text:
What causes the lowering of vapour pressure in volatile/nonvolatile solvent mixtures?
"Based on Figure 13.18, you might think that the reason volatile solvent molecules in a solution are less likely to escape to the gas phase, compared to the pure solvent, is that the solute molecules are physically blocking the solvent molecules from leaving at the surface. This is a common misconception. Design an experiment to test the hypothesis that solute blocking of solvent vaporization is not the reason that solutions have lower vapor pressures than pure solvents."

The explanation (in the main chapter, not as an answer to this question) given in Chemistry: The Central Science (13th Edition) by Brown, LeMay, and Bursten (abbreviated as CTCS) is: "A solution consisting of a volatile liquid solvent and a nonvolatile solute forms spontaneously because of the increase in entropy that accompanies their mixing. In effect, the solvent molecules are stabilized in their liquid state by this process and thus have a lower tendency to escape into the vapor state. Therefore, when a nonvolatile solute is present, the vapor pressure of the solvent is lower than the vapor pressure of the pure solvent, as illustrated in ▲ Figure 13.18."
However, this is, I believe, the only textbook which gives this explanation for lowering in vapour pressure on adding a non-volatile solute. The other (reputed) textbooks I've read, like Zumdahl and NCERT Grade XII Chemistry, all give the explanation that CTCS calls a "common misconception". So, which explanation is correct? If possible, please give an experiment also.

Wow! Thanks for asking this question. I'm also following the Grade XII NCERT Chemistry textbook and I was also skeptical about the explanation given in that

I think that the second explanation is the correct one. A simple experiment can prove it: if you place a porous cover over the pure solvent, its vapour pressure will not change. In alternative, you can place a few corks and observe the same result (no variation in the vapour pressure). In both cases surface sites are blocked, but the vapor pressure will obviously remain unaltered. Therefore the reason cannot be that the solute molecules are hindering the solvent molecule escape by occupying the solution surface.
The force that is driving the solvent molecules to the vapour phase is the difference between the entropy of the liquid and the vapour phases: when the molecule escape the liquid there is an entropy gain. Adding a non-volatile solute will increase the entropy of the liquid phase without having an effect on the entropy of the vapour phrase (given that the solute is non-volatile). This will lower the difference in entropy between the two phases, which is the driving force that make vaporisation happen, therefore there will be a lower number of solvent particles in the vapour phase and a consequent lower vapour pressure.
Textbooks like Chemistry: The Molecular Nature of Matter and Change (Silberberg, Amateis) or Physical Chemistry for the Biosciences (Chang) provides the same explanation. In addition, there is a short paper/article on this topic[1] where the author highlights how only very few freshman chemistry texts give an adequate explanation of this phenomenon by making use of a entropy-based argument.
     1 Vapor Pressure Lowering by Nonvolatile Solutes, G. D. Peckham, J. Chem. Educ., 1998, 75 (6), 787

This is similar to adsorption on a surface: van der Waals and other forces (e.g. hydrogen bond) bind the solute to the solvent, just as a gas may be bound to a solid. The lowest energy state depends on the strength of the solute-solvent bonds (adhesion) and the solute-solute bonds (cohesion) as opposed to random molecular motion causing evaporation. See http://en.wikipedia.org/wiki/Van_der_Waals_force and http://www.chem.purdue.edu/gchelp/solutions/eboil.html

But if we consider an ideal solution (just assuming) then this logic cannot be applied.

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text:
Lewis Acid Catalysts
I am studying for the International Chemistry Olympiad and some of questions involve synthesis of complicated organic compounds and i have no idea how to do them.
The setting of the questions is usually like this: they give you a rough idea of the end product however some of the groups are replaced with R. They replace all the other intermediates with A, B, C ... They also give the reactants and catalysts for each step.
Most of the time the catalysts are lewis acids. I wondering if you were able to provide mechanisms and general resulting reactions of these catalyses so i could predict what the product will be.
Also, is $\ce{NaBH4}$ a lewis acid?

Welcome to chemistry.se! I am afraid, I do not quite understand what you are asking. I might be better, if you could provide us with a specific question, an example of what you are struggling with. In its current form, this question might be too broad and is likely to be closed. You also might want to consider to ask two separate questions for the two problems you face.

Ask yourself: can NaBH4 accept anymore electrons? Remember the octet rule.

Can you provide a specific example of a question, so we know the exact context?

Hi, thanks for reading my question and sorry for being really confusing. I cant write the question down as it is really long and i cant draw the structures here. Here is the link to the question- file:///C:/Users/pc/Downloads/volume3-icho41-45.pdf

It is the 7th question from the 2011 paper on page 1272.

@the The "link" you posted is a pathname on your computer. It will not help us and we cannot help you.

@bon The edit you suggested was not really necessary, I approved it, but generally I would consider this to be too minor.

Sorry, i am not the best when it comes to computers. However it doesn't really matter, all i want to know is general mechanisms for reactions involving lewis acid catalysts.

Firstly, $\ce{NaBH4}$ is not a Lewis acid because it cannot accept any more electrons. Aqueous $\ce{NaBH4}$ is a commonly used reducing agent often for reduction of carbonyls to their respective alcohols.
The mechanism of Lewis acid catalysis depends on the reaction but often Lewis acids are used to generate cationic electrophiles such as in the Friedel-Crafts reaction.

Thank you for answering my  question. However is there any other reaction mechanisms involving lewis acid catalysts?

@the any reaction where you want a cationic electrophile could quite easily involve a lewis acid. are you thinking of any specific example?

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text:
Can you have addition polymerisation with monomers other than alkenes?
Every example of addition polymerisation I have seen so far has used alkenes as the monomers. I was wondering if it is possible to have some other molecule. I don't think the following molecule can exist due to stability reasons, but as an example could you not use something like
$$\ce{H-N=N-H}\,?$$
Surely all you need is the presence of a double bond?

You're right, the monomer needn't be an alkene. Non-alkenes with double bonds can also form addition polymers. In fact, the monomer needn't always have a double bond either! Surprised? Take a look at this:

(Image source: chemwiki.ucdavis.edu)

the linked reaction is a condensation polymerisation not an addition polymerisation

What molecule is removed in this "condensation" reaction?  Thought so ... this is an addition!

$\ce{N2H2}$ is a very unstable molecule, prone to disproportionation into hydrazine and nitrogen. $\ce{N-N}$ bonds except one in $\ce{N2}$ molecule are unstable and generally unfavorable. So, no there are no polymers with nitrogen as a sole component of the main chain.
Several classes of compounds of 3-row elements demonstrate tendency for oligomerization, namely $\ce{(HP)_{n}}$ derivatives usually exist as a ring. It is possible to obtain metastable at room temperature plastic sulfur, containing linear sulfur chains terminated by free radicals.
However, generally when in search of easily polymerizing molecule that produce main chain containing elements other than carbon, one usually have to look into compounds producing chains with two elements. The most common and known one is formaldehyde $\ce{CH2=O}$, producing $\ce{HO-(CH2-O)_{n}-H}$ chains. A less known, but interesting example is hexachlorotriphosphazene, the source material for production of Polyphosphazenes.
In  general, metastable double bonds are rare for elements other than carbon (sometimes with nitrogen or oxygen partners), so usually reactions intended to produce such bonds, say, $\ce{Si=Si}$ usually end with olygomer instead, typically of ring structure. Moreover, single homoelemental bonds are usually much weaker for elements other than CHNOB, so typically when a chains containing other elements is occurred, it is usually hetero-atomic chain, produced by means other than olygomerization of double-bond containing monomer. For example, an extremely wide-used family of polymers (polysiloxanes, or silicones) is produce by hydrolisys of dialkyldicholorsilanes $\ce{R2SiCl2}$.

To my knowledge, diazene (diimine, $\ce{N2H2}$) does not polymerize but simply decomposes (to hydrazine and nitrogen). Even in the polymerization azobenzene-substituted vinylethers (consider azobenzene as diazene in with both hydrogen atoms are replaced by phenyl rings), the $\ce{N=N}$ double bond does not participate in the polymerization.
But you are right, a lot of copolymers consists of monomers in which the polymerization does not take place at $\ce{C=C}$ bonds.
Think in the reaction of $\ce{\alpha,\omega}$-diols

with phosgene ($\ce{COCl2}$) to polycarbonates
with diisocyanates to polyurethanes

or the reaction of dianhydrides with diamines to polyimides.

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text:
Why do gummy bears explode when added to hot potassium chlorate?
This link shows that a gummy bear explodes when in contact with heated potassium chlorate, $\ce{KClO3}$. But what in a gummy bear creates this reaction?
Also, do other foods (fruit, icing sugar...) react as violently with potassium chlorate?

Never, never, never repeat this experiment on your own. Potassium chlorate is known to form mixture with violent and sensitive behavior with almost everything combustible and many incombustible things.

Also, be prepared to throw away the test tube afterwards.

Potassium chlorate is a source of oxygen. After heating, it decomposes to $\ce{O2}$ and $\ce{KCl}$:
$$\ce{4 KClO3 → KCl + 3 KClO4}$$
$$\ce{KClO4 → KCl + 2O2}$$
The gummy bear is mainly composed of sugar and other carbohydrates. Those carbohydrates will react with oxygen, combustion occurs. For example, glucose will react in this manner:
$$\ce{6O2 + C6H12O6 -> 6CO2 + 6H2O}$$
If there is any material present which does not burn, such as $\ce{H2O}$, the temperature will not rise as high. For gummy bears the reaction works spectacularly because they are mainly carbohydrates (>70%).
An apple, for example, has only ~13% carbohydrates – unless you dry it, of course. On the other hand, this video on YouTube is an example of how sugar itself reacts violently with potassium chlorate.

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text:
Is shielding of electrons really best described as shielding or is it really Coulombic repulsion between electrons?
For example, take the 2s electron in lithium - is the reason that it's ionisation energy is lower than would be expected if "shielding" didn't occur at all because the 1s electrons shield the attractive coulombic force from the nucleus or because the 2s electron is repelled by the 1s electrons? Is it possibly a mixture of the two?
To me, shielding seems to be a model to help conceptualize what is actually happening; am I wrong about this? What is actually going on?

The shielding model works as follows:
 (source)
So it is a mixture of attraction and pushing from protons and non-valence electrons, in accord with the Coulomb's law. The prediction of the effective core charge is not an easy task - you will have to use quantum chemistry methods (e.g. Hartree-Fock) to predict those, like its done in this reference.
I think its best viewed as a pragmatic model to boil down the higher, not very graphic, quantum theories, to one number, the effective nuclear charge:
$$ Z_\text{eff} = Z - S$$

Why do electrons that are outside electron of interest have no effect? Surely they also repel the electron on interest TOWARDS the positively charged nucleus.

Electrons in the same shell do have an effect on ionisation energy, but electrons in a higher shell are "spread out" and have only a small effect. You have to picture them as an electron cloud - which will push the electron of interest from all directions and therefor cancels itself for the most part.

Its actually Coulombic repulsion between electrons and attraction between electron and protons.
It can be said shelding if you are considering the last electron. But if u consider any electron in the lower shells the electron in the upper shells will push the electron towards the nucleus and it won't be called shilding.


text:
Why is the units of kcat 1/s?
I understand that $k_\text{cat}$ measures the turnover number of an enzyme. This measure is therefore a quantity of molecule conversions per unit of time. I suspect that my problem is more that of a lack of maths but why is the unit expressed in $\mathrm s^{-1}$? Why not just seconds?

The turnover frequency is a rate. (It's actually the kinetic rate of the reaction in saturating substrate concentration, normalized to the amount of enzyme.)
It might be helpful to think about something like a factory that makes a certain number of widgets in a certain number of time. How would you naturally describe how fast the factory could make widgets? Most people would express how fast the factory works in terms of "widgets per hour". It's the same with enzymes - it's most natural to describe how fast the enzyme works in terms of "turnovers per second". It's just that the "turnover" is implied and is viewed as "dimensionless", so it's left off, leaving you simply with "s-1" as a unit.
(Note that even though the "turnover" is left off, it's still very important to know it's there, and to know what, exactly, one "turnover" is. In complicated reactions there is sometimes different reasonable ways to count what "one turnover" is, and different definitions could result in a 2, 3, 4, 5, etc. fold difference in value. Those vexing numeric differences become easy to explain once you realize this researcher is using the formation of one molecule of this product as "one turnover", but that researcher is using the disappearance of one molecule of that substrate as "one turnover".)
Naively, one could invert the ratio to come up with a value in time, rather than in 1/time. For example, if a factory makes 5 widgets per hour, it takes them 0.2 hours to make a widget ... sort of. One issue you run into is that that's the time on average. In a stochastic process like enzyme turnover, it's highly unlikely that any single turnover will take exactly the time specified. Some will take longer, some will be quicker. Only for a large number of turnovers in aggregate does that time value have meaning. Saying that "a turnover takes 5 s" gives a false impression about individual turnovers in a way that saying "this enzyme works at 0.2 turnovers per second" doesn't.
Additionally, in the case of multi-step reactions, saying that "a turnover takes 5 s" may be incorrect. The entire pathway may take much more that that. It's just that once the pathway has reached steady state it's only the rate limiting step which takes 5 s.  For example, take a pathway of 20 steps, 19 of which take 4 s each, but one which takes 5 s. It's going to take (19*4 + 5) = 81 seconds for any given substrate molecule to start at the beginning of the pathway and make it to the end. However, once the pathway is full and each step is handing off products to the next step, the pathway will be producing product at a rate of 1 every 5s. (This, of course, doesn't apply to enzymes where all reactions take place at the same active site, but it's a consideration in the general case, which explains why 1/s is the preferred unit for reaction rates.)

This is a good answer and gets a +1 from me.  But "(It's actually the kinetic rate of the reaction in saturating substrate concentration.)" doesn't actually seem to be correct.  Somewhere in there we'd need to multiply by a concentration, presumably the concentration of catalyst, to get a rate.  Also, the idea in the last paragraph is sound but the math doesn't work out.  The "rate" of a 20-step process with 19 4-second steps and 1 5-second step is given by $\nu_{total} = \frac{1}{\sum{\frac{1}{\nu_i}}} = \frac{1}{19\frac{1}{4} + \frac{1}{5}}=\frac{20}{99}$.

@CurtF. You're right that there is a factor related to the amount of enzyme - I'll edit to clarify. On your second point, I don't quite see where you're getting that equation from. It's likely being misapplied if it's saying that adding additional steps in series to a reaction that takes 5 seconds per turnover can somehow make the combined reaction (all of which passes through the obligate 5 per second step) take only 4.95 seconds per turnover.

You were actually describing the turnover frequency of a catalyst, and frequency has the unit hertz which is 1/s. Turnover number is a dimensionless value: number of molecules of reagent that a catalytic site can convert before becoming inactivated.

Is the description on wikipedia, that it has different meanings in enzymology and other chemical fields, wrong then?

I think that it is. I don't see why it would be differently defined in enzymology.

In enzymology, the turnover number has no relationship to the enzyme becoming inactivated.  Instead, it is the reaction rate per mole of enzyme, in the extrapolation of infinite substrate concentration.

As you say this represents "conversions per unit of time".
The mathematical expression for "per time unit" is 1/(time unit), and seconds is a particular time unit.

Okay, but why does it have to be 1/(time unit). I.e. why does it have to be (eg.) 5 conversions per second = 5 1/s. Why not simply 5 s. I don't understand the requirement for the '1 divided by'. Thanks

Compare to a situation where you are paid 5 dollars per hour for your work.  You work 8 hours and expect 40 dollars at the end of the day.  But your boss says, "no your pay for today is 5 hours, not $40".

Vmax=kcat * ([E total])
If the units for Vmax are M/s i.e. s-1 and you divide Vmax by the concentration of total enzyme which has units in M, the Ms will cancel out, leaving you with 1/s or s-1


text:
Expressions for constant volume and constant pressure heat capacities
My lecture handout says:
$$C_V = \frac{\mathrm{d}q_V}{\mathrm{d}T} = \left(\frac{\partial U}{\partial T}\right)_{\!V}$$
I understand that $C_V = \mathrm{d}q_V/\mathrm{d}T$ and that $\mathrm{d}U = \mathrm{d}q_V$, but what's the need for the partial derivative? Why can't you just substitute and get:
$$C_V = \frac{\mathrm{d}U}{\mathrm{d}T}$$
Likewise for the constant pressure heat capacity:
$$C_p = \frac{\mathrm{d}q_p}{\mathrm{d}T} = \left(\frac{\partial H}{\partial T}\right)_{\!p}$$

To complement entropid's answer, let's look at the formulation of $U$ that is slightly more useful:
$$\begin{align}
U &= q + w \\
\mathrm{d}U &= \mathrm{d}q + \mathrm{d}w \\
&= C\,\mathrm{d}T - p\,\mathrm{d}V
\end{align}$$
At constant volume, $\mathrm{d}V = 0$, so $p\,\mathrm{d}V = 0$
$$\begin{align}
\mathrm{d}U &= C_V\,\mathrm{d}T - 0 \\
C_V &= \frac{\mathrm{d}U}{\mathrm{d}T}
\end{align}$$
However, this notation assumes that $U$ is only a function of $T$, as in $U(T)$. It is not. In this version, $U$ is a function of $V$ and $T$: $U(V, T)$. Note that we do not need $p$ as a variable, since it is dependent only on $V$ and $T$ as well - for an ideal/perfect gas $p(V,T)=\frac{nRT}{V}$. Thus, we need partial derivative notation so that we know we are not trying to take the total derivative of $U$:
$$C_V = \left(\frac{\partial U}{\partial T}\right)_{\!V}$$

The difference in internal energy ($\mathrm{d}U$) is a function of many different variables:
$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V \!$$
Therefore you must specify – via a partial derivative – with respect to which of the variables you are doing the derivative, otherwise you would need a total derivative, which is not what the equation requires.

But since $dU=dq_V$ why can't you just substitute this in?

$dU=dq_v$ only at constant volume. You need the partial derivative notation to specify that you are at constant volume. Likewise for constant pressure.


text:
What is the difference between "reaction pathway" and "reaction mechanism"?
What is the difference between "reaction pathway" and "reaction mechanism"?
Are these expressions synonyms?
Some textbooks authors use both without making clear to the learner if they are synonyms or not.

Usually with the reaction mechanism the sequence of events (electron transfers) is indicated, while the term reaction pathway usually refers to the reaction coordinate diagram associated with the reaction, i.e. the change in energy with an extra notion on the transition states.

Just to expound, a reaction pathway deals with the potential energy landscape for a given reaction mechanism.

As far as my knowledge is concerned when we talk about Reaction Mechanism then we mean the Steps through which the reactants are passed and reach to the Product. Say for example the making of water from Hydrogen and Oxygen which involves 40 steps and at the end we get what we know H2 +1/2 O2 giving H2O
Another example can be the chlorination of Methane to form CCl4.
While in Reaction pathway we talk about the inside of the reactions involved like talking about electrons movement .

No, Khair Muhammabd. Go through the reaction mechanism. Actually, the reaction mechanism is about sequence and detailed understanding of elementary reactions(that are here called elementary steps in the term of Composite Reaction) for conversion of reactants to the desired products.
It is accompanied with the formation of Reaction Intermediate at each elementary step(except last one). Where the overall reaction can be obtained by adding all elementary reactions together. So as the overall rate of reaction can be obtained by adding the rates of all elementary steps.
But what happens, every composite reaction has one elementary step (Mostly 1st step) that is carried out very fast as compared to other steps. Say, T1=1sec for 1st step. and T2=0.0000000001 sec reaction.
So the total time of the composite reaction will almost be equal to first step (T(total)=T1) while T2 is neglected. Basically the first step is called Rate dertermining step(RDS).
So overall reaction rate will be taken equal to the rate of RDS.
R(r*n)=R(RDS).
While I know nothing about Reaction Pathways, still going through it.


text:
Calculating the Gibbs Free Energy of Mixing
This is a Physical Chemistry Question and comes from the Atkins Physical Chemistry book.

Suppose that $2.0~\mathrm{mol}$ $\ce{H2}$ at $2.0~\mathrm{atm}$ and $25~^\circ\mathrm{C}$ and $4.0~\mathrm{mol}$ $\ce{N2}$ at $3.0~\mathrm{atm}$ and $25~^\circ\mathrm{C}$ are mixed at constant volume. Calculate $\Delta_\text{mix}G$. What would be the value of $\Delta_\text{mix}G$ had the pressures been identical initially?

The answers are given but I don't understand how to get to $-9.7~\mathrm{kJ}$ for the first part of the question (when initial pressures are different. The gases are initially at different pressures but does that make a difference. Can I still use Raoult's Law?

Let’s start with the last question: Calculate $\Delta_\text{mix}G$ when the pressures are equal in both compartments before mixing. Most physical chemistry textbooks treat this case (including Atkins), and the expression of $\Delta_\text{mix}G$ is:
$$\Delta_\text{mix}G= nRT(x_1\ln x_1 +x_2\ln x_2)$$ In this exercise, $1$ denotes hydrogen, $2$ denotes nitrogen $n=6\ \text{mol}$, $x_1=\frac{2}{6}$ and $x_2=\frac{4}{6}$. So, by substituting these values in the above expression, we find: $$\Delta_\text{mix}G= 6\ \text{mol}\times 8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}}\times 298\ \text{K}\times(\frac{2}{6}\ln\frac{2}{6} +\frac{4}{6}\ln\frac{4}{6})= -9462\ \text{J}$$ As $\Delta_\text{mix}G<0$ at constant temperature and pressure, this means that the mixing of two ideal gases is a spontaneous process .

Let’s now treat the first question with two different pressure in the two compartments. The initial Gibbs energy is $$G_\text{i}=n_1(\mu_1^0 +RT\ln p_1)+n_2(\mu_2^0 +RT\ln p_2)$$ Where $n_1=2\ \text{mol}$, $n_2=4\ \text{mol}$, $p_1=2\ \text{atm}$, $p_2=3\ \text{atm}$. $$G_\text{i}=2\ \text{mol}\times(\mu_1^0 +RT\ln2)+4\ \text{mol}\times(\mu_2^0 +RT\ln3)$$
When the partition is removed, and each gas occupies the volume $V_1+V_2$.
The partial pressure of hydrogen is $$p^\prime_1=\frac{2\ \text{mol}\times RT}{V_1+V_2}$$
The partial pressure of nitrogen is $$p^\prime_2= \frac{4\ \text{mol}\times RT}{V_1+V_2} $$  On the other hand: $$V_1=\frac{2\ \text{mol}\times RT}{2\ \text{atm}}=1\frac{\text{mol}}{\text{atm}}\times RT$$ $$V_2=\frac{4\ \text{mol}\times RT}{3\ \text{atm}}=\frac{4}{3}\frac{\text{mol}}{\text{atm}}\times RT$$ i.e.  $$V_1+V_2=\frac{7}{3}\frac{\text{mol}}{\text{atm}}\times RT$$
So $$p^\prime_1=\frac{2\ \text{mol}\times RT}{(7/3\ \text{mol/atm})RT}=\frac{6}{7}\ \text{atm}$$ $$p^\prime_2= \frac{4\ \text{mol}\times RT}{(7/3\ \text{mol/atm})RT}=\frac{12}{7}\ \text{atm} $$ The final Gibbs energy is $$G_\text{f}=n_1(\mu_1^0 +RT\ln p^\prime_1)+n_2(\mu_2^0 +RT\ln p^\prime_2)$$  $$G_\text{f}=2\ \text{mol}\times(\mu_1^0 +RT\ln\frac{6}{7})+4\ \text{mol}\times(\mu_2^0 +RT\ln\frac{12}{7})$$ The Gibbs energy of mixing is the difference between $G_\text{f}$ and $G_\text{i}$
$$\Delta_\text{mix}G= 2\ \text{mol}\times RT\ln\frac{6/7}{2} +4\ \text{mol}\times RT\ln\frac{12/7}{3}$$  $$\Delta_\text{mix}G= 2\ \text{mol}\times RT\ln\frac{6}{14} +4\ \text{mol}\times RT\ln\frac{12}{21}$$  $$\Delta_\text{mix}G= 2\ \text{mol}\times 8.314\frac{\text{J}}{\text{mol}\cdot\text{K}}\times 298\ \text{K}\times(\ln\frac{3}{7} +2\times\ln\frac{4}{7})= -9744\ \text{J}\approx{-9.7\ \text{kJ}}$$
In this case, the value of $\Delta_\text{mix}G$ is the sum of two contributions: the mixing itself and the changes of pressure of the two gases to their final total pressure.

I hope it’s clear now.

Why isn't zero? I mean at constant $T$ and $p$ it follows that $dG=-SdT + Vdp=0$.


text:
Why is it possible, that an equilibrium shifts?
So in class, I've learned about Equilibrium. So it is basically when the rate of the reverse and the forward reaction are the same. So I was like that sounds pretty easy, I can understand this. And then suddenly there was a twist (there always is in class). The teacher was like: so if you increase or add a stressor, the equilibrium will shift to stop it. How can equilibrium shift, I thought it was a concept not a "thing" to move up or down. Is it because it shifts to maintain the Equilibrium Constant?
I do not understand this. If this question has already been asked then please point me to that. I have even searched online and could not find an explanation in layman's terms.

related: http://chemistry.stackexchange.com/q/5486/4945 Have a look also at Le Chatelier's principle

Let's do a simple generic example. We have a reaction:
$$\ce{2A <=>[k_1][k_{-1}] B}$$
Experimentally we determine that the rates of the forward and reverse reactions are:
$$\text{Rate}_1=k_1[\ce{A}]^2$$
$$\text{Rate}_{-1}=k_{-1}[\ce{B}]$$
At equilibrium:
$$\text{Rate}_1=\text{Rate}_{-1}$$
$$k_1[\ce{A}]^2=k_{-1}[\ce{B}]$$
Since the rate constants are constants we can combine them into the equilibrium constant to generate the law of mass action:
$$K_{eq}=\frac{k_1}{k_{-1}}=\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}$$
Now, if $k_1=2k_{-1}$, we have a value of the equilibrium constant, which gives us the relationship between $[\ce{A}]_{eq}$ and $[\ce{B}]_{eq}$ that defines equilibrium:
$$K_{eq}=\frac{k_1}{k_{-1}}=\frac{2k_{-1}}{k_{-1}}=2$$
$$K_{eq}=\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=2$$
$$2{[\ce{A}]_{eq}^2}=[\ce{B}]$$
In this case, only concentrations of $\ce{A}$ and $\ce{B}$ that satisfy $2{[\ce{A}]_{eq}^2}=[\ce{B}]$ constitute equilibrium. Other combinations of concentrations are not at equilibrium and therefore the forward and reverse rates are not equal.
Let's see how this works. An equilibrium situation is $[\ce{A}]=1\text{ M}$ and $[\ce{B}]=2\text{ M}$ because:
$$\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=\frac{2}{1^2}=2$$
The rates are:
$$\text{Rate}_1=k_1[\ce{A}]^2=2k_{-1}[\ce{A}]^2=2k_{-1}(1^2)=2k_{-1}$$
$$\text{Rate}_{-1}=k_{-1}[\ce{B}]=k_{-1}(2)=2k_{-1}$$
However, what if $[\ce{A}]=[\ce{B}]=2\text{ M}$?
$$\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=\frac{2}{2^2}=\frac{1}{2}\ne K_{eq}$$
$$\text{Rate}_1=k_1[\ce{A}]^2=2k_{-1}[\ce{A}]^2=2k_{-1}(2^2)=8k_{-1}$$
$$\text{Rate}_{-1}=k_{-1}[\ce{B}]=k_{-1}(2)=2k_{-1}$$
The concentrations, when plugged into the mass action expression, do not match the relationship defined by the equilibrium constant. More importantly, the rates are no longer equal. In this example, the rate of the forward reaction is now 4 times the rate of the reverse reaction. This is how equilibrium is reestablished. The forward reaction is going faster, but as it does so, $\ce{A}$ is consumed (slowing down the forward reaction) and $\ce{B}$ is produced (speeding up the reverse reaction) until the rates are equal again.
We can even do a bit of algebra to figure out what the concentrations of $\ce{A}$ and $\ce{B}$ will be once equilibrium is reestablished. Let's have $x$ represent the change in concentration of $\ce{B}$, so the change in the concentration of $\ce{A}$ is $-2x$.
$$K_{eq}=2=\frac{[\ce{B}]}{[\ce{A}]^2}=\frac{2+x}{(2-2x)^2}=\frac{2+x}{4-8x+4x^2}$$
$$2(4-8x+4x^2)=2+x$$
$$8-16x+8x^2=2+x$$
$$6-17x+8x^2=0$$
$$x=0.44695..., 1.6781...$$
In our case, we must use $x=0.44695$ since the other value would lead to a negative concentration for $\ce{A}$. Thus:
$$[\ce{A}]_{eq}=2.00-2(0.44695)=1.11$$
$$[\ce{B}]_{eq}=2.00+0.44695=2.45$$

so what you are saying is that no matter what the stressor is, Equilibrium will return to maintain the equilibrium constant, either by alleviating either side of the equation. So just to make it clear, the equilibrium constant is essentially just a rate of the products over the reactants? Am I right. Its not about the concentrations but the rates. So what happens when equilibrium shifts, does the rate change? So what happens is like the equilibrium constant changes and then returns back to it.

Very close. Perturbing an equilibrium situation does disrupt the rate balance. The equilibrium constant is the ratio of the rate constants. The equilibrium constant does not change. What does change is the reaction quotient. For more information about the difference between equilibrium constant and reaction quotient, see http://chemistry.stackexchange.com/questions/4977/le-chateliers-principle-how-can-the-equilibrium-shift-while-kc-remains-consta/4982#4982


text:
How to selectively convert benzene to p-nitrotoluene?

I converted benzene into toluene by Friedel-Crafts alkylation reaction.
Then by nitration I got two compounds o-nitrotoluene and p-nitrotoluene.

Is it possible to get only p-nitrotoluene?

I don't think there is a more effective (read: selective) way to  do this conversion. Maybe check pubchem for another route, they have quite some references there.

Solubility differences or other things

I'd doubt that exclusive nitration of toluene in para position is possible in homogenous solution.
But imagine to stuff a toluene molecule into a very narrow "tube". This tube will block any attack at C-2 (and C-3), while a nitrating agent may approach C-4.
Exactly this has been done in the vapour phase nitration of toluene with nitric acid over a zeolite (ZSM-5). See: Kalbasi, R.; Ghiaci, M.; Massah, A. Highly selective vapor phase nitration of toluene to 4-nitro toluene using modified and unmodified H3PO4/ZSM-5. Applied Catalysis A: General 2009, 353 (1), 1–8. DOI: 10.1016/j.apcata.2008.10.013.

I think It can be done. But it's a rather long process.First prepare nitrobenzene from benzene by nitration.Then, convert it to aniline by adding a reducing agent(preferably Sn/Hcl).Then, Convert the NH2 group to NHCOCH3 by adding ethanoic anhydride in acetic acid(Which is a weaker activating group than NH2 but a stronger one as compared to CH3).Then, Add the methyl group by friedel-crafts alkylation. Then, convert NHCOCH3 back to NH2 by hydrolysis.Then, convert aniline to diazonium chloride by adding NaNo2 in Hcl. Finally convert diazonium chloride to Nitro group by Adding NaNo2/Cu. U get p-Nitrotoluene. But, I am not too sure this is acceptable.

This suffers from the same problems of selectivity (in the friedel-crafts step) as the OP's synthesis but it is far longer so will be much less efficient.

I don't like the first process because practically we always have more than 1 molecule in a sample. Thus, After friedel crafts alkylation of benzene, After some time as the concentration of reactant decreases and that of product increases, Methyl Chloride may react with toluene(Since it is more activated than benzene)to give p-methyl toluene. Thus, p-nitrotoluene is not formed.

This is true but your synthesis suffers from the same problem in the alkylation step.

Your proposed Friedel-Crafts alkylation of aniline suffers from exactly the same problem...

I don't get it.. How come?? NH2 group has a stronger activating nature than CH3 group.Thus, CH3Cl should react with Aniline and not toluene(As Aniline is more activated than toluene).

F-C alkylation of aniline wouldn't  work at all, as amino group would react with catalyst.

Right. I messed that up. Have now edited the answer.

The product of your alkylation also has the amino group, do you think it just magically disappeared?

no, it does not, look carefully... before alkylation, NH2 is changed to NHCOCH3...after alkylation, NHCOCH3 is changed to NH2 only due to the problem pointed out by Mithoron


text:
Separating gold and silver from gold foil using hydrochloric acid and hydrogen peroxide
I used  $\ce{HCl}$ and $\ce{H2O2}$ to recover gold foils by dissolving all other metals. My solution I assume contains a large amount of silver. How do I drop the silver out of the  $\ce{HCl}$  and peroxide solution.)

Hot oxidizing acids, such as nitric acid od sulfuric acid, are the more canonical methods to separate gold from less rare metals.

My solution I assume contains a large amount of silver.

If so, shouldn't you have noticed a precipitate of silver chloride ($\ce{AgCl}$) under these conditions?

I would have used aqua regia but was trying to leave the foils solid. I should have used just nitric acid.I would hate to waste the dissolved metals in the HCL. Wanted to recover the silver and copper.Am I correct to assume if I added the nitric acid the silver chloride would participate.

Probably not since the silver is dissolved as $AgCl_2^-$.

authors: Αγγελική Καρακώστα


text:
Can a change in internal energy always be expressed as the product of the constant volume heat capacity and the change in temperature?
It makes sense to me conceptually that $dU=C_VdT$ but is that always the case because I have seen that result derived thus:
$$dU=U(V,T)=(\frac{\partial U}{\partial T})_VdT+(\frac{\partial U}{\partial V})_TdV$$
(Not the typical expression for internal energy $dU=TdS-pdV$
But for a perfect gas $(\frac{\partial U}{\partial V})_T=0$ so for a perfect gas internal energy is only a function of $T$: $U=U(T)$. Thus, it follows that;
$$dU=(\frac{\partial U}{\partial T})_VdT=C_VdT$$
However, is this only true for a perfect gas? I ask this because, the knowledge of this equality is required to show that entropy is a state function and I was wondering whether the proof that uses the above argument is only valid for a perfect gas or whether it's valid more generally?
I will show you the proof now:
$$dU=dq-pdV=C_VdT$$
$$dq=C_VdT+pdV$$
$$\frac{dq}{T}=\frac{C_V}{T}dT+\frac{nR}VdV$$ (using pV=nRT)
Then if you differentiate appropriately you can show it's an exact differential. However, is this only valid for a perfect gas?

The statement in the title question "internal energy [can] always be expressed as the product of the constant volume heat capacity and the change in temperature" would only apply to a calorically perfect gas.
If you instead said "an infinitesimal change in internal energy can be expressed as the product of the constant volume heat capacity and an infinitesimal change in temperature", that is only true for a thermally perfect gas.
For real gas, the heat capacities depend upon both temperature and pressure.

For a real gas (and for single phase liquids and solids of constant composition), the effect of temperature and molar volume on molar internal energy is given by:
$$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_T\right]dV$$where, in general, Cv is a function of both temperature and pressure.  Note that, for an ideal gas, the term in brackets is zero and Cv becomes dependent only on temperature.


text:
If hydrogen bonding in water was weaker, what happens to H+ ion concentration?

Water ionization becomes much less evident if the hydrogen bonds are just a few percent stronger but pure water contains considerably more $\ce{H+}$ ions if they are few percent weaker.

I found this line in some article. You can even copy/paste this line on search bar of Google. I didn't understood the latter part of this statement which is "but pure water contains considerably more $\ce{H+}$ ions if they are few percent weaker". The author is indicating to which of the following three situations?

only $\ce{H+}$ ion concentration will increase
both $\ce{H+}$ and $\ce{OH-}$ will increase
$\ce{H+}$ will increase and $\ce{OH-}$ will decrease that is the equilibrium constant remains same.

Sorry could you rephrase the section in the middle as it is very unclear and confusing. Also what do you mean by 'H='

The article by Martin Chaplin 'Water's Hydrogen Bond Strength' (arXiv:0706.1355 [cond-mat.soft], no evidence of peer review) is discussing how changes to the properties of water would alter life.
If the hydrogen bonding in water was stronger, there would be less ionization (less $\ce{H+}$ and less $\ce{OH-}$).
If the hydrogen bonding in water was weaker, there would be more ionization (more $\ce{H+}$ and more $\ce{OH-}$).


text:
Why is Gibbs free energy more useful than internal energy?
$$\mathrm{d}U<T\,\mathrm{d}S-p\,\mathrm{d}V=
\left(\frac{\partial U}{\partial S}\right)_{\!V}\,\mathrm{d}S
+\left(\frac{\partial U}{\partial V}\right)_{\!S}\,\mathrm{d}V$$
So for a change to occur at a constant entropy and volume, the internal energy must decrease. I get that this statement isn't useful because you can't control entropy in a lab. Furthermore, given that:
$$\mathrm{d}G<V\,\mathrm{d}p-S\,\mathrm{d}T=
\left(\frac{\partial G}{\partial p}\right)_{\!T}\,\mathrm{d}p
+\left(\frac{\partial G}{\partial T}\right)_{\!p}\,\mathrm{d}T$$
this shows that a spontaneous change at a constant pressure and temperature is accompanied by a decrease in Gibbs free energy – far more useful as these variables can be controlled.
However, since, for a spontaneous process, $\mathrm{d}U$ must decrease at a constant entropy and volume which is expressed by $\mathrm dU<T\,\mathrm dS-p\,\mathrm dV$. However, expressions for $\mathrm dS$ and $\mathrm dV$ can be expressed as a function of $p$ and $T$ making this inequality a function of pressure an temperature.
\begin{align}
\mathrm{d}U &<T\,\mathrm{d}S-p\,\mathrm{d}V\\
S&=S(p,T)\\
V&=V(p,T)\\
\end{align}
Sub in the total derivatives and factorize:
$$\mathrm{d}U<\left[T\left(\frac{\partial S}{\partial T}\right)_{\!p}
- p\left(\frac{\partial V}{\partial T}\right)_{\!p}\right]\,\mathrm{d}T
+ \left[T\left(\frac{\partial S}{\partial p}\right)_{\!T}
- p\left(\frac{\partial V}{\partial p}\right)_{\!T}\right]\,\mathrm{d}p$$
Thus, $U=U(p,T)$
This implies that for a spontaneous change at constant pressure and temperature the internal energy must decrease however this is not the case clearly. Does making the substitutions invalidate the inequality? Where is the fault in my reasoning?
So, what's the need for Gibbs free energy. Surely a spontaneous change at constant pressure and temperature is accompanied by a decrease in internal energy? There must be a fault in my reasoning, but where and why?

Thanks for the edit - I see now what you meant.
The short answer for why we "need" Gibbs energy is that the internal energy for a spontaneous process at constant temperature $T$ and pressure $p$ does not necessarily decrease. Internal energy is the thermodynamic potential (wiki link) for constant $S$ and $V$ - this makes sense if you think about it:
If the internal energy decreases, then heat must be released by the system, and/or work must be being done on the surroundings. However, real spontaneous processes happen where heat is absorbed and work is not done on the surroundings - the reason is that if the net entropy increase is greater than the enthalpy decrease, the process will still be spontaneous.
The equation for $\mathrm{d}U$ includes entropy terms, but not in a way that reflects spontaneity - the equation for Gibbs free energy does.
The other reason we "need" Gibbs energy is that it also tells us how much "thermodynamically useful" work we can get out of a process.
Mathematical Explanation:
The two inequalities that you started with only represent spontaneous process under very specific conditions:
$$\mathrm{d}U < 0$$
is true for spontaneous conditions only when $S$ and $V$ are constant. The reason for this is that we define spontaneity as the direction in which total entropy of the universe increases - it is at the root of our concept of "spontaneous" vs. "nonspontaneous."
For example, to arrive at this expression for $\mathrm{d}U$, we start with the Clausius inequality:
$$\mathrm{d}S - \frac{\mathrm{d}q}{T} \geq 0$$
This is just a way of saying that the net change of the entropy of the universe must be greater than or equal to zero for any process, and that the net entropy change of the universe is given by the change for the system ($\mathrm{d}S$) plus the change in surroundings, $-\frac{\mathrm{d}q}{T}$.
To turn this into something useful, we make a restriction: constant volume. This lets us assume no $pV$ work, and we can therefore write:
$$\mathrm{d}S - \frac{\mathrm{d}U}{T} \geq 0$$
and
$$T\,\mathrm{d}S \geq \mathrm{d}U$$
since $\mathrm{d}U = \mathrm{d}q$ at constant $V$.
If we then assume no entropy change for the system, we can also write:
$$\mathrm{d}U \leq 0$$
In other words, the restriction of constant $S$ and $V$ are built-in into the inequality. You can substitute in other thermodynamic variables to get $\mathrm{d}U$ in terms of $T$ and $p$, but it doesn't tell you anything about spontaneity under conditions other than constant $S$ and $V$.
Gibbs energy is useful because the inequality $\mathrm{d}G < 0$ is true for spontaneous processes under constant $T$ and $p$ conditions, which as you mentioned, are easy to achieve in a lab.
The other thermodynamic potentials are derived in similar ways - by holding different variables constant and seeing what happens when change in entropy is greater than zero.

What do you mean by "The equation for dU includes entropy terms, but not in a way that reflects spontaneity - the equation for Gibbs free energy does."? I have shown that the internal energy can be expressed in terms of pressure and temperature alone. Am I missing the point somewhere?

I think what is happening is that you are assuming that a spontaneous change will be accompanied by a decrease in internal energy - this is only true at constant S and V, not constant T and P. You have an equation for U(T,P), but it won't tell you anything about spontaneity - just how U changes with T and P. I'll update my answer with some math to see if it explains it more clearly.

Yes please, I am just unsure as to why my equation for U=U(T,p) does not help to indicate spontaneity.

I added a more mathematical explanation - I hope this clears it up for you. Also - I noticed a small error in your dU equation - dV in the second term should be dp.

@thomij how is it possible to get G from this? Or is it more complex?

I have managed to confuse myself again - I follow your argument and it makes sense but I can't disprove what I have said in the question. I have made edits to make it clearer as to what my confusion is. Please can you help.

The problem is that your inequality for dU < 0 only describes a spontaneous process when S and V are constant. In other words: your math and logic are fine, but your starting premise is incorrect. Or maybe another way to say it is: your equation for U(T,p) is right, but it doesn't mean what you think it means. You won't be able to disprove it, because it is a true statement - it's just a different true statement than you think it is.

"how is it possible to get G from this? Or is it more complex? " - I don't know of a straightforward way to get from U at constant S and V to G. You can probably do some substitutions to get a function of G, but at constant S and V it won't be useful. The standard approach is to use the Clausius inequality, then assume constant pressure. When you do this, the left hand side of the inequality becomes dH - TdS, which is how dG is defined.
If you are interested, Physical Chemistry by Peter Atkins has a great explanation for this (in chapter 4 in my very old 6th edition)

This is oversimplifying. Changes in Energy, enthalpy, and entropy, are based on the First Law and are the changes for a fixed amount of reaction, usually one equation unit, from a given state to completion. The Gibbs and Helmholtz free energies combine the opposing effects of energy change and entropy change to give the change between a given state and EQUILIBRIUM.


text:
Practical safety of storing potassium
Potassium ($\ce{K}$) seems to be the transitional alkali metal in terms of safety. Lighter than $\ce{K}$ and we can store under oil with no worries. Heavier than $\ce{K}$ and only ampules will work. $\ce{K}$ is such that even under oil it can absorb enough oxygen to form superoxides that are contact explosives. But this is the theory: does anyone have any experience with extended proper storage of potassium?
I ask for two reasons. First, there are lots of online videos showing $\ce{K}$ under oil that the scientist then proceeds to cut, smash, and otherwise manhandle with no fear of detonation. Second, my $\ce{K}$ arrived tarnished, which I have stored under oil, and then vacuumed sealed in a bag, which has shown zero change in visual oxidation. So while I get that leaving an open jar of $\ce{K}$ under oil is problematic, I'm less clear on well stored $\ce{K}$.
So I'd appreciate any comments from practitioners that use $\ce{K}$ regularly. What situations are genuine dangers? Thanks!

@venture: Please undelete your answer. There is nothing wrong with it.

You should have sufficient reputation to cast an undelete vote. If I remember correctly, it needs 3, mine is already on it.

I actually don't have enough reputation--need 4k.

I have never worked with potassium personally, but have experience with other alkali metals.   You are spot on about potassium superoxide.  I have direct knowledge of an explosion that occurred when potassium metal was stored for an extended period of time in mineral oil.  I would therefore be extremely cautious if you intend to store it in this manner.  Factors such as length of storage time, permeability of your container seals and storage temperature can all affect the amount of superoxide that forms.  Based on the evidence I have seen, it is not safe to assume that there will be no transport of oxygen through a mineral oil barrier.  The only way I would work with this metal is in an inert glovebox, period.  That is my own preference, of course, and I always tend to err on the side of caution.

Right, all of this is what I know. And yet, every demonstration I can document seems to only be storing under oil, whereas I am doing that and using inert gas and then sealing the bottle in a vacuum bag (poor man's glove box). So it either suggests that poorly stored $\ce{K}$ is a bit more resistant to manhandling that we might think, or that lots of what we see on YouTube is rather freshly prepared. The source I bought my $\ce{K}$ from said they had stored for an extended period with no ill effects, which suggests 1). But what you say reads true.

I would suspect that the risk increases with the amount of superoxide that forms on the metal surface and at some point reaches a critical level that, if ignited, imparts sufficient energy to start the metallic substrate to ignite. Cutting could be a trigger, and I've read accounts of self ignition. I have personally worked with $K$ but have never experienced autoignition or explosions.

The following excerpt regarding the management of alkali metals is from:
Management of time-sensitive chemicals (I): Misconceptions leading to incidents by Jim Bailey, David Blair, Lydia Boada-Clista, Dan Marsick, David Quigley, Fred Simmons and Helena Whyte
"Slow reactions of alkali metals and their alloys with oxygen to form oxides and superoxides have been well documented*. Even when these metals are stored under mineral oil, oxygen can dissolve in the mineral oil and react. Potassium, under these circumstances, forms yellow or orange coating that can explode or catch fire upon cutting. NaK, a eutectic mixture of potassium and sodium, can undergo a similar reaction. Lithium stored under dry nitrogen can react with the nitrogen to form a nitride. Formation of the nitride is autocatalytic and can eventually lead to autoignition of the lithium. Fires of alkali metals are extremely dangerous and difficult to extinguish."

Management of time-sensitive chemicals (II): Their identification, chemistry and management,Jim Bailey,David Blair,Lydia Boada-Clista,Dan Marsick,David Quigley,Fred Simmons,Helena Whyte: Chemical Health and Safety, Volume 11, Issue 6, November–December 2004, Pages 17–24

I've come to the realization that if you plan to keep chemicals around you need to know all the risks and provide effective management including a plan for periodic inspection.

I have used potassium metal in glass containers stored in mineral oil for 30 years.  You are correct that there is residual oxygen in mineral oil that will "rust (oxidize)" the metal, but I have never seen the yellow/orange superoxide layer that some have described.  If you need it unreacted, then it needs to be stored so as to not come in contact with oxygen. If you are using it to remove residual oxygen from a glove box, then you probably don't want to even open it until you get ready to use it and then open it in the glove box.  If you are using it to demonstrate periodic reactivity, then storing it in mineral oil in a sealed glass container has worked well for me over hundreds of demos for three decades with no issues, even with potassium that has been stored this way for three or more years.


text:
What causes electrons to move from zinc to copper?
Here is something that no video in youtube about electrochemistry can explain me about galvanic cells.
Suppose we have $\ce{Zn}$ metal immerse in $\ce{Zn^{2+}}$ in one side,  then $\ce{Cu}$ metal immerse in $\ce{Cu^{2+}}$ in the other side. Then, if we link the $\ce{Zn}$ metal to $\ce{Cu}$ metal with some conductor wire, electrons will flow from the $\ce{Zn}$ metal to the copper wire, but this will stop quickly without a salt bridge.
My question is why do the electrons move from $\ce{Zn}$ metal through the wire to the $\ce{Cu}$ metal in the first place? What causes that?

possible duplicate of Why is it important to use a salt bridge in a voltaic cell? can a wire be used?

Your question is answered quite well in that duplicate question, but I should add that Zn should ideally never be in contact with Cu2+ in a useful cell (i.e. Cu2+ should not cross the salt bridge) because the Cu2+ will be directly reduced at the Zn electrode instead of the electrons passing through the external circuit.

I had read that answer. The problem is that it starts with "The electrons flow from the anode to the cathode. So, the oxidation ...", but it doesn't answer why the electrons flow from the anode to the cathode with the salt bridge, but not withouth it.

As it says, "The electrons move thrown the wire (and your device, which I haven't included in the diagram), leave unbalanced positive charge in this vessel." So it will work vanishingly briefly, but because one electrode is releasing cations into solution and the other is removing cations from solution, a charge imbalance arises in each half-cell that opposes further electron transfer. Redox reactions will occur until this imbalance matches the cell voltage. (it doesn't take much—I don't know if it's even possible to measure it)

This is also pretty much the same thing.

Yes, but why will the electrons move thrown the wire in the first place ?   It doesnt answer it.

Because it's more energetically favourable for the zinc electrode to be reduced, basically. Another relevant link. Spontaneous chemical reactions are driven by moving a system from a higher energy state to a lower energy state. You're probably familiar with this as ∆G's for reactions at constant pressure and redox reactions are the same, only expressed as a voltage: ∆G=-nFE because voltages are more convenient when dealing with redox reactions.

Hmm. Let's forget about the Zn solution , the Cu solution and the salt bridge. If we simply take Zn metal and connect it to Cu metal through a wire, i think there will be a voltage potential difference between them generated ( like LDC3 mentioned )  . Correct ? So, eletrons will flow from the Zn to the Cu ( even if this flow will stop very briefly ) ?

Yes, it's known as the Volta potential, though it's pretty complicated stuff if you haven't studied materials chemistry.

Omg, it's really clear now ! I only have two last questions . What is happening in the salt bridge after the initial flow of electrons is that the cations of the salt are being attracted to the Cu2+ solution with extra electrons and are being neutralized by it ( receiving their extra electrons ) , and the anions are being attracted to the Zn2+ solution with lack electrons and also being neutralized by it ( giving their extra electrons to the Zn2+ solution ) ?

No redox reactions happen with the salts in the salt bridge, the ions just migrate to maintain charge balance. The side where the reduction occurs has more negative ions (since the cations are being reduced), so cations migrate from the salt bridge to replace them. Likewise, the other side has more positive ions since the electrode is oxidizing to produce cations, so anions from the salt bridge migrate to keep balance. Nothing is being "neutralized" as such. The only place where electron transfer occurs is at the electrode-solution interfaces.

I see. So, i'm guessing that the ammount of extra electrons or missing electrons in the solutions of Zn2+ and Cu2+ are directly related to the Fermi level of the Zn metal and the Cu metal, respectively. If we don't have a salt bridge, the Fermi level of the metals will become equal soon enough and the flow of current will stop.

The Fermi levels are always equal because the two electrodes are in equilibrium (they're connected to each other). Probably should have done this in chat, but last one: the charge imbalance in a galvanic cell is in the solution, not the electrodes. If we pretend the reduction half-cell has 100 Cu^2+ and 100 SO4^2-, if we reduce one copper ion to neutral Cu, there is one extra SO4^2- left in the solution, so it has a net negative charge. This charge makes it harder to reduce a second Cu^2+, eventually stopping the reduction, if a couple of K+ are added from the salt bridge, the charge goes away

What i mean is that if we don't have a salt bridge, the fermi level of the Zn will become equal to the fermi level of the Cu soon enough, so current will stop flowing .  Anyways, you were very very very helpful.   Thanks a bunch from a guy really newbie in Chemistry .

There is no reaction since the electrons will only flow one way in the wire. When you connect the zinc metal to the copper metal with a wire, there is a voltage potential generated between them, just like in a thermocouple.
From Wikipedia:

Any junction of dissimilar metals will produce an electric potential related to temperature.

In this case, we don't care about the temperature effects.
When a salt bridge is placed between the solution, then the electrons are carried by the anions in the solution from one electrode to the other electrode.
So the zinc metal releases electrons when it dissolves, which travel through the wire easily, to the copper. The $\ce {Cu^{2+}}$ in solution grabs the electrons from the copper electrode, adding copper metal to the electrode. Since there is now more negative ions in this solution, the anions move across the bridge to the zinc electrode. The anions attack the zinc metal, causing the zinc to loose electrons when it dissolves.
I'm not sure what starts the reaction, nor why the zinc dissolves easily.

I think you just mentioned what really starts the reaction.  Whenever we connect two dissimilar metals with wire, electrons will flow from the least electronegative one to the most electronegative one.   Correct ?

@nerdy There is a potential, the electrons won't flow unless there is a circuit for them. In thermocouples, the metals are usually fused together and the potential from this junction is monitored.

Also, i didn't understand 100%. The potential is based on the temperature difference ? So, if they are with the same temperature, no potential difference betweem them is guaranteed ?

@nerdy No, when using thermocouples to measure temperature, you have 2 thermocouples, one where you know what the temperature is. You measure the difference in activity to determine the temperature since it is nearly linear. There is always a potential difference until you cool it way down. For example, if there is a potential of 1.27V at 30°C and you know that the potential changes 0.07V for each degree, what temperature would the junction need to be at to give a potential of 2.75V?

I see. There is a potential between the Zn metal and the Cu metal . So, if we connect them through a wire, electrons will flow from Zn metal to the Cu metal, right ?       What do you mean by a circuit ? Isn't a wire enough ?

@nerdy Because of the potential, the electrons can only flow one way. Zinc will give up electrons, but it must go into solution. Without a salt bridge, the solution get positively charged and will stop the reaction. The salt bridge allows anions to balance the charge and to take electrons away from the copper electrode.

Could you say that electrons start moving from zinc to copper because copper has higher electronegativity than zinc?

Electron move from zinc to copper as zinc is more reactive metal than copper and thus more readily loses the electron to turn into zn2+ ion . Thus electrons move from zinc to copper


text:
Is there an authoritative list of chemical reactions/transformations?
Is there somewhere where I can find an authoritative list of chemical reactions, energy requirements, known catalysts, etc.? If not, how do you know if and how to transform substance A to substance B? What references do you use?

Welcome to Chemistry.se! Find out more about this site and take the [tour] and further information can be obtained in the [help]. I retagged your question to fit it into a wider range - the books tag is for literature in general, so please don't be confused.

Well, you are asking about whole several branches of chermistry - organic/inorganic synthesis/technology - more like a liberary than list, but what's google for :) - you can always try searching on internet


text:
Can I change glass transition temperature (Tg) and impact resistance with additives?
I have three different HDPE polymers with different melting index but they don't have any elasticity. Physical form of polymers is granules. Can I change glass transition temperature (Tg) and impact resistance with additives?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

No. Buy a suitable Polyethylene.

The glass transition temperature depends on the cooling rate and molecular weight distribution and could be influenced by additives:
Most liquid plasticizers are low molecular weight organic materials with a glass transition temperature in the range of $125-225^0\text{C}$. When this kind of additives is added to an organic polymer, the result is a weight averaging of the glass transition temperature between that of the polymer and that of the plasticiser if they are miscible. The reduction of the the glass transition temperature of a polymer by the plasticiser is usually about   $5^0\text{C}$ per weight percent of plasticiser.
Impact modifiers ( a kind of additives) can affect the impact resistance of a polymer. Impact modifiers are key additives for
increasing flexibility and impact strength
to meet physical property requirements of
rigid polymers. All impact modifiers are elastomeric or
rubbery in nature, with a lower modulus
than the host polymer. The dispersed
rubber phase acts to absorb or dissipate the
energy of impact in order to stop craze or
crack propagation.

Plasticisers don't work in PE, because they do not dissolve. You could try blending with some low Mw LDPE.


text:
Why was atomic mass scale changed from Oxygen - 16 to Carbon - 12?
Why was unified atomic mass scale introduced and why was Oxygen - 16 replaced by carbon - 12 for standardizing atomic scale?

The mass scale has changed over time, largely due to different isotopes of the "baseline." Not surprisingly, there's a good Wikipedia article on the matter.

In the 20th century, until the 1960s chemists and physicists used two different atomic-mass scales. The chemists used a "atomic mass unit" (amu) scale such that the natural mixture of oxygen isotopes had an atomic mass 16, while the physicists assigned the same number 16 to only the atomic mass of the most common oxygen isotope (O-16, containing eight protons and eight neutrons). However, because oxygen-17 and oxygen-18 are also present in natural oxygen this led to two different tables of atomic mass. The unified scale based on carbon-12, $\ce{^12C}$, met the physicists' need to base the scale on a pure isotope, while being numerically close to the chemists' scale.

In short, Dalton suggested $\ce{^1H}$ as the basis of the mass scale, but Ostwald pushed later for $\ce{^16O}$. Unfortunately, no one knew about isotopes yet.
The problem was that physics stuck to isotopically pure O-16 for their mass scale, and chemistry used "oxygen" (i.e., with some O-17 and O-18 present). There's 0.028% difference between those scales. That may not sound like much, but with molecules of ~200-500 amu, it starts to add up.
Using C-12, it's isotopically pure (making physics happy) very close to the previous chemistry "O-16 + O-17 + O-18" scale.
From IUPAC:

In April 1957 at the bar in the Hotel Krasnapolski in Amsterdam, Nier suggested to Mattauch that the $\ce{^12C}$ = 12 mass scale be adopted because of carbon's use as a secondary standard in mass spectrometry. Also, $\ce{^12C}$ = 12 implied acceptable relative changes in the atomic weight scale, i.e., 42 parts-per-million (ppm) compared to 275 ppm for the $\ce{^16O}$ = 16 scale (which would not acceptable to chemists). Enthusiastically, Mattauch made a worldwide effort in the late 1950s to publicize the $\ce{^12C}$ = 12 scale and obtain the physicist's approval, while Wichers obtained the chemist's approval.

And since it was neither side's original choice it neatly sidesteps the whole "we won and you lost" issue.

Not sure what I'm misreading, but I don't get why carbon is more "isotopically pure" versus oxygen... C-13 is 1.1% versus O-18 at 0.2%.

Carbon per se isn't isotopically pure. The new standard is C-12, not "carbon."

Once isotopes were understood, the best compromise was to use isotopically pure C-12, since it minimized differences in masses, particularly in chemistry.

The atomic masses were determined by mass spectrometry Each nuclide is essentially isotopically pure. The main reason was to not obsolete all the chemical data in the literatures. When the change was implemented, there were great advances being made in mass spectrometers and the physicists were happily remeasuring atomic masses and coud just as easily compare to C-12 as to any other nuclide

The changes in the physical scale were large, the chemical scale almost zero.

@jimchmst - I think you should turn your comments into a separate answer since this is a pretty active question.


text:
Why do simple molecular substances have weak intermolecular forces and why do giant covalent substances have strong intermolecular forces?
Why do simple molecular substances have weak intermolecular forces and why do giant covalent substances have strong intermolecular forces?
I understand that in simple molecular substances the atoms within a molecule are held together by strong covalent bonds, but the intermolecular forces between molecules are weak.
But I don't see what makes giant covalent substances have stronger intermolecular forces? What makes them stronger and have higher melting points?
I thought I understood but I think I must have gone wrong somewhere and now I am very confused. Help would be much appreciated. Also please explain simply, chemistry does not come naturally to me so I might get even more confused.

Are you sure you're not comparing apples to oranges? Ask yourself this: in a giant covalent substance, what does an intermolecular force mean? Is there more than one molecule to interact with one another? Think about the differences between intermolecular and intramolecular forces.

Are you defining "giant covalent substances" as larger molecules, such as octane in comparison to methane? Or using the phrase to mean large covalent networks, such as diamond?

In simple molecules the intermolecular forces are, in order of increasing strength, dispersion forces, permanent dipole interactions and hydrogen bonding (which has significant covalent character but is generally considered to be an intermolecular force).
By contrast giant covalent repeating structures such as diamond and $\ce{SiO2}$ are not molecular in the same sense as they can theoretically be infinitely large. Therefore they do not really have intermolecular forces but they are simply held together by covalent bonds between the atoms in the structure.

So giant covalent structures don't have intermolecular forces because they aren't molecules?

@francesca it depends on how you define molecules but very large strcutures like diamond, graphite, $\ce{SiO2}$ etc. are so large that intermolecular forces are not really relevant because there may only be one (or a few) 'molecules'

You are forgetting the vast number [>10^23] of atoms involved a g.at.wt. of diamond or graphene possibly has 10^10 to 10^15 "molecules". Just a guess.

Molecules are made of fixed numbers of atoms joined together by covalent bonds, and can range from the very small (even down to single atoms, as in the noble gases) to the very large (as in polymers, proteins or even DNA).
The covalent bonds holding the molecules together are very strong, but these are largely irrelevant to the physical properties of the substance. Physical properties (melting points, boiling points, solubility in water,...) are governed by the intermolecular forces - forces attracting one molecule to its neighbors - van der Waals attractions or hydrogen bonds.
Molecular substances tend to be gases, liquids or low melting point solids, because the intermolecular forces of attraction are comparatively weak. You don't have to break any covalent bonds in order to melt or boil a molecular substance.
The value of the melting or boiling point will depend on the strength of the intermolecular forces. The presence of hydrogen bonding will lift the melting and boiling points. The larger the molecule the more van der Waals attractions are possible - and those will also need more energy to break.

Two factors matter: the nature of the bonds and the scale of the interactions
The simple case is that, for many giant covalent substances, all the bonds are strong covalent bonds. Diamond, where all the carbons in the substance are connected to each other by strong covalent bonds, is a good example.
The more complex case relates to how strong the weak interactions between individual molecules (van der Waals forces) are in molecules of different size. Iodine, a substance containing small molecules containing two iodine atoms, is volatile because it has weak forces between the molecules. But polyethylene, a polymer is a fairly solid and non volatile substance, contains long strands of alkane-like chains held together by the same forces.
But the strength of those weak forces between the components making up the substance, depend on the surface area of the components. Iodine molecules have a small surface area, so the forces are weak; polyethylene chains have a large surface area, so the attractive forces are strong. This is also why geckos can walk up walls: same forces at play but the structure of their fingers and toes creates huge surface area maximising the overall strength of the weak interactions.
The surface area also explains why the shape of the molecule matters. Compact molecules have smaller forces than their isomers of the same size that are less compact. So the compact 2,2 dimethyl butane (sometimes called neopentane, C(CH3)4) is far more volatile than its isomer pentane, a 5-carbon long chain.

Far more volatile? maybe more volatile.  Same mass not the same size.

In covalent molecules, there’s a theory that the bigger the molecule, the higher the melting point and boiling point.
However, the main reason why giant covalent structures have high melting and boiling point is that it is the strong covalent bond that must be overcome in order to melt or boil the giant covalent structure.
Let me use graphite as an example, although there is weak intermolecular bonds between the layers of carbon atom (which is actually graphene). However, breaking the weak intermolecular bonds between layers of carbon atom will only allow layers of carbon atom to slide over each other, which make graphite slippery and soft. In order to melt or boil graphite, you have to break the strong covalent bonds.
In contrast, for simple molecular structures, the reason why it have low melting and boiling point is because it is the weak intermolecular forces that must be overcome in order to melt or boil it.
I understand it’s so complicated, but as long as you review it for a few times, you will remember it.

Is it a theory or a simple observation? Evaporation and decomposition are not the same. Evaporation returns the same molecular composition on condensation. Decomposition in elements seems to do the same but the vapor composition is not molecularly the same as the condensed phase. That most elements have only one common allotrope it seems that they might be evaporating. If allotropes are possible they can be interconverted. Taking carbon as the example decompose graphene recompose diamond.


text:
How to prove that entropy is a state function?
Consider the following steps:
\begin{align}
\mathrm{d}U &= \mathrm{d}q - p\,\mathrm{d}V\\
\mathrm{d}U &= C_V\,\mathrm{d}T\tag1\\
\end{align}
Side question: Is equation $(1)$ only true for a perfect gas or all substances?
\begin{align}
C_V\mathrm{d}T &= \mathrm{d}q - p\,\mathrm{d}V\\
\mathrm{d}q &= C_V\,\mathrm{d}T + p\,\mathrm{d}V\\
\end{align}
Divide by T:
$$\mathrm{d}S = \frac{C_V}{T}\,\mathrm{d}T + \frac pT\,\mathrm{d}V$$
The proof requires a substitution of $\frac pT=\frac{nR}V$ because when it is then differentiated with respect to $T$ it equates to zero and so does $\frac{C_V}{T}$ when it is differentiated with respect to $V$ which shows it is an exact differential. However, if you do not make the substitution you do not get zero so the two differentials do not equal each other. This would suggest that entropy is not a state function. Am I going wrong somewhere?

$dU=C_VdT$ is true only for a thermally perfect gas. The corresponding equation for a real gas is given here: http://en.wikipedia.org/wiki/Internal_energy#Changes_due_to_temperature_and_volume

Is it valid for solids?

the full equation is valid for all materials

Ok thanks. Do you have any ideas about the second part to the question?

sorry Rob I don't have time right now, but see: http://pubs.acs.org/doi/pdf/10.1021/ed063p846 is you can

I can't seem to be able to access that document but thanks anyway - you've helped me a lot with many questions.

I don't understand what you mean by saying "if you don't make the substitution you do not get zero".  The relevant point is that, for ideal gases, $$p = nRT/V,$$ so $p/T$ is constant with respect to $T$.  Therefore $$\frac \partial{\partial T}\left(\frac pT\right)=0$$ for an ideal gas, whether or not you use the substitution to simplify the process.
If you don't use the substitution, you need to use the quotient rule to do the partial derivative properly, and you should still end up getting zero if you do it correctly.  Could that be where your confusion is coming from?

I see, thank you very much. I didn't spot that you'd need to use the quotient rule.


text:
Why is dG = dH − TdS?
I get that $G=H-TS$ because then: $$\begin{align}\mathrm dG&=\mathrm dH-T\,\mathrm dS-S\mathrm dT\\&=T\,\mathrm dS+V\,\mathrm dp-T\,\mathrm dS-S\,\mathrm dT\end{align}$$
Therefore, by cancelling: $\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$ which is the equations for $\mathrm dG$. However, I can’t get this result from using $\mathrm dG=\mathrm dH-T\,\mathrm dS$.
$$\mathrm dH=S\,\mathrm dT+V\,\mathrm dp$$
$$\mathrm dG=\left(S\,\mathrm dT+V\,\mathrm dp\right)-T\,\mathrm dS$$
This does not give me the right equation. Where am I going wrong?

Why is dG=dH-TdS?

It isn't. That would only be true at constant temperature. In general:
$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S - S\,\mathrm{d}T$$

If temperature is the same but not constant (initial and final state have same temperature) can we say $ΔG=ΔH-TΔS$?

If you write $\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S$, you drop off one term of the differentiation of $G$ which is $S\mathrm{d}T$.
By definition $G = H - TS$. So, $$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S-S\,\mathrm{d}T$$
Then, you should write $\mathrm{d}H = T\,\mathrm{d}S + V\,\mathrm{d}p$ (not as you have written $\mathrm{d}H = S\,\mathrm{d}T + V\,\mathrm{d}p$ ).
Finally, $$\mathrm{d}G = T\,\mathrm{d}S + V\,\mathrm{d}p - T\,\mathrm{d}S - S\,\mathrm{d}T$$
 $$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T$$


text:
how to use di-tert-butyl dicarbonate (diBOc) as a solvent?
how should we use di-tert-butyl dicarbonate(diBOc) as solvent?
In reference, diBOc was in the form of a liquid but when I checked the bottle it was in solid state.
diBOc should be stored at 8-10 ℃. So how is it possible for it to be used as liquid?
Is it OK to melt it at RT? Wouldn't melting, make diBOc have structural differences?

Di-tert-butyl dicarbonate melts around room temperature (m.p.=23°C).  It does not decompose at this or even slightly higher temperatures.  For example, it is typically purified by distillation under reduced pressure at temperatures up to around 65°C.  At higher temperatures it will decompose to isobutene, t-butyl alcohol and carbon dioxide.

In any case it is OK to melt di-tert-butyl dicarbonate and use it between room temperature and up to around 40-60°C.  Just make sure that when you open the bottle and use it that you do so in a dry environment as water will decompose di-tert-butyl dicarbonate.


text:
Is reducing nitrogen oxides with ammonia practical in a gasifier?
Sorry if this is an outright stupid question. I am no chemist.
For nitric oxide:
$$
\ce{2NH3 + 2NO -> 2 N2 + H2 + 2H2O}
$$
For nitrogen dioxide:
$$
\ce{2NH3 + 2NO2 -> 2N2 + H2 + 2H2O + O2}
$$
or
$$
\ce{4NH3 + 2NO2 -> 3N2 + 2H2 + 4H2O}
$$
As you can see, I'm trying to find if it is practical to take two toxins and convert them into non-volatile substances. Also, of course, harvesting this hydrogen gas would also be preferred; you could just add it back into the resulting Syngas. And I suppose ammonia doesn't have to be used. It just seemed convenient and the above reactions (producing useful Hydrogen and other non-volatiles) somehow came into my mind.
And it seems for every $\ce{O}$ (or $\ce{N}$) in the $\ce{NO_{x}}$ it can make another $\ce{O2}$ (or $\ce{N2}$, respectively). Or double the ammonia, to effectively double up on the first reaction.
If its the case that one reaction is more practical than another (as it usually is). I have noticed that conversion is simple:
$$
\ce{2NO + O2 -> 2NO2}
$$
readily happens in air and at 150C it reverses:
$$
\ce{2NO2 -> 2NO + O2}
$$
Most of this leads to gasification. Gasifiers stratify the combustion process to extract producer gas (Called Syngas. It is $\ce{H2 + CO}$, and it burns clean.). If a gasifier is built a certain way, the only pollutants could be $\ce{NO_{x}}$, and since some gasifiers can use biowaste, then ammonia is not hard to reconstitute on-site in some applications. Plus, I think it might be possible to extract ammonia directly from the pyrolysis stage when using garbage in certain gasifiers. As well, I think Ammonia is in the resulting trace gases sometimes depending on the gasifier/fuel setup.
There are some resources to be had. Gasifiers also have to typically filter out sulfur and mercury from the resulting Syngas. If those can help at all here and/or could also be included in this detoxification. As well, there are usually sources of heat (up to 1100C), tar, char (hot carbon), and hot or cold ash (burnt charcoal), $\ce{H2 + CO}$, $\ce{CO2}$, and other trace gases like $\ce{CH4}$ or whatever, depending on the gasifier and fuel. These resources could be utilized if needed, but most already have a role to play. Carbon dioxide, methane and tar are used to make more Syngas. The wasted byproducts are: sulfur, mercury, cold ash, and the "whatever" trace gases. I believe sulfur is in the form hydrogen sulfide. There is also probably substantial unsued hot Char (hot reactive carbon), but it burns quick, and only exists in the heart of the machine, then once all the heat is extracted it turns to ash.
Actually, I wonder if I could pump the $\ce{NO_{x}}$ (or other pollutants) back into the pyrolysis stage, if they would might reform into something else. Less toxic, eventually.
Of course, I feel like I'm barking up the wrong tree. As $\ce{NO_{x}}$ gases are closely scrutinized by municipalities, any practical reduction would probably be used already and emission of $\ce{NO_{x}}$ gases into the environment wouldn't be a problem, but that is assuming way too much for humans. So, I have to ask:
Do any chemists see a way, using these resources (or other practical external ones), to detoxify the nitrogen oxide from the resulting burnt Syngas fumes?

use of ammonia to for NOx reduction is in widespread practice http://en.wikipedia.org/wiki/Selective_catalytic_reduction

Oh wow! Darn I googled left and right, off and on, for a few days and never came up with that link. lol! At least I see that it's not barking up the wrong tree, specifically. Now hows about the practical way to use it here? I guess I got a lot more reading up ahead...

Also, SCR uses a different reaction, and doesn't produce Hydrogen gas. With temperatures up to 1100C and some other wasted byproducts at our disposal, I wonder if Hydrogen could be made at this point instead.

Please tell me how this question is unclear? I find it quite clear and the graceful edits help to attune this question.  Also, I would think an unclear question would have a negative vote score... not positive. Even DavePhD seemed to understand my question. I even give very specific formulas that I would like to achieve. But I do not know if they are possible.


text:
Should the volume of a solute be factored into the volume of a solvent when preparing a substance of a certain molarity?
Upon being asked to prepare a solution of a given molarity from a solution with high concentration I was confused whether or not I should factor in the volume taken from the high concentration substance into the volume of the solvent. I think that an example could illustrate what I am asking more appropriately. Lets say I have to prepare a 5 M substance in 50 ml of water. The substance is in a container that is 10 M so I must calculate the number of moles and then the amount of solute to extract from the 10 M container.  $$M = {m\over{L}} $$ $$5 M = {m\over{.05}} $$ $$moles = .25$$ given that the solution's concentration is 10 M $$10M={.25\over{L}}$$ $$L = .025$$ Now in order to achieve the 5 M in the 50 ml of water, do I pour 25 ml of water and then the 25 ml, which then adds to 50 ml? or do I pour the 50 ml of water and then the 25ml resulting in a total volume of 75 ml?

The only volume that is important in the denominator of your equation is the total volume of solution; neither the volume of solute or solvent are explicitly relevant.
Pragmatically, there is not likely to be much of a volume change when taking a concentrated solution and diluting it.
So with reference to your question, you should add enough solvent to get the proper final volume.  Explicitly, add the 25 mL of stock solution and 'dilute to mark'-- that is add enough solvent so the total solution is 50 mL.  The added solvent might not be exactly 25 mL to make this final volume.

You only add the $\ce{25 mL}$ water. After all you want to dilute to a certain molarity. You start from a 10 M solution and want to go to a 5M solution. Neglecting the volume of the dissolved compound you simply add another equivalent of solvent to dilute by a factor 2, i.e. $\ce{25 mL}$ water in this case.

This answer is explicitly untrue, but may be pragmatically true.  The volume of solute may be negligible, but if it isn't this approach will lead to improper concentration.

If you want the final concentration to be any more precise than "about 5M," this method is definitely incorrect. This is why you fill to volume using a volumetric flask (as @Lighthart wrote in their answer). The total volume is all that matters here, not the volume of solvent. Never assume the volume of solute is negligible.


text:
Do molecules with carboxyl groups have properties of carboxyl's sub-functional-groups?
We're learning about functional groups of different organic molecules, and there are some I'm confused about: carboxyl, carbonyl, and hydroxyl. I was looking online but I couldn't find a direct answer
I know that the carbonyl group is a "-CO", the hydroxyl group is a "-OH", and the carboxyl group is a "-COOH".
We learned that carbonyl groups make molecules a ketone or aldehyde, the hydroxyl group makes molecules alcohols, and the carboxyl group makes molecules carboxylic acids.
Knowing that that a carbonyl group is just a carbon double-bonded to an oxygen, and a carboxyl group is a carbonyl group bonded to a hydroxyl group, then should carboxyl molecules also have all the properties of hydroxyl and carbonyl groups (such as being an alcohol), on top of its new ones? Or, does it only have the properties associated with themselves?
Thanks.

Carboxyl groups do share some properties with hydroxy and carbonyl groups such as the ability to engage in hydrogen bonding but in terms of chemical reactivity they are very different.
The reason for this lies in the structure of the carboxyl group. Instead of the carbonyl and the hydroxyl groups being completely independent of each other, there is an orbital interaction between the three $p$ orbitals on the carbon and the two oxygens which creates a $\pi$ system across the three atoms that tends to stabilise the group, particularly when it is deprotonated. (I couldn't find any decent pictures of these $\pi$ orbitals but if anyone can please edit them in)
The most obvious property that carboxyl groups have which hydroxyl and carbonyl groups do not display (hydroxyls do slightly but nowehere near as much as carboxyls) is acidity. This high acidity is caused by the fact that the carboxylate ion which forms when carboxyl groups donate a proton is stabilised by the electron delocalisation $\pi$ system. This can also be expressed using resonance structures as shown below.

The strength of the acidity can be seen in the $pK_a$ values of carboxylic acids compared to alcohols.$$Ethanoic~acid~pK_a = 4.74$$$$Ethanol~pK_a = 16$$
Remebering that $pK_a$ has a logarithmic scale we can see that carboxylic acids are over $10^{10}$ times more acidic than their respective alcohols.
An excellent introduction to many other properties and reactions of carboxylic acids can be found here: https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid1.htm


text:
Chemical Composition Changes in Liquids
Water is $\ce{H2O}$ in all it's states: solid(ice), liquid, and gas(water vapor).
In your drink, when you add artificial flavoring and coloring, does it change the chemical composition of the water? What makes the flavor and color change?
Are you still drinking $\ce{H2O}$, with dense amounts of another liquid, or do they blend together?

Not really sure what you are after here, but maybe this will help ... artificial flavors and colors are generally present at very low concentrations, dissolved in the water.  "Dissolved" means that the individual additive molecules are completely surrounded by water molecules, which are interacting with it to stabilize it in its dissolved form.  Since their concentration is low, each such additive-water aggregate is pretty far away from any others, so what you have is not very different from water, in a chemical sense at least.  Note also that the water molecules around a given additive molecule are not static .. they are constantly exchanging with the "bulk" pure water solvent around them.
In some cases, an additive may not be very water-soluble ... in such a case a surfactant solubilizer molecule may be added to help the additives dissolve in the water.  The surfactant molecules have one end that likes to interact with the insoluble additives (the hydrophobic end), and one end that likes to interact with water (the hydrophilic end).  They aggregate themselves into larger spherical structures called micelles, where the hydrophobic parts of the molecules are protected on the inside (along with the additive molecules), and the hydrophilic parts are on the outside, exposed to the water.  These micellar aggregates can be large enough to scatter light, and so drinks with these additive-solubilizer combinations will often appear cloudy.  An example of this is the citrus flavoring in sodas like Mountain Dew and Squirt ... in those cases, brominated vegetable oil (BVO) is added as a surfactant to dissolve the citrus flavor molecules.
(Aside: BVO is not exactly good for you, and several soft drink manufacturers have said that they will eliminate it from their formulations, which means maybe I can start drinking Fresca again)


text:
How to accurately calculate the electromotive force for various conditions in a salt water battery?
This question is on the same topic as this one but more complex.
So, let's consider the cell with $\ce{Zn}$ and $\ce{Cu}$ electrodes inside the $\ce{NaCl}$ solution.
On the $\ce{Zn}$ electrode we have
$$\ce{Zn + 2Cl^- -> ZnCl2  + 2e^-},$$ and on the copper one
$$\ce{H3+O + e^- -> 1/2 H2 + H2O}.$$
For both reactions we can find standard electrode potentials, but they are given for normal conditions, and $1~\mathrm{M}$ concentration of $\ce{Zn^{2+}}$ and $\ce{H3+O}$. To calculate the EMF for non-$1~\mathrm{M}$ concentration we use the Nernst equation. In this case we have something like
$$E=E_0 - \frac{\mathcal{R}T}{n\mathcal{F}}\ln{\frac{[\ce{Zn^{2+}}]}{[\ce{H3+O}]}}.$$
But really in the initial $\ce{NaCl}$ solution we do not have any $\ce{Zn}$ ions (so their concentration is very close to zero and unknown to us). Of course, in few microseconds some $\ce{Zn}$ from electrode will be dissolved creating some finite potential. But in the experiment when putting $\ce{Zn}$ electrode into the same solution we always observe the approximately same EMF value, $\approx 0.8~\mathrm{V}$. How to predict theoretically this value?
Furthermore, my question is how to calculate the EMF of such cell as precisely as possible for different experimental conditions, e.g. different temperature?
I have made an experiment showing that the EMF of a 'cell' with $\ce{Zn}$ and $\ce{Cu}$ electrodes into $\ce{NaCl}$ solution is decreasing when heating the $\ce{NaCl}$ solution.
It's strange, let me explain why. The Nernst equation tells us that EMF is increasing with $T$ when $[\ce{Zn^{2+}}]$ is less than $[\ce{H3+O}]$ (because $\log$ is negative), and otherwise EMF is decreasing with $T$ ($\log$ is positive). But in my opinion the concentration of $\ce{Zn}$ ions is always less than $[\ce{H3+O}]$ because we do not have $\ce{Zn}$ in the initial solution. So then logarithm is always negative so EMF is growing with $T$.
In this case, why do I observe the EMF decreasing for $\ce{Zn-Cu}$ pair?
And the last question. When we talk about $\ce{Zn-Cu}$ pair we assume that zinc dissolves; and the hydrogen gas is created on the copper electrode.
But why do I have the EMF when putting into a $\ce{NaCl}$ solution the $\ce{Al}$ ($\ce{Fe, Cr, etc}$) electrodes instead of $\ce{Cu}$, and remain instant the $\ce{Zn}$ electrode? I am not sure that hydrogen is created in such pairs: all these metals have a negative standard electrode potential. Measured EMFs of all these ($\ce{Zn-Al, Zn-Fe, Zn-Cr}$) pairs are significantly different, so the second metal is important in these cases. What reactions are going on in such pairs?

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I suspect that the reason you see a decay in the cell potential of $\ce{Zn/1M~ NaCl/Cu}$ with increasing temperature is due to increased corrosion of zinc. In $\ce{NaCl}$ solutions the anodic corrosion reaction is dissolution of $\ce{Zn}$ as $\ce{Zn^{2+}}$ and the cathodic reaction is reduction of dissolved oxygen to form an insoluble zinc hydroxide (Ichiro Suzuki, Corrosion Science 1985,25 (11), 1029-1034). The insoluble hydroxide passivates the metal surface. So, you have two effects which would contribute to EMF decay: 1) increased $\ce{Zn^{2+}}$ concentration and 2) passivation of the electrode.
One last thing which may or may not be helpful to you - if you are planning on utilizing the EMF from this cell to drive any kind of current you must account for the hydrogen overpotential on the copper electrode. It takes a lot of extra voltage to make hydrogen on pretty much any metal except platinum. You may find that you have a cell that looks good at open circuit (i.e. high potential with no current flow) only to see it crumble to nothing when you try connecting it to a load.

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Thank you, now I understand there are two parallel reactions on the cathode: reduction of oxygen to a zinc hydroxide, and reduction of hydrogen ions to gaseous hydrogen.
So, now the question is how to estimate the emf decay theoretically. Is the zinc ions' concentration linear increasing with T or more complicated?

Thanks for the edit to clean up my original answer. The concentration of zinc ions would be determined by the corrosion rate of the zinc electrode at a given temperature. In the event that the zinc electrode becomes passivated, you might expect to observe parabolic kinetics. On the other hand, if the cathodic corrosion reaction were primarily shuttled to the copper electrode, there may be little passivation for either electrode and you might find linear corrosion kinetics. I would suggest that you do some digging in the literature on zinc corrosion kinetics.

Thank you. But if the cathodic corrosion reaction is primarily shuttled to the copper electrode, we will have some EMF due to it. The standard electrode potential for $$O_2+2H_2O+4e^−\rightarrow 4 OH^−$$ is +0.401. So due to the zinc corrosion we must have EMF of 0.40+0.76=1.16 V at the normal conditions and 1M concentrations?

The Nernst equation for the corrosion cell is: $$E_{cell}=E_{cell}^{o}-\frac{nF}{RT}ln\left ( \frac{\left [Zn^{2+}  \right ]^{2}\left [ OH^{-} \right ]^{4}}{\left [ O_{2} \right ]} \right )$$ The cell potential you gave would be correct for standard state conditions, but the solubility of oxygen in water at room temperature is only about 10 ppm.  The other species are also present at concentrations much lower than standard state.  As the temperature is increased, the solubility of oxygen would decrease and the cell potential would be shifted in the negative direction.

This could be yet another effect causing the EMF of your cell to decay with increasing temperature.


text:
Is diazotization of o-phenylenediamine to benzotriazole reversible?
Is the synthesis of benzotriazole from o-phenylenediamine with acetic acid and sodium nitrite a reversible process? Not even specifically to o-phenylenediamine but any derivative.
Phenylenediamines (usually para), their derivatives and generally aromatic amines form various colour dyes on old colour photographic films that I like to experiment with as a hobby.

Yes, a common synthesis of benzotriazole starts with o-phenylenediamine as shown in the following figure.

Image Source
The first step in the reaction involves straightforward diazotization of one of the amino groups (exactly like what's done in the first part of the Sandmeyer reaction) to produce the ortho-amino diazonium ion. Instead of eliminating nitrogen as in the Sandmeyer case, the molecule loses a proton and then the diazomium ion internally captures the ortho-imino group to yield the triazole product.
The reaction is not reversible, the benzotriazole is very stable and the intermediate diazonium ion is relatively high energy making the back reaction unlikely under normal conditions.  The reaction will also work with substituted o-phenylenediamines as long as the substituent doesn't interfere with the diazotization step.  So for example, alkyl- or halo-substituted o-phenylenediamines will produce the corresponding alkyl- or halo-benzotriazoles.
Intramolecular cyclization to produce a triazole will only occur if the amino groups in the starting diamine are ortho to one another.  If m- or p-phenylenediamines or non-ortho naphthyl diamines are used, then products resulting from intermolecular azo coupling will occur. See an article by Freeman [1] for some examples.
Reference

Freeman, H. Aromatic Amines: Use in Azo Dye Chemistry. Front Biosci 2013, 18 (1), 145. DOI: 10.2741/4093. (Open Access)


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Why use a heating block?
Is a heating block necessary for heating small chemistry samples? I figure it helps a ton (not to mention it stabilizes too) in getting heat to the reaction but is there anything else it does?

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You have already mostly answered the question yourself ;)
A heating block, made from a material with a high thermal conductivity, helps to keep (a lot of) samples distributed over a larger area (several vials, Eppendorf tubes, etc.) at exactly the same temperature even if the primary heat source is punctual or just a layer of heating wires at the bottom of the instrument.