $25\ \mathrm{^\circ C}$. When the temperature is $50\ \mathrm{^\circ C}$, the $K_\mathrm w$ can be calculated to be somewhere around $10^{-12}$ using Vant Hoff's equation, but this is so weird.

MY ATTEMPT
If we will solve for $K_\mathrm w=10^{-12}$, we get $\rm pH=6$. Will the water become acidic due to increase in temperature?

The effect is correct; however your estimated $\mathrm{pH}$ is a bit low. The actual $\mathrm{pH}$ of neutral water at $50\ \mathrm{^\circ C}$ and normal pressure is about $\mathrm{pH} = 6.6$; i.e. the autoprotolysis constant of water is about $K_\mathrm w \approx 5\times10^{-14}$.

People in this discussion don't understand the difference between "neutrality" and solution acidity. It is best to think of the Bronstead definition of acidity in discussions like this one. In that context it is known that methanol is more acidic than water and a solution of pure acetic is neutral, as well as more acidic than either methanol or water. (yes acetic acid has an autoprotolysis constant that happens to be about 14. So does methanol, ammonia, sulfuric acid and any other compound that self dissociates to a solvated hydrogen ion and the conjugate base form of the solvent)

A solution of hot water is more acidic than cold water because it has a higher activity of dissociated hydrated hydrogen ions, yet it is neutral. For more reasons that I can easily type you can not test these facts with pH paper.

TL;DR
$\rm pH=6$ is the neutral pH at $\rm 50\,^{\circ} C.$
Long Version
Actually there is nothing wrong with your calculations. It seems that you are confused about the definition of a solution being acidic. If something is acidic, this means that the concentration of $\ce{H+}$ ions (technically $\ce{H3O+}$ ions) is greater than the concentration of $\ce{OH-}$ ions. At room temperature (25 degrees Celsius) a solution is neutral if its pH is 7. This means that for a solution to acidic, its pH must be lower than 7 at 25 degrees.
When we increase the temperature to 50 degrees, the pH of a neutral solution is actually 6, not 7. You can check this as if you were find the total concentration of $\ce{H+}$ ions and $\ce{OH-}$ ions, you will find that they both equal $10^{-6}$. Since there concentration is equal, the solution must be neutral. So for a solution to be acidic at a temperature of 50 degrees, its pH must be lower than 6, not 7.
Explanation
Now you might wonder what the reason for this. Consider the following equation which is the auto-disassociation of water $$\ce{2H2O + heat \leftrightharpoons H3O+ + OH-}$$ As you can see, this reaction is actually endothermic. This means that when you increase the temperature, due to Le Chatelier's principle, the equilibrium will shift to the right; hence the concentration of $\ce{H3O+}$ and $\ce{OH-}$ ions will increase. This is indicated by the increase in the $\rm K_w$ value, as more products are formed.
Therefore since there is a greater of concentration of $\ce{H3O+}$, a solution which is neutral will have more $\ce{H3O+}$ ions at 50 degrees than it would when it is 25 degrees. Therefore its pH will be lower than 7 despite being neutral.

So we can conclude that when we increase the temperature  as there is increase in Kw both H3O+ and OH- ions concentration increases making it neutral.
One more doubt. Why do we write H3O+ and H+ interchangeably as if both mean the same?

@AliHasan H+ is easier to write but H3O+ is the actual form of H+ in water

Yes, we just use H+ for convenience, however is reality it is the hydronium ion, H3O+, that exists. Also higher hydrated protons do exists in solutions such as $\ce{H5O2+}$ and $\ce{H9O4+}$

S0 we get pH=6. That means if we are boiling water and at that time we put a litmus paper in it, it's colour will change. Ph=6 at 50C and at 100C it will become more less. And as we increase temperature the pH value will tend 1 being more and more acidic.
What I think is that water will still remain neutral as OH and H+ concentration will be same but a pH paper should most probably turn red. But that does not happen.

@AliHasan That is because that the pH would not linearly decrease with temperature, but is rather like an exponential curve. Meaning that if heating the water to 50C decreases the pH, heating it by another 50C won't decrease the pH to 5 but to something like 5.5 (I don't exactly). So for the litmus paper to actually become red, the temperature has to extremely high (but the water would have evaporated to steam well before you could even reach the required temperature).

@Nanoputian as Ali Hasan mentioned if concentration of OH- and H+ are same at any temperature then how can the pH of water change?

@AbhishekPallipparagopakumar I think you are confused. The concentration of H+ is always equal to the concentration of OH- in water at any temperature. However the concentration of H+ and OH- are different at different temperatures, however the concentration of H+ and OH- are still equal each other. Since pH is only dependent on the concentration of H+, the pH will also change with a change in temperature.

@Nanoputian your comment was a bit unclear...could you make it a in simple language.If H2O dissociates then equal amount H+ and OH- will be generated then how do you say at a temperature there will be more H+ than OH- as one molecule of water dissociates into one H+ and one OH-.

@AbhishekPallipparagopakumar What I mean is that at 25C the concentration of H+ is equal to $10^{-7}$ which is equal to the concentration of OH- as the concentration of H+ must always equal the concentration of OH- in pure water (otherwise it would be charged). If I increased the temperature of the water to 50C, the concentration of H+ and OH- will be still be equal to each other (so the water is still neutral). however the concentration of H+ will be something like $10^{-6}$ which is also equal to the concentration of OH-.

So what I am saying is that the concentration of H+ in pure water at 25C and 50C are different to each other. however at each temperature, the concentration of H+ and OH- are always equal to each other. Is this clear now?

@Nanoputian yup thats a good explanation......but is it necessary k[OH-]*k[H+] should be always equal to 10^(-14)..........

@AbhishekPallipparagopakumar Well firstly it is [OH-]*[H3O+], not k[OH-]k[H3O+]. But yes, at 25C, [OH-][H3O+] is always equal to $10^{-14}$, which is why our pH scale goes from $0-14$ and pH + pOH = $14$ at 25C.

@Nanoputian aah yes....got it...


text:
Why is urobilin yellow?
The yellow color of urine is due due a compound known as urochrome, or urobilin. Why is this chemical yellow? There isn't any coordination complexes. What else could cause this color?
I have not learned organic chemistry yet so please make it simple.

Many organic molecules, including food dyes, are colored because they contain chromophores. A chromophore is a region in the molecule where the energy difference between two different molecular orbitals falls within the range of the visible spectrum. Light that hits the molecule can thus excite the electrons in the chromophore, resulting in the emittance of a particular wavelength of light.
$\hspace{1cm}$
In urobilin (seen above) specifically, the tetrapyrrole moiety, as a result of its conjugated pi-bond system, is the chromophore that gives this compound its characteristic yellow color.

What is the tetrapyyrole moiety?

"Tetra" meaning four, "pyrrole" meaning an aromatic, 5-memered ring containing 4 carbons and 1 nitrogen, and "moiety" meaning a part of a molecule. These 4 pyrroles constitute the main body of urobilin.

The color is probably due to the fact that the middle two pyrroles are conjugated, further reducing the HOMO-LUMO gap.  Regular pyrrole is a colorless liquid.


text:
Water: What to use H3O+ or H+?
Why can we use $\ce{H+}$ and $\ce{H3O+}$ interchangeably?
I have seen in many places that first the reaction is written including Hydronium ion but then in bracket its written that we can write it as $\ce{H+}$ for simplicity. I am not clear with this.
$$\ce{H2O +H2O <=> H3O+ + OH-}$$
OR
$$\ce{H2O(l) <=> H+(aq) + OH- (aq)}$$

Since your solvent is itself water, it makes no difference whether you use $\ce{H+}$ or $\ce{H3O+}$.
$\ce{H3O+}$ is basically the hydrated form of $\ce{H+}$. If you know, the oxygen atom in water contains two lone pairs. When it donates one of the lone pairs to the hydrogen atom which doesn't have any electrons, you get $\ce{H3O+}$.
So,
$\ce{H3O+}$ is not $\ce{H+}$
$\ce{H3O+}$ is $\ce{H+(aq)}$
This means that the aqueous form of $\ce{H+}$ is represented as $\ce{H3O+}$

In all cases, acids yield protons ( or hydronium ions H3O+) and bases yield OH- (hydroxide) ions in aqueous solutions.
The H3O+ ion is considered to be the same as the H+ ion as it is the H+ ion joined to a water molecule. The proton cannot exist in aqueous solution, due to its positive charge it is attracted to the electrons on water molecules and the symbol H3O+ is used to represent this transfer.
The equation can be written as:
H+ + H2O(l) → H3O+(aq).
This is hydrolysis as it is involving water as a reactant.
Consider the first equation in the question , the ionisation equation of water:
H2O(l) + H2O(l)→H3O+(aq) + OH-(aq)
The  H3O+ is the conjugate acid of H2O. So H3O+ is used as a shorthand for a proton in aqueous solution. In a non-aqueous solution the proton would form a different structure.
The second equation:
H2O(l) → H+(aq) + OH-(aq)
Shows that H2O is made up of equal parts H+ and OH- ions and is amphoteric (can be an acid or a base) having a deprotonated form  (OH-). The ionic component is at a very low concentration and a water molecule is generally considered covalent with a dipole moment favouring a slight positive charge.
The H3O+ ion concentration in pure water at 25° C is 10^-7 dm^-3. This can be written as:
[H3O+] = 10^-7
where the symbol [ ] means the "molarity of" (units in moles dm^-3).
The number of H3O+ and OH- ions formed by the ionisation of pure water must be equal ( from the equation):
[H3O+] = [OH-] = 10^-7).
This shows that pure water is neither acidic or basic, it is neutral. The product of [H3O+] = [OH-] is the ionic product of water.
[H3O+][OH-]=10^-7 × 10^-7 = 10^-14
shows that in aqueous (water) solutions, whether acidic, basic or neutral,  the product of the ion concentrations equals 10^-14.
Acidic solutions contain more H3O+ ions than OH- ions. For basic solutions it is the reverse.
Therefore a water solution is :
Neutral when [H3O+]= 10^-7.
Acidic when [H3O+] > 10^-7.
Basic when [H3O+] < 10^-7.

What? I don't see how this answers the question about when to use $\ce{H+}$ or $\ce{H3O+}$. Have a look at the duplicate question to see a more in depth answer.

Please visit this page, this page and this one on how to format your posts better. Alternatively, visit this chatroom for further formatting guidance.

This post has triggered and automatic flag because you have edited it so many times, please avoid this in the future. The community would also very much appreciate if you could take a little time and read through the links that inɒzɘmɒЯ.A.M has posted. Edit summaries should comment on the things performed not work as hidden comments.

It is a correct answer. I also respect all of what your saying except for @Bon saying I copied it from the duplicate answer.


text:
Do hydroxymethyl-substituted cyclopropylmethyl cations undergo ring expansion?

In the image, what would happen after step 2?

Will a double bond be formed between carbon 3 and 4? If this is the case, then we would get benzene in the end.
But, if the positive charge in shifted to carbon 1, then won't it be a more stable carbocation?
Also, if a double bond were to be formed between carbon 1 and 4 now, it would be more substituted than a double bond between carbon 3 and 4.

In spite of all these factors, the double bond is formed between carbons 3 and 4, and benzene is formed.
Kindly explain how this process takes place.And, please tell me how to name this compound.

I'm suspicious of making a primary carbocation. It's possible that you would get a concerted shift, but even next to a cyclopropane, that looks bad...

The following takes place-

And then H3O+ can be assumed to give H+ and H2O.
Try it out and you'll get to the product.
If the positive charge is shifted to carbon 1, it would form a more stable alkene since it is more substituted. But after formation of C1-C4 double bond, no further reaction will occur since after expansion + charge will be on a double bonded carbon which would be highly unstable. Since Benzene is more stable than the product which will be formed after shifting of + charge to Carbon 1, and since heat is also applied, there will be enough energy to form more stable benzene and thus the reaction takes place in such a manner.

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text:
The usage of 'limiting composition'
I am reading wikipedia page on Molybdenum bronze and they mention this:

Blue bronzes, with limiting composition $\ce{A_{0.3}MoO3}$ that is,
  $\ce{A_3Mo10O30}$.

I used to see these fractional numbers (e.g. 0.3) and always wondered whether technically speaking they are correct or not. I understand that, as the wikipedia page says, they are equivalent to the 'full' numbers (e.g. 3) but still I am not sure whether in literature one of them is preferred or more common than the other?

These are called non-stoichiometric compounds or berthollides. Because the ratio of the fractional component is not fixed, but can vary over a small but continuous range, it would be pointless to try to express the formula as least common multiples (LCM).
For example, iron(II) sulfide can have a ratio from about $\ce{Fe_{$0.8$}S}$ to $\ce{FeS}$. So in LCM terms, the formula varies continually from $\ce{Fe4S5}$ through $\ce{FeS}$, including $\ce{Fe9S10}$, $\ce{Fe19S20}$ and so forth, all with similar physical and chemical properties.


text:
Temperature dependence of reaction enthalpy

Estimate $\Delta_\mathrm{c}H^\circ(500\mathrm{~K})$ for methane by using the data given on the temperature dependence of heat capacities.
$$\ce{CH4(g) + 2 O2(g) -> CO2(g) + 2H2O(g)}$$

My working
First, heat capacities as functions of temperature.
$$\begin{align}
C_{p,m}(\ce{CO2}) &= 44.22 + 8.79\times10^{-3} T - \frac{8.62\times10^{5}}{T^2}\\
C_{p,m}(\ce{H2O}) &= 75.29 + 0 + 0 \\
C_{p,m}(\ce{O2}) &= 29.96 + 4.18\times10^{-3}T - \frac{1.67\times10^{5}}{T^2} \\
C_{p,m}(\ce{CH4}) &= 35.31
\end{align}$$
Note: a function of temperature dependence wasn't given in the data table for methane, this value is at $298 \mathrm{~K}$. I used this because I did not know what else to do.
Then, I used Kirchhoff's Law:
$$\Delta H(500\mathrm{~K}) = \Delta H(298 \mathrm{~K}) + \int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\Delta C_{p,m}\,\mathrm{d}T$$
I computed $$\Delta C_{p,m} = \sum_{\mathrm{products}}\nu C_{p,m}-\sum_{\mathrm{reactants}}\nu C_{p,m}$$ (here, I used the molar heat capacities values listed above, adding the functions after multiplying by the appropriate stoichiometric coefficients) and obtained:
$$\Delta C_{p,m} = 16.15 + 0.43\times10^{-3}T - \frac{5.28\times10^{5}}{T^2}$$
Therefore:
$$\begin{align}
&&\int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\Delta C_{p,m}\,\mathrm{d}T &= 2.581 \mathrm{~kJ~mol^{-1}} \\
&&\Delta H^\circ(298\mathrm{~K}) &= -890 \mathrm{~kJ~mol^{-1}} \\
&\Rightarrow & \Delta H^\circ(500\mathrm{~K}) &= -887.4 \mathrm{~kJ~mol^{-1}}
\end{align}$$
However, my result is incorrect. The correct result is $-803.07\mathrm{~kJ~mol^{-1}}$.
Could someone please help me out on this?
As a reference, the textbook is Physical Chemistry 10th ed by Atkins.

Note to others reading: this is Exercise 2C.9(a) on page 105 of the text.

I fixed some $\LaTeX$ issues. There's a couple of things I want to nitpick on. First is the lack of units in your expressions for heat capacity - I know it's all in $\mathrm{J~K^{-1}~mol^{-1}}$ but yeah. Second thing is that you shouldn't include the differential $\mathrm{d}T$ in the equation when you're finding $\Delta C_{p,m}$. The differential changes the meaning of the equation

Now on to the important part which is the question itself! I redid the whole question and got a value of -799.76 kJ/mol. Oh well... There are a couple of things I would question. The first is your ΔH at 298 K. I used values of $\Delta H_f$ from the appendix and got a standard reaction enthalpy of -802.34 kJ/mol (remember to use H2O(g) and not H2O(l)).

The next is your calculation of $\Delta C_{p,m}$. 1) For H2O you should use the heat capacity of steam, not water; Table 2B.1 only gives that of water (it says liquids). So I would take the value from Table 2C.5 instead. Next, for the values that you do take from Table 2B.1, don't forget the $10^{-3}$ coefficient in $b$ and $10^5$ coefficient in $c$ (look at the header row).

@orthocresol: Hi! Thanks for the cleaning up the latex. And I have taken note of everything that you mentioned, and fixed the values in the question to reflect that. It was quite stupid of meto ignore the exponents in the coefficients. The enthalpy at 298K comes from 2c.4, the enthalpy of combustion. (am I not allowed to use that?) If I do, then I get the result stated above. If use -802.34 kJ/mol , however, I do indeed get -799.76 kJ/mol

Yeah, which is why I am a little confused too haha. Where did you get the numerical answer from? Regarding the 298 K enthalpy - I think the combustion data in 2C.4 has H2O(l) as a product instead of H2O(g). If you calculate the rxn enthalpy using heats of formation then using H2O(l) instead gives the value -890.36 kJ/mol.

http://fdslive.oup.com/www.oup.com/orc/resources/chemistry/pchem10e/student/solutions/pchem10e_answers_exercises.pdf (their website)

It's been a few days already but I went to consult the solutions manual. I'll type up an answer.

All enthalpies are taken to be at $298\mathrm{~K}$ if not specified.
The enthalpy of combustion that is shown in Table 2C.4, $\Delta_\mathrm{c} H^\circ = -890\mathrm{~kJ~mol^{-1}}$, corresponds to $\Delta H^\circ$ for the reaction:
$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$
which is not the reaction you are interested in (since $\ce{H2O}$ is in the liquid state). Therefore, to calculate the enthalpy of the reaction, you have to use $\Delta_\mathrm{f} H^\circ$ values from Tables 2C.4 and 2C.5:
$$\begin{align}
\Delta H^\circ &= 2\Delta_\mathrm{f} H^\circ(\ce{H2O(g)}) + \Delta_\mathrm{f} H^\circ(\ce{CO2(g)}) - \Delta_\mathrm{f} H^\circ(\ce{CH4(g)}) \\
&= -802.34\mathrm{~kJ~mol^{-1}}
\end{align}$$
as I said in the comments.
Now, even though the question said to use the temperature dependence of heat capacities in Table 2C.1 (in the appendix), the solution manual irritatingly takes the values from Problem 2C.7, on page 106. Here, the heat capacity is expressed in the form $C_{p,\mathrm{m}} = \alpha + \beta T + \gamma T^2$:
$$\begin{array}{c|ccc}
\mathrm{Molecule} & \alpha / (\mathrm{J~K^{-1}~mol^{-1}}) & \beta / (\mathrm{mJ~K^{-2}~mol^{-1}}) & \gamma / (\mathrm{μJ~K^{-3}~mol^{-1}}) \\
\hline
\ce{CH4(g)} & 14.16 & 75.5 & -17.99 \\
\ce{CO2(g)} & 26.86 & 6.97 & -0.82 \\
\ce{O2(g)} & 25.72 & 12.98 & -3.862 \\
\ce{H2O(g)} & 30.36 & 9.61 & 1.184
\end{array}$$
Therefore, we have:
$$\begin{align}
\Delta C_{p,\mathrm{m}} &= 2C_{p,\mathrm{m}}(\ce{H2O(g)}) + C_{p,\mathrm{m}}(\ce{CO2(g)}) - 2C_{p,\mathrm{m}}(\ce{O2(g)}) - C_{p,\mathrm{m}}(\ce{CH4(g)}) \\
&= \left[21.98 - \left(75.27 \times 10^{-3}\right)(T/\mathrm{K}) + \left(27.262 \times 10^{-6}\right)(T/\mathrm{K})^2\right] \mathrm{~J~K^{-1}~mol^{-1}}
\end{align}$$
and integrating from $298\mathrm{~K}$ to $500\mathrm{~K}$ (I got lazy and used my graphing calculator):
$$\int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\!\!\Delta C_{p,\mathrm{m}}\,\mathrm{d}T = -731 \mathrm{~J~mol^{-1}} $$
and:
$$\begin{align}
\Delta H^\circ (500~\mathrm{K}) &= \Delta H^\circ (298~\mathrm{K}) + \int\limits_{298\mathrm{~K}}^{500\mathrm{~K}}\!\!\Delta C_{p,\mathrm{m}}\,\mathrm{d}T \\
&= (-802.34 - 0.731) \mathrm{~kJ~mol^{-1}} \\
&= -803.07 \mathrm{~kJ~mol^{-1}}
\end{align}$$
Now that's irritating, because if we use whatever data we can find from the appendix, we get the answer $-799.76\mathrm{~kJ~mol^{-1}}$, as mentioned in the comments. However, I suppose there's two things that can be learnt:

Do not assume that the combustion reaction gives $\ce{H2O(g)}$, especially if it is given for $T = 298\mathrm{~K}$.
Check the units in the heading of the table.

Your procedure itself is perfectly sound, which is a good thing.

What is ∆r Cp in general khirchoff equation.....is it like the difference of Cp of product and reactant in a reaction

@Abhinav yes, but you have to weight it by the stoichiometric coefficient. This is explained in (any edition of) Atkins' Physical Chemistry.


text:
How many stereoisomers of 1-methylidene-2-(4,4,6-trimethylhepta-2,5-dien-1-yl)-1,2-dihydronaphthalene are there?

What is the total number of stereoisomer of this compound?

2
4
8
16

As there is only 1 chiral carbon, and 2 out of 4 C=C double bonds that contribute to geometric isomerism (cis/trans), I think there are 2^(1+2)=8 stereisomers. But the answer says 4. Can someone please explain the reasoning? Thanks!

Welcome to Chemistry Exchange, Matthew. Well, how many stereo-centres do you think there are?

There is 1 chiral carbon.
While there are 4 C=C double bonds, only 2 contribute to stereoisomerism.
So, 2^(1+2)=8.
But the answer says it is 4.

Think again about whether cis-trans isomerism is really possible in both of your cases.

Is it the one in the cycleohexene that does not have cis-trans isomerism? But why?

The trans cyclohexene diastereomer would be extremely unstable.

The question would benefit further, if you said which double bonds you consider stereogenic.

The text is correct. You are partly correct about the alkenes, but mistakenly assume 1) all are stereocenters. 2, it is incorrect to assume that even if stereogenic, alkenes have two stereocenters simply because they have 2 carbons. What makes them stereogenic is that there are geometric differences relative between the two carbons. Thus stereogenicity comes from that relationship, not the number of carbons. 3) While everyone know what you mean by cis and trans, those are not acceptable terms in organic chemistry unless you are referring to sides of rings or in relative terms only.
Looking at just olefins and going from left to right:
Next, the endocyclic alkene must be cis. You cannot place an E olefin in a ring smaller than 8 eight carbons but that would make the name change...this is asking for configurational (stereo) isomers...not constitutional variants.
Next, the exocyclic alkene is terminal. It has two of the same group (H) on the a carbon, it is therefore not a stereogenic center. How do you know? Switch the two hydrogens...does anything change? no, so it can't be.
Next, the E-olefin next to the quaternary carbon is the only stereogenic alkene...it could be Z. So that is a stereocenter (both carbons are)...though that would be incredibly high energy (the largest substituents (and neopentyl is enormous) are sterically congested).
Finally, the last olefin double bond is not a stereocenter. There are two of the same group on one of the carbons. Transposing them changes nothing.
But we're not finished. There is a chiral center. Do you see it? The maximum and actual number of stereoisomers is two. 2^2 is 4, the answer is 4. Here they all are.


text:
How to quantify the amount of paracetamol present
What is the most accurate way of determining the number of moles of paracetamol from a solution, mixed with acid or base?
I do not have access to more advanced laboratory methods like liquid chromatography, but can use methods such as titration and spectrophotometry.

http://www.rsc.org/learn-chemistry/resource/download/res00000925/cmp00001273/pdf

@orthocresol - to combat potential link rot, how about you summarize that link as an answer?

Alright - here we go. Paracetamol reacts with $\ce{Fe^3+}$ in a 1:1 ratio to give $\ce{Fe^2+}$ - exactly how I am not sure. You can then use a spectrophotometric method (e.g. reaction with $\ce{K3[Fe(CN)]6}$ to form Prussian blue) to quantify the amount of $\ce{Fe^2+}$ present.

will the above method still work if there solution already consists of other compounds such as 4-aminophenol and ethanoic acid?
what my experiment is trying to find out is whether pH affects the rate of hydrolysis of paracetamol.
so i will be using different acidic and basic solutions to hydrolyse as much paracetamol as possible in a fixed amount of time. After that, can i still apply the method you guys have mentioned?


text:
Percentage error. How to find percentage error?
The Question

Calculate the percentage dissociation of $\ce{H2S(g)}$ if $0.1\ \mathrm{mol}$ of $\ce{H2S}$ is kept in a $0.5\ \mathrm L$ flask at $1000\ \mathrm K$. The value for $K_c$ for the reaction $\ce{2H2S(g) <=> 2H2(g) + S2(g)}$ is $1\times10^{-7}$.

MY ATTEMPT
Initially there are $0.1\ \mathrm{mol}$ of $\ce{H2S}$ and nothing for $\ce{H2}$ and $\ce{S2}$
Now at equilibrium let us assume $x$ moles dissociated so
$0.1 - x\ \mathrm{mol}$ of $\ce{H2S}$, $x\ \mathrm{mol}$ of $\ce{H2}$, and $x/2\ \mathrm{mol}$ of $\ce{S2}$
Now I know that degree of dissociation = $\alpha$ = amount dissociated/ initial amount in our case $x/0.1=\alpha$ so $x = 0.1\alpha$. So I substitute.
now we have $0.1-0.1\alpha$ of $\ce{H2S}$, $0.1\alpha$ of $\ce{H2}$, and $0.1\alpha/2$ of $\ce{S2}$
Thus
$$K_c = \frac{\left[0.1\alpha/2V\right](\ce{S2})\left[0.1\alpha/V\right]^2(\ce{H2})}{[0.1-0.1\alpha/V]^2(\ce{H2S})}$$
Substituting value of $V$ and solving I reach here
$$\frac{0.1\alpha^3\alpha^3}{( 0.1 - 0.1\alpha)^2}$$
Now my doubt. I assumed $0.1\alpha$ very small in comparison to $0.1$ so I neglected it and assumed the whole $0.1 - 0.1\alpha$ to be $0.1$ only. Now my doubt is that what is the percentage error in my assumption.
On further solving I get percentage of degree of dissociation as $1\,\%$ and that is correct. My method is correct, my only doubt is how to calculate percentage error in my assumption.

Skipping to your question about

$$\frac{0.1\alpha^3\alpha^3}{( 0.1 - 0.1\alpha)^2}$$

I'll point out two things.
(1) Significant figures come into play here. The Kc is only given to one significant figure as is the molarity of the solution. So each of those quantities has a +/- 50% error. Remember that in Chemistry 0.1 m isn't the same as 0.1000 m.
So the molarity of the solution is 0.1 +/- 0.05. If $\alpha < 0.1$ then the assumption that $0.1\alpha$ is negligible is reasonable.
(2) You can solve the equation for alpha exactly by trial and error. Solve as you did for $\alpha$, then use a factor 10 times larger and one 10 times smaller. One number should be too big and the other too small as compared to Kc. Calculate the average of the two values and test that number. That average will be too big or two small. Discard one side of the bracket. So you're creating an ever shrinking bracket about the "true" value. 6-8 iterations should get you very close.
Now with the "true value" and the estimate from the assumption that $0.1\alpha$ was negligible you can calculate the "real" percentage error for $\alpha$.


text:
Enthalpy of mixing in a regular solution
In a mixture of $A$ and $B$ where $A$ and $B$ are randomly mixed and the average coordination number is $z$. If there are $n_B$ molecules of $B$ in the mixture these will be in contact with a total of $zn_B$ molecules, of which $x_Azn_B$ contacts are with molecules of A. This is the total number of BA contacts. Before mixing, these contacts would have been BB contacts in pure B and so the change in energy on mixing is, for the B molecules:
$$x_Azn_B(H_{AB}-H_{BB})$$
Similarly for A:
$$x_Bzn_A(H_{AB}-H_{AA})$$
This is the point where I get confused. Upon adding together the following expression is somehow achieved:
$$\Delta H_m=\frac{zn_An_B}{2(n_A+n_B)}(2H_{AB}-H_{AA}-H_{BB})$$
Its really just the fraction at the start of that expression I am confused by. However, now the bit I really do not understand. Differentiating with respect to $n_i$ gives $\mu_i$? And apparently this gives $\beta x_j^2$. I have absolutely no idea how. Please can someone explain what's going on here?


text:
Stereo selective reaction products

Pyruvic acid on reduction with yeast gives (-) lactic acid while with $\ce{NaBH4}$ it reduces to (+) lactic acid. How?

I know this has something to do with stereoselectivity. But I don't know which one would be a correct answer.
It would be more helpful if you could explain a bit about stereo-selective reactions. I know what it is, what I want to know is, how to find which of the products will be stereochemically preferred one, just like in this question.

There is no way that borohydride reduction of pyruvic acid can give anything other than racemic lactic acid.

Enyzmatic reduction (as would occur in yeast) could easily be a stereoselective process - probably via selective delivery of a hydride from NADH.
As the other respondent notes, there is simply no way that combining achiral reagents (pyruvic acid and NaBH4) can yield anything other than a racemic product.

That is the exact reason, I'm asking this question for.I have read it in  O P Tondon's  organic chemistry for competitions book.

As for the answer, how may I know or predict ,the product formed after stereoselectivy. Is there a way to predict what should be the major product or something?

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


text:
Difference between chemical potential and gibbs free energy
I know that the chemical potential is the molar Gibbs free energy and that it is an intensive variable and that the gibbs free energy is extensive. However, they have identical units: $\mathrm{J~mol^{-1}}$. I don't understand why $G$ is extensive if it has $\mathrm{mol^{-1}}$ in its units.

It's a matter of notation - technically, $G$ has units of $\mathrm{J}$. But we commonly speak of $\Delta G_r$ of a chemical reaction which has units of $\mathrm{J~mol^{-1}}$. The reason is because $\Delta G_r$ is by definition a difference of molar Gibbs energies: $\Delta G_r = \sum_J \nu_J G_{m,J}$ (or chemical potentials if you prefer). The stoichiometric coefficients $\nu$ are dimensionless and $G_m$ has units $\mathrm{J~mol^{-1}}$. I find it a bit irritating that this notation (same with $\Delta S$ and $\Delta H$) is used in thermochemistry, but that's just how it is.

I know that the chemical potential is the molar Gibbs free energy and that it is an intensive variable and that the gibbs free energy is extensive.

The chemical potential is related to the Gibbs energy in one of its definitions:
$$\mu_i = \left(\frac{dG}{dn_i}\right)_{T,P,N_{j\neq i}}$$
So it is the change in Gibbs energy when one component changes amount while the other components are of constant amount. The dimensions are energy per amount of substance (with units of e.g. J/mol).
The term "Gibbs energy" is used for a lot of things, and the dimensions could be energy, or energy per amount of substance. The Gibbs energy of reaction $\Delta_r G$, arguably the most important Gibbs energy for a chemist, is defined as:
$$\Delta_r G = \left(\frac{dG}{d\xi}\right)_{T,P}$$
where $\xi$ is the extent of reaction with dimensions amount of substance (i.e. amounts of all reactants and products change as $\xi$ changes, according to the stoichiometry specified in the reaction equation). As a consequence, the dimensions of chemical potential and Gibbs free energy of reaction are the same.
For other uses, the term "Gibbs energy" might refer to the Gibbs energy of a given system, and would have dimensions of energy. $G$, like the enthalpy $H$, does not have a defined reference point that is zero, so there is never a specific value associated with $G$. You can, however, use $\Delta G$ to describe the change in Gibbs energy from a defined state to another defined state. This $\Delta G$ would have dimensions of energy, and would be extensive (if you duplicate the system and look at the total, it would have a $\Delta G$ that is twice as large).
It would be easier if the symbols would indicate which quantities are per amount of substance (intensive) and which are not (extensive). Some textbooks use bold typeface for all molar quantities, but introductory textbooks are usually not rigid in terms of dimensions at all.

I don't understand why G is extensive if it has $\pu{mol−1}$ in its units.

The Gibbs energy of reaction is not extensive. However, if you multiply all coefficients by a constant factor, you also have to multiply the Gibbs energy of reaction by the same factor. For example,
$$\ce{H2 + 1/2 O2 -> H2O}\ \ \ \ \Delta_r G^\circ = \pu{−237.13 kJ mol−1}$$
$$\ce{2H2 + O2 -> 2H2O}\ \ \ \ \Delta_r G^\circ = \pu{−474.26 kJ mol−1}$$
The first is equal to the Gibbs energy of formation of water, and the second avoids fractional coefficients. The two are different because the extent of reaction is linked differently to changes in the amount of the reactants and products.

authors: Арина Аракелян


text:
Can the change in internal energy be nonzero if temperature is constant?
I have learnt that internal energy, $U$, is a state function and it only depends on temperature... So if $\Delta T = 0$ then $\Delta U = 0$.
However when I was studying exothermic and endothermic reactions it was written in my textbook that at a constant temperature and pressure, $\Delta U$ is negative for exothermic reactions. How is that possible? Shouldn't $\Delta U$ be zero since the temperature is constant?

The internal energy of every chemical species differs from that of every other chemical species.  So, if you have a combination of species, even under ideal gas conditions, the combined internal energy depends on the number of moles of each chemical species.  If the number of moles of the various chemical species changes, for example if reacting species convert into product species,  the internal energy of the combination changes.  To keep the temperature constant, you would have to remove or add heat, depending on if the reaction is exothermic or endothermic.

TL;DR: Do not just memorise thermodynamics equations! And if you have an issue with the equations $\Delta U = 0$ or $\Delta H = 0$ for an isothermal process, read on...

The first problem
You said that an exothermic reaction corresponds to $\Delta U < 0$. This is not true. It is defined by $\Delta H < 0$.
However, we should still discuss this issue because $H$ is also a state function and in some cases $H$ is dependent only on temperature.

How to study thermodynamics
This kind of issue in thermodynamics frequently crops up here, and it is a very common mistake amongst students to indiscriminately use equations that they have learnt.
So, you have an equation that says $\Delta U = 0$ for an isothermal process, i.e. one at constant $T$. The important question here is not "what is the equation" or "what is the answer"! Instead, you should be asking yourself "how do I derive this result", and the answer will naturally follow. This should really apply to everything you do - how can you expect to apply a formula that you do not actually understand?
Furthermore, if you are able to actually understand how an equation comes about, then you don't need to memorise it - you can derive it. For example, what's the point of memorising $w = -nRT\ln(V_2/V_1)$, along with the conditions, if you can simply obtain the equation from the very definition of work $đw = -\int p \,\mathrm{d}V$ by substituting in $p = nRT/V$ and integrating? What's more, the fact that you actually substituted in the ideal gas law should tell you something about the conditions that accompany the equations - for one, it's only applicable to ideal gases. What about when you reach the equation $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$ and happily memorise it, only to find out later that there are actually three more equations that look exactly like it? Are you going to memorise all four? No - you need to know where they come from.

Internal energy
In general, the internal energy $U$ of a substance is a function of at least two variables, for example $T$ and $p$. However, it can be shown using statistical mechanics that for an ideal gas,
$$U = n\left(\frac{3NRT}{2} + U_0\right)$$
Here, $N$ is the number of atoms in one molecule of the gas (for example, $N = 2$ for $\ce{H2}$, and $N = 3$ for $\ce{H2O}$). $U_0$ is the molar internal energy at absolute zero (this contains terms such as electronic energy), and crucially, this depends on the exact identity of the gas. This result is one example of the equipartition theorem in action (although the theorem itself is more powerful).
We do not need to concern ourselves with its derivation now. The point is that: this equation can only be applied to a fixed amount of a fixed ideal gas at a time. For example: if I have one mole of $\ce{H2}$ at $298\mathrm{~K}$, it is going to have a different internal energy from one mole of $\ce{Ar}$ at $298\mathrm{~K}$, even though the amount of substance and the temperature are the same.
What does this mean?
Firstly, it means that for anything that is not an ideal gas, you cannot assume that $U = U(T)$ and therefore you cannot assume that $\Delta T = 0$ implies $\Delta U = 0$. Let's say I compress $10\mathrm{~g}$ of water by reversibly increasing the pressure from $1\mathrm{~atm}$ to $10\mathrm{~atm}$, at a constant temperature of $323\mathrm{~K}$. Is the change in internal energy equal to zero? No, because water is not an ideal gas.
Secondly, it means that if the amount of ideal gas changes in any way, $U$ is going to change. Let's say I have a balloon filled with $1\mathrm{~mol}$ of argon gas, and I add a further $2\mathrm{~mol}$ argon gas into the balloon, all at a fixed temperature of $298\mathrm{~K}$. Is the change in internal energy equal to zero? No, in fact it will triple because you are changing $n$ from $1\mathrm{~mol}$ to $3\mathrm{~mol}$!
Lastly, it means that if the chemical composition is changing in any way, you are automatically not allowed to say that $\Delta T = 0$ implies $\Delta U = 0$. And this is the case even if all reactants and products are ideal gases. Why not? Let's look at this reaction, and let's assume that everything there behaves as an ideal gas, and let's say that the reaction vessel is kept at a constant temperature of $300\mathrm{~K}$.
$$\ce{H2 (g) + Cl2 (g) -> 2HCl (g)}$$
Let's write an expression for the internal energy of the reactants. On the left-hand side, we have:
$$U_\mathrm{reactants} = \left[\frac{3}{2}RT + U_0(\ce{H2})\right] + \left[\frac{3}{2}RT + U_0(\ce{Cl2})\right]$$
On the right-hand side, we have:
$$U_\mathrm{products} = 3RT + 2U_0(\ce{HCl})$$
So, is $\Delta U$ equal to zero? No; even though the multiples of $RT$ cancel out with each other, the values of $U_0$ for the reacting species are different, which makes
$$\Delta U = 2U_0(\ce{HCl}) - U_0(\ce{H2}) - U_0(\ce{Cl2}) \neq 0.$$

Enthalpy
Let's go back to our (closed) system where there is only one ideal gas and it's not undergoing any chemical reaction. Within this system, we can safely say that if $\Delta T = 0$, then $\Delta U = 0$. So:
$$\begin{align}
\Delta H &= \Delta (U + pV) \\
&= \Delta U + \Delta (pV) \\
&= \Delta (pV) \qquad \\
&= \Delta (nRT) \\
&= nR \Delta T \\
&= 0
\end{align}$$
We haven't made any assumptions here apart from the ideality of the gas (the fact that $n$ is constant came from the fact that it is a closed system). So, for an ideal gas, $\Delta T = 0$ implies $\Delta H = 0$.
However, we did use the fact that $\Delta U = 0$. Therefore, what I said about the cases above where $\Delta U \neq 0$ also applies here.
Regarding the terms endothermic and exothermic, they are used to describe chemical reactions. And for chemical reactions, there is absolutely no way you can say that constant temperature implies $\Delta U = 0$ or $\Delta H = 0$. That's why we can actually use the terms endothermic and exothermic. If every single chemical reaction had $\Delta H = 0$, life would be rather plain!

Internal Energy is a function of temperature only in ideal gases, but for real life situation it is a function of temperature and even volume of the thermodynamic system, for example when solid water (ice) melts to liquid water then during melting the temperature of the "water system" is constant yet its internal energy of system increases, because it takes heat from surrounding so its internal energy will increase but keep a thermometer in contact with the water, you will see no change in temperature. This internal energy is now added to the KE of the molecules of water and to break the Hydrogen bonds b/w water molecules. (internal Energy of system = total KE of system + Total PE due to interaction b/w constituents of system), whereas in ideal gas there is no type of interaction b/w gas particles so no PE in ideal gas but only KE of
ideal gas contributes to its internal energy, and as KE of an ideal gas is a function of temperature of ideal gas, hence for ideal gas internal energy depends only on temperature.
For your question enthalpy change is -ve for exothermic reactions not internal energy change, so in your book it was a misprint, btw in chemical reaction we assume the gas evolved to be ideal, then only can we say const temperature gives const U of system, but you know the world isn't always so ideal (real life state change and volume change considerations can completely change the ideal picture, so for the time being better not venture there and first get basic definitions and conditions for an equation correct).

For exothermic reactions and endothermic reactions...as the reaction occurs, heat is evolved and absorbed by the reaction system respectively. Which means that there is a total change in the heat energy of the system and the change is suffered by the internal energy of the system. Let us consider the example:  C+O2--->CO2...Now as this reaction occurs, the [externally] temperature remains a constant...that is we don't add heat nor do we take heat purposely from the system but then as the reaction proceeds and as it's an exothermic one, heat energy is evolved in the system which is not the problem of the constant temperature that was preset for the system...instead as the reaction proceeds, whatever energy (released by the breaking and creation of new bonds in the resultant compound) is indeed added to the total internal energy of the system.
Therefore, [delta] U can be non zero especially when the system id=s undergoing a reaction.Hope you could understand that...?! Please let me know.


text:
Why do closo-boranes have a 2− charge?
In a question I was solving, it was mentioned that:

[...] closo-boranes are compounds having the general formula $\ce{B_nH_n^2-}$ [...] an example is $\ce{B6H6^2-}$.

Why do these boranes possess a (2−) charge? I am aware that in electron-deficient compounds such as these, there are different types of bonds between boron atoms (such as three-centre two-electron bonds), but I don't know why this should lead to a negative charge.

wikipedia://Polyhedral skeletal electron pair theory ; also hybridization model usually performs poorly outside CHNOF compounds.

@permeakra the wiki didn't help me too much, other than increasing my knowledge, it didn't had anything related to my question i.e. didn't mentioned anything about the charge on closo-boranes

@ChinmayChandak read it again, in particular pay close attention to $B_6H_6^{2-}$ orbital diagram

@permeakra I read it once again, and as I said, I didn't find anything related to the charge on closo-borane, how is there a 2- charge on closo-borane ?

@ChinmayChandak Charge equals number of electron in the particle minus number of electrons in the atoms composing particles. Count electrons on orbital diagram and read corresponding text. In short: when the boran is composed, it is built from n*2p orbitals of BH fragments each coming with one electrons and forming n bonding orbitals (so there is no charge here) and n sp orbitals of BH fragments, forming one large bonding orbitals. Howevere, since BH fragments do NOT give electrons to this orbital, 2 electrons must come from elsewhere, so 2- charge. Seriously, pay attention to electron counts.

The application of MO theory to (electron-deficient) cluster compounds such as boranes is generally summarised in a set of rules, known as Wade's rules. We can look at the closo structure $\ce{[B6H6]^2-}$ in more detail to see where the two negative charges come from.
In closo boranes, each BH fragment points outwards from a vertex of a polyhedron. For $\ce{[B6H6]^2-}$, it turns out that this polyhedron is the familiar octahedron, which has six vertices.
What we must do now is to count how many electrons and how many orbitals each BH fragment contributes to the entire cluster. This can be done fairly easily:

Here we have neglected the boron 2s orbitals. One 2p orbital on boron has the correct symmetry to overlap with the hydrogen 1s orbital, forming bonding (blue) and antibonding (red) orbitals; the other two 2p orbitals remain non-bonding (gold).
So, each BH fragment contributes two valence electrons to the cluster. Since there are six BH fragments, the total valence electron count is (so far) 12. Saying this implies that the boron 2s electrons aren't valence electrons, which is, in a sense, true; they do not participate much in the delocalised bonding in the entire cluster.
If we use the three bonding and nonbonding orbitals to form properly symmetrised MOs across the whole cluster, it turns out we get seven bonding MOs across the entire cluster. I'm not keen to ChemDraw these MOs, so I've taken a picture from Housecroft's Inorganic Chemistry (4th ed.):

The lowest energy MO ($\mathrm{a_{1g}}$) comes from an in-phase combination of all six bonding fragment MOs. The three next lowest ($\mathrm{t_{1u}}$) come from overlap of two bonding fragment MOs with four nonbonding 2p orbitals on the other boron atoms. Three more ($\mathrm{t_{2g}}$) come from "sideways", or "tangential", overlap between four boron 2p orbitals.
In general, it turns out that for a polyhedron with $n$ vertices (at least, where $n \geq 6$ — see comments), there will be $n+1$ bonding MOs which can be made using the fragment MOs at each vertex.
Now, in order to get a stable cluster, we need to be able to fill all these $n+1$ bonding MOs with electrons. In the case of $n = 6$, we have 7 MOs to fill, and so we need 14 electrons. However, as established previously, we only have 12 valence electrons!
The solution is therefore to tack on two extra electrons, and that's why the closo borane requires a (2−) negative charge. If it didn't have this negative charge, it wouldn't be able to fill all of the bonding MOs in an octahedron, and it's quite likely that it would choose to adopt a different geometry that is no longer closo.

Interesting, but does this apply to all such forms?  What happens if we try a tetrahedron?  If it's isoelectronic to white phosphorus, then the borane anion would have four negative charges.

@OscarLanzi truthfully, it's been a long time since I've studied this, and I've forgotten all of it by now, so I have no idea what would happen. My first instinct is to agree - yes it would need 4 negative charges. One would have four orbitals in a tetrahedral arrangement which (just like in methane) would form one $a_1$ and three $t$ bonding orbitals; the BH units provide 4 electrons so need 4 more to fill them up. Searching for $\ce{B4H4}$ brings up some interesting discussions, eg. https://www.sciencedirect.com/science/article/pii/S2210271X1730316X though I merely skimmed.

I'm not entirely sure why $\ce{B4H4}$ is unstable, whereas $\ce{B4Cl4}$ isn't. I would perhaps hypothesise that the tetrahedral cage is just too small — having the more electronegative chlorine on the outside helps to reduce the electron density inside the cage. But that's just wild speculation.


text:
Difference in work done by a gas in reversible and irreversible isothermal expansion
Why there is difference in work done by gas in isothermal expansion when gas expands through reversible and irreversible process?

In a reversible expansion, the gas expands with the minimum pressure while in an irreversible process, the pressure is greater than a minimum, so pV of irreversible expansion is always greater than that of reversible process, hence the difference

For a massless piston, the force per unit area exerted by the gas on the inner face of the piston is exactly equal to the external pressure exerted by the surroundings on the outer face of the piston, irrespective of whether the expansion is reversible or irreversible (by Newton's 2nd law).  The work done on the surroundings is thus always equal to the integral of the external pressure integrated over the volume change.
However, for an irreversible expansion, the force per unit area exerted by the gas is comprised of two components: (1) the local gas pressure at the piston face (as determined by applying the ideal gas law locally) and (2) viscous stresses that are proportional to the rate at which the gas is expanding.  Furthermore, in an irreversible expansion, the gas pressure and temperature are typically not even uniform within the cylinder.
In a reversible expansion, which is carried out very slowly, the viscous stresses are negligible (since the rate of expansion is negligible), and the gas pressure and temperature are uniform throughout the cylinder. Thus the external pressure is equal to the uniform pressure of the gas calculated from the ideal gas law.  Consequently, for a reversible expansion, one can use the pressure calculated from the ideal gas law in place of the external pressure to calculate the work done.
To get a better feel for how this all plays out, particularly with regard to viscous stresses and viscous dissipation, please see the following link:  https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/.  This article points out the close analogy between the transient behavior of an ideal gas experiencing an irreversible expansion or compression, and that of a spring and damper connected in parallel. It demonstrates that the physical mechanism responsible for the difference between the irreversible work and the irreversible work is viscous dissipation of mechanical energy to thermal energy (yes, even in the ideal gas region, gases exhibit non-zero viscosity).


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Why does DMF (dimethylformamide) have 3 peaks in its 1-H spectrum?
I get that the carbons can't rotate around the nitrogen due to its partial double bond character, but how does this result in the hydrogens being in different environments? As such would this mean all non rotatable bonds would result in hydrogens all with different peaks?

this mean all non rotatable bonds would result in hydrogens all with different peaks - they do. Alkenes for example.

If you understand the partial double bond character induced by the N-CO moeity, then you can draw out the molecule as shown below.

The three signals you get are from the CH and the two CH3 groups. When the energy barrier for rotation is high to prevent averaging (fast rotation), you can see that the two methyl groups are non-equivalent: one is trans to the C-H, and the other is cis to it. This is a very common characteristic of many amide-containing molecules. The energy barrier to rotation is highly dependent on steric interactions. The number of possible conformers arising from even small peptide molecules increases very rapidly, when you consider the combination of cis/trans type orientations of the amide bonds.
Now, as you start to heat this system (or observe at a lower magnetic field - that may be for another question), you will start to see signal averaging, as the two methyl groups begin to exchange on the NMR timescale. As this rate of exchange becomes faster, you will get signal coalescing and finally complete signal averaging to the point where the two CH3 groups are equivalent. This is seen below, with the bottom spectrum being at low temperature (and/or high magnetic field) and the top spectrum being at high temperature (and/or low magnetic field).

Signal averaging via rotation about bonds happens even in single bonds. If you cool a CH3 group down far enough, you will be able to slow down the rotation of the C-C bond enough to be able to distinguish all protons as non-equivalent.

So it's all about speed of rotation then. Last bit of answer clarified it nicely, thanks a lot.


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Is it harmful to mix Hydrogen Peroxide with oven cleaning spray?
My mother recently stumbled upon this mix when she was trying to remove stains from some laundry. When just the peroxide didn't work, she sprayed on some Easy Off, and the stain disappeared. Now she uses this method regularly, and I'm worried that it might be toxic or something. I just want to make sure she isn't poisoning herself.

It's good to note that all chemicals can be toxic or harmful. It's not a good idea to go mixing reactive chemicals like those , I'm not sure what easy off is but oven cleaners normally pretty reactive. Sounds like a bit of overkill would probly damage the clothes or washing machine.

I'd say to use things that are designed to go in the washing machine,on clothes and not harmful to skin that is not corrosive like the oven cleaner!.

I wonder if this is just used on white cottons. I'd expect Easy Off to be KOH or NaOH, really quite nasty stuff. When I was working for IBM in a manufacturing plant I was always amused by the thought of trying to get  such oven cleaning approved on the production line. The operator would have been in a bubble suit! You never wanted to ask if a process procedure was safe. ;-)

Chemically, we are talking about mixing hydrogen peroxide with sodium hydroxide. They will react to form sodium hydroperoxide, but the latter is not as good an oxidizing agent as hydrogen peroxide, and a weaker base than sodium hydroxide. Probably the added base helps to dissolve some components of the stain. I don't see this as risky, especially if it is on a small scale (spraying laundry), and uses 3% peroxide, rather than mixing them in bulk. Note: With hydrogen peroxide, you have to watch out because some metal ions catalyze its decomposition to oxygen and water.

Sodium hydroperoxide is a better nucleophile than sodium hydroxide which will facilitate the stain removal.

I accidentally stumbled across this question. I once had a coworker clean a toilet with bleach and 409. The tragedy was he knew better and yet no one was injured. The man was smart but his thin stress management had a way of making things less clear headed when making everyday decisions.
I had to do OSHA 10 and OSHA 30 for work many years later. I realize such training is not much but it is something that made me want to check carefully as I make decisions. I have watched different people not check or measure twice to cut once or some such decision making process.
That said, in my opinion oven cleaners are harcore effective. I can't say what the oven cleaner in question was made of because I wasn't there. I want to comment on what @iad22agp wrote because it gave me the chills.
If a cleaner made with lye or sodium hydroxide is used or within splashing distance it sounds like extra care should be made to keep hydrogen peroxide very far away. I am not a chemist but I can read MSDS cautionary info. According to the California Fatality Assessment and Control Evaluation report below, posted by the CDC/NIOSH, a "mixture of sodium hydroxide and hydrogen peroxide produces a strong oxidizer called sodium peroxide and can result in an exothermic mist from the onset of high temperatures and heat from the decomposition of incompatible mixtures causing a severe thermal hazard."
California FACE Report #09CA004: A Chemist Dies from Burns Caused by Mixing Chemicals (pdf)
NIOSH: California Case Report: 09CA004 (html)
The man who died must have experienced a horror show as he passed... I realize the FACE report is not an MSDS for either product but it seems important to consider when cleaning with chemicals in the way described by the OP's question. I realize that industrial chemicals are more dangerous than domestic cleaning products but being extra careful beats dying every time.

To be fair though, the container in that report was $\pu{50%}\ \ce{H2O2}$ while household cleaners will typically be $\pu{3%}$.

Indeed, "I realize industrial chemicals are more dangerous."


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What is the mechanism for the double bond migration in 4-methylcyclopent-2-en-1-one?
The problem I was asked to answer:

Please explain using only enolisation and protonation steps:

My work so far:

Seems completely wrong... can someone please correct me?
EDIT I just proposed a new mechanism:

After Step 2, re-protonate alpha to the carbonyl (the double bond has now migrated to the beta-gamma position).  Now enolize the carbonyl - remove the proton next to the methyl group (the top one), then push the electrons all the way back, protonating the gamma position.  Job done!


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Negative Ohm reading from a Electrochemical reaction?
The purpose of my experiment is to determine which salt is the best electrolyte. The setup is a tarnished spoon (Ag2S) is placed on top of thin plastic pieces that separate it from a container coated with aluminum foil. A solution of 1.0 M Salt and .2 M baking soda is heated and then poured into the container.  The following electrochemical takes place 3 Ag2S(s) + 2 Al(s) → 6 Ag(s) + Al2S3(s), the redox reaction being  Al + 3 Ag+ → Al3+ + 3 Ag. A multi-meter set to 200 ohms is attached, the positive end to the foil, and the negative end to the spoon. The picture included is my set up.
My logic was I could take the resistance measurement, take its reciprocal, and then find the conductance. I thought whichever salt would give me the best conductance would be the best electrolyte. However, I've only gotten readings of Negative ohms. My readings have been -190 ohms and then it steadily increased. The problem there can't be a negative conductance. From research I've found this, "conductivity is negative, it means that the resistance is negative, right ? This may occur if you have created a generator in your experiment, for example through a non-desired oxydoreduction phenomenon between two metals. If a current is produced in your experiment due to such reaction between 2 metals of different oxidation potentials, and if you material acts as en electrolyte, your apparatus may interpret it as a negative resistance."
How could I use the negative Ohms reading? Should I be collecting data in a different way? Help!

How could I use the negative Ohms reading?

You're using the meter the wrong way. A meter is set up to have a particular low voltage and measures how much current flows. Since your electrochemical cell has a voltage and creates its own current the meter reading goes whacko.

Should I be collecting data in a different way?

If measure current that would give you some idea. In a better controlled analytical chemistry experiment a special electrode pair with two flat plates is used so that the geometric configuration stays exactly the same.


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How to interpret the rate constants of in-vivo racemization of thalidomide?
The article by Eriksson et. al. found that thalidomide undergoes chiral inversion in vivo. It found the mean rate constants too, $0.17~\mathrm{h^{-1}}$ (R to S) and $0.12~\mathrm{h^{-1}}$ (S to R).
I have no idea what those numbers mean. Does that mean 17% of the R-thalidomide becomes S-thalidomide every hour?
Reference
Eriksson, T.; Bjöurkman,S.; Roth, B.; Fyge, Å; Höuglund, P; Chirality 1995, 7 (1), 44-52. (PubMed)

Woo. It does look better. I noticed you italized R and S. Was it with asterisks?. Had you used $R$ instead, would it become bold and italic?

The $ signs invoke mathematical mode, where you can typeset formulae with special commands. A short description is given in the second link above.  It takes some time to learn but it can be incredibly helpful. The * belong to the markdown set that is used on the site. Visit this page to learn more. This one is much easier to learn. If you have any questions that are not part of the [help], then visit [meta].