text:
Why do some reactions require specific pressures to happen?
For example, carbon monoxide reacts with hydrogen to synthesize methanol in the presence of some catalysts, but the pressure needs to be $\pu{50 atm}$ and the temperature needs to be $\pu{523 K}$.
Why the pressure? What does the pressure adds to the reaction?

ChemGuide has a good introductory article here.
The effects of increasing pressure and temperature are, to an extent, equivalent. Increased pressure leads to increased collisions and increased collision strength between molecules, allowing the (usually high) activation energy barrier to be overcome at a noticeable rate; at standard temperatures and pressures, for example, collisions between $\ce{CO}$ and $\ce{H2}$ are far too infrequent and involve far too little energy for the formation of $\ce{CH3OH}$.

Remark. How do we achieve high pressures and temperatures? We could pressurize a steel cylinder and use it as a reaction vessel, for one, but a novel approach involves sonochemistry. Ultrasound waves in solution lead to cavitation, the formation and rapid collapse of bubbles, which are highly pressurized and at high temperatures. The use of ultrasound can even open up new reaction pathways not usually available; for example, radical species are often transient, but the forcing conditions in these bubbles allow them to survive long enough to react.

I think I got it. so, it's just an optimized way of achieving the reaction. We could achieve the same reaction at 283K and a much higher pressure or at 1 atm and a much higher temperature, is this it? temperature and pressure are somewhat equivalent in a chemical system? what is the ratio T/P that accounts for pressure change equivalence to temperature change

High temperature and high pressure can both serve to accelerate a reaction based on kinetic considerations, yes; in other senses, they may not be similar. I only speak of qualitative considerations of high temperature and high pressure; I'm not sure you can quantitatively describe their relationship.
Note also that thermodynamic considerations, as highlighted by sixtytrees, make this more complex. By Le Chatelier's principle, high pressures favor reactions where there are less moles of gas in the product, whereas high temperatures favor endothermic reactions.

Almost, but not quite. Can you achieve the same reaction at room temperature, but higher pressure? Yes, you can. In diamond anvil (used for high pressure experiments) you can achieve the same result. It chemically doable, but expensive. However, just increasing temperature and decreasing pressure will not help.

Here is why. High temperature will accelerate both direct and back reaction. But low pressure would shift equilibrium to dissociation. So at high T and low P you will very fast arrive at 1% MeOH, 99% of 2H2+CO.

Engineers keep 1500 C temperature when making steel. They would have increased temperature of MeOH synthesis if it was helpful. Increasing pressure in large volumes is hard. That is why they don't go above 50 atm.

So, where do I look for the ideal pressure and temperature of a reaction? Is this experimentally obtained and recorded in a table of some kind? Temperature is a variable I can fully understand in system, because it accounts for the difference in enthalpy of the reactants and the products, but I don't know how to fit pressure in the equation, and since the purpose of my inquiries is to gather knowledge for making an Open Source piece of software (a Minecraft Mod), I need a good approximation of those values. How can I translate pressure into enthalpy of formation? What's the ratio?

The reaction you are talking about is an equilibrium. Pressure is added to shift equilibrium towards methanol.
CO+2H2 <-> MeOH
Consider an equilibrium pressure. In one second number of MeOH molecules dissociate = number of molecules formed. MeOH doesn't need a partner to dissociate. But CO heeds two partners (2 H2 molecules) to turn into MeOH. If you decrease the pressure MeOH will dissociate at the same rate. But collisions of CO with H2 will be less frequent. So less MeOH will form.
Here [Y] is the concentration of Y (in moles/L)
Let system be at equilibrium at given pressure:
K$_{eq}$ = [CO]*[H$_2$]$^2$/[MeOH]
Increase pressure by a factor of two. What happens to concentration of MeOH?
K$_{eq}$ = (2*[CO])(2[H$_2$])$^2$/(x*[MeOH])
Left parts are equal, so right parts are equal too.
[CO][H$_2$]$^2$/[MeOH] =  (2[CO])(2[H$_2$])$^2$/(x*[MeOH])
After dividing by [CO], [H2], [MeOH] we have 1=2*2$^2$/x
X=8. We have 8 times more MeOH after increasing the pressure by a factor of 2. If there was no reaction the amount of MeOH would just increase by 2. So we have 4 times more MeOH produced by just doubling the pressure.
This was mostly the thermodynamics aspect (how much MeOH we have when the reaction finishes). There is also kinetics (how fast reaction goes initially). Kinetics increases as pressure$^3$. Temperature also increases the speed of the reaction.
Pressure is needed to use the Le Chatelier's principle. This is one of a dozen fundamental principles of chemistry that you should look into. The higher the concentration of the reagent the more product will be produced. If the pressure is low formation of gasses that will spread around is preferred. At higher pressures formation of gas doesn't increase entropy (measure of disorder) by much and enthalphy (measure of heat) starts playing a role.
Note that a lot of processes depend on pressure. Water can vaporize by decreasing pressure or condensed by increasing it. Human's breathing requires high (1 atmosphere or so) pressure.
For MeOH synthesis you have 4 limiting cases:
Low P, low T: no reaction and if you wait forever you have little product at equilibrium time
Low P, high T: reaction goes fast, but against you MeOH falls apart rapidly
High P, low T: very high yield of MeOH if you wait forever, but reaction is somewhat slow
High P, high T: reaction is fast, but how much product you get depends on ratio of P and T. P should grow faster than T to keep equilibrium favorable.

Equilibrium is dictated by Gibbs energy G. Gibbs energy is just G = H - TS here H is measure of heat produced/consumed (enthalpy) T is temperature and S is measure of chaos/disorder produced.
Crystal has high order, so it has low S (enthropy), but formation of crystals is exothermic (favorable enthalpy). Gas is different. It has a lot of entropy (S is big), but little enthalpy of formation (it takes energy to turn crystal into gas). If you have high T (lots of energy around) then S becomes important. If you have low T, then H becomes important. This is why at high temperature you have gas and at low temperature you have crystals.
Now, enthropy of gas increases as you increase the volume it can occupy (gas feels it has less freedom when it is locked in a tight cell, but more freedom in bigger volume. So, if pressure is low 1 molecule MeOH dissociates three molecules form, they fly away from each other oad hardly ever meet. S increased a lot.
At high pressure they have little space and don't gain much enthropy. At the same time coming together is allows then to release some heat H (enthalphy).

Pressure doesn't add enthalpy H. It takes away entropic benefit S for MeOH to fall apart. Pressure says "dear gases, I know you love freedom and don't want to marry each other. But I will push you so close to each other that you will hit each other with elbows all the time. So yo uhave no freedom to lose when you form MeOH".

I did not fully understand your reply, specially this statement:  At higher pressures formation of gas doesn't increase entropy (measure of disorder) by much and enthalphy (measure of heat) starts playing a role.

so, basically pressure is needed to increase the yield of the reaction without the need to increase the temperature, is that it? So if pressure is so much more efficient than temperature at adding the needed enthalpy of formation, what dictates the point where pressure is preferable over temperature for a chemical compound systesis in lab?

Since when do you measure pressure in mol/L?

@Martin I measure concentration in mol/L. At a given temperature pressure is defined by concentration of ideal gas. Don't put your words in my mouth.

@sixtytrees You state ' Here [Y] is the concentration of Y (in moles/L)' and then continuing with argument of ' Increase pressure by a factor of two.' without changing the notation.

@Martin Rules of demagogue, right.  Don't you know that at a given temperature pressure is of partial ideal gas defines its concentration? This is not a homework question, so I don't want to give unnecessary details.

You should really read up on the be nice policy. Calling me a demagogue is not nice. Apart from that I am not trying to appeal to anyone. I think your answer is bad as it contains too many words, and not enough explanation. This is my personal opinion and I have downvoted such answers before this one and I will continue to do so. I am sorry if you take that personal, it is not. The content you are writing is ambiguous and hence I downvoted it. On another note, you might want to have a look at how to use MathJax , so that you posts don't fall apart on different displays.

This is old but I wanted to add a quantitative / physical perspective here.
As you know, increasing pressure has a similar effect on reactions as increasing pressure. Recalling the ideal gas law,
$$pV = nRT$$
We have that $p$ is proportional to $T$. In other words, if you hold all other variables constant, increase in $p$ leads to increase in $T$, and vice-versa.
Pressure can be expressed as force per unit area, or energy per unit volume:
$$ p = \frac{\vec{F}}{A} = \frac{E}{V}$$
So it follows that increasing pressure increases force on molecules. Force effects the acceleration of particles by $\vec{a} = \frac{\vec{F}}{m}$, and velocity follows: $\vec{v} = \int \vec{a} dt$.
Temperature can be expressed using the velocities of particles (by a classical Boltzmann distribution),
$$ T = \frac{\sum_i^N m_i\vec{v_{i \alpha}}^2}{k_b N}$$
(Note the numerator is kinetic energy). So you can see that increasing $p$ and $T$ have the same physical result: faster motion of particles. Faster particles means more collisions, and more collisions means faster reactions, or occurrence of a reaction where it could not happen otherwise, because an energy-barrier is overcome ($p \propto E$ and $T \propto E_k$).


text:
Iodine Titration with Free SO2 in wine
To determine the free $\ce{SO2}$ in wine, a common method used is called the Ripper Titration Method.
The equation is as follows: $$\ce{SO2 (aq) + I2 (aq) + 2H2O (l) -> 4H+ (aq) + SO4^2- (aq) + 2I- (aq)}$$
This method involves titrating an Iodine solution into a sample of wine. I have found that triiodide is made from reacting $\ce{I-}$ with $\ce{I2}$. However, if it is the triiodide we are titrating, how is the iodine reacting with the wine?
Is it triiodide ($\ce{I3-}$) that is being titrated with the wine sample or iodine ($\ce{I2}$)?

@Nij States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

@Martin-マーチン just fitting form based on how I would write for P&P. I'll check those recommendations for other details later, thanks.

Reaction goes along the lines $\ce{3 I2 -> 2 I3- -> 6 I-}$. The beginning of titration is $\ce{I2}$. The end of titration is $\ce{I-}$. $\ce{I3-}$ forms as an intermediate, but it can be ignored in titration.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

I have down voted this because while it isn't completely wrong, it contains not explanation and should have probably better be left as a comment.

@martin, Downvoting something that is not wrong (or that is honestly right) is an interesting practice. What do you disagree with?

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text:
Is Q in the Nernst equation [Reduction]/[Oxidation] or [Oxidation]/[Reduction]?
The Nernst equation is given by :
$$E_{\text{cell}} = E_0 – \frac{RT}{nF}\ln{Q}$$
I have seen  ${Q}$ given as both ${[\text{oxidation}]/[\text{reduction}]}$ and ${[\text{reduction}]/[\text{oxidation}]}$. When does each form apply?
Also, for the following cell:
$${\ce{Cu}|\ce{Cu2+}(\pu{0.001 M})|\ce{Cu2+}(\pu{1 M})|\ce{Cu}}$$
${[\text{oxidation}]}$  is 0.001, as the half reaction would be $\ce{Cu -> 2e- + Cu2+}$  because $\ce{Cu}$ is losing 2 electrons and becoming oxidised, right?

It depends on the sign before the log term. The way you have written it is reduction/oxidation e.g.
$$E_{\text{Ox/Red}} = E^0_{\text{Ox/Red}} - \frac{RT}{nF}\ln\left(\frac{a_{\text{Red}}}{a_{\text{Ox}}}\right)$$
where a is the activity. Note the ratio is the opposite way round red/ox . (If you write Q the opposite way up then there is a positive sign before the log)
A derivation is given below for the Daniel cell:  $$\ce{Zn(s) + Cu^{2+}(aq)-> Zn^{2+} (aq) + Cu(s)}$$
At the anode (where oxidation occurs) the left half cell is  $$\text{Red}\rightarrow \text{Ox}^{n+}+ne^{-}$$
and at the cathode $$ \text{Ox}+ne^-\rightarrow \text{Red}^{n+}$$
The free energy $\Delta G_{\text{Rev}}$ is determined from electrochemical potentials for the species.
$$\Delta G_{\text{rev}} = \mu_{\ce{Zn^{2+}}} +  \mu_{\ce{Cu}}  -\mu_{\ce{Cu^{2+}}} - \mu_{\ce{Zn}}$$
$$ \Delta G_{\text{rev}} = \Delta G^0_{\text{rev}} +RT\ln\left(\frac{a_{\ce{Zn^{2+}}}}{a_{\ce{Cu^{2+}}}}\right)$$
where the last equation is obtained from definition of chemical potential in terms of activities, $\mu = \mu^0+\ln a$. (The activity of each metal is 1) If the reaction is carried out reversibly then the work done is equal to charge time potential difference through which the charge is moved. Therefore $$ \Delta G_{\text{rev}} =-nF\Delta \phi$$ In a reversible reaction we change potential difference to E the 'electromotive force'(emf). Thus we can write
$$-2FE=\Delta G^0_{\text{rev}}+RT\ln\left(\frac{a_{\ce{Zn^{2+}}}}{a_{\ce{Cu^{2+}}}}\right)$$
For standard conditions the activities $ a_{\ce{Zn^{2+}}}=a_{E}=1$ and then we can write $$ E=E^0-\frac{RT}{2F}\ln\left(\frac{a_{\ce{Zn^{2+}}}}{a_{\ce{Cu^{2+}}}}\right)$$ When there are n electrons exchanged and the reaction is $$\text{Ox}^{n^+}+ne^-=\text{Red}$$ then the Nernst equation is $$ E=E^0-\frac{RT}{nF}\ln\left(\frac{a_{\text{Red}}}{a_{\text{Ox}^{n+}}}\right)$$

I'm familiar with everything in your answer except that $\mu$ in the expression for $\Delta G$. Can you kindly tell what is? I would have expected it to be $\Delta G_{\text{rev}} = G_{\ce{Zn^{2+}}} +  G_{\ce{Cu}} - G_{\ce{Cu^{2+}}} - G_{\ce{Zn}}$ instead (as $\Delta$ signifies a difference). Thanks!

$\mu$ is the chemical or the increase of  free energy  when  one mole of a component is added to a large quantity of solution so that the overall composition is not affected. Mathematically  for species $i$, $\displaystyle \mu_i=\left (\frac{\partial G}{\partial n_i} \right)_{T,P,n_j}$ which is also the partial molar free energy  sometimes written as $\bar G$

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text:
A steel sample was analysed for nickel

A steel sample ($\pu{0.3500 g}$) was dissolved in nitric acid and the resulting solution made up to $\pu{500 ml}$. The absorbance of this solution measured by atomic absorption was found to be $0.410$. A standard $\pu{6 ppm}$ solution of $\ce{Ni}$ gave an absorbance of $0.522$ under the same conditions. Assuming linearity between absorbance and concentration, calculate the $\% \ce{Ni}$ in the steel sample. (8 marks)

Seems so simple but is confusing me.
A steel sample ($\pu{0.3500 g}$) was dissolved in nitric acid and the resulting solution made up to $\pu{500 ml}$.The absorbance of this solution measured by atomic absorption was found to be $0.410$.
A standard 6 ppm solution of Ni gave an absorbance of $0.522$ under the same conditions.
In the final diluted sample,
$$\frac{\pu{6 ppm}\ \ce{Ni}}{0.410\times0.522}=\pu{28.03476311 ppm}\ \ce{Ni} \tag{1 }$$
In the first solution
$$\frac{\pu{28.03476311 ppm}\ \ce{Ni}}{\pu{500 ml}}=\pu{0.05606952622 mg}\ \ce{Ni} \tag{2}$$
Supposing the density of the first solution to be near that of water:
$$\pu{1000 g}\times\pu{0.05606952622\times10^{-6} g}=\pu{5.60695\times10^{-5} g} \tag{3}$$
In the steel sample:
$$\frac{\pu{5.60695\times10^{-5} g}}{\pu{0.35 g}}=0.000160198=0.016\ \%\ \ce{Ni}\tag{4}$$
$2$nd attempt:
\begin{align}
\frac{\pu{6 ppm}}{0.522}&=\frac x{0.410}\\
\implies \frac{\pu{6 ppm}\times0.410}{0.522}& =x\\
x&=\pu{4.7126ppm}\ \ce{Ni} \tag{6}
\end{align}
This means you have $\pu{4.7126 mg}$ nickel per $\pu{1 kg}$ of steel (ppm is one millionth, and thus, mg per kg). Then one tenth of that amount of steel $(\pu{0.1 kg}/\pu{1 kg})$ contains $1/10$ of the $\ce{Ni}$. that is $\pu{0.47126 mg}$
Since it was marked up to $\pu{500 ml}$
In the steel sample:
\begin{align}
\pu{0.47126 mg}\times\frac{500}{50}&=\pu{4.7126 mg ml-1}\\
\frac{4.7126}{100}&=0.047126\\
\frac{\pu{0.047126 g}}{\pu{0.35 g}}&=0.134645\\
&=13.46\ \%\ \ce{Ni} \tag{7}
\end{align}
I feel that I am missing a step in calculations. I would prefer to read a resource of working this out myself than finding the answer.
I feel more confident in my second attempt of finding a correct answer.
This question is from a past paper.

Your first attempt goes wrong in your first step; this is corrected in your second attempt, but then you choose to work with 50 mL, which comes out of nowhere. It may help if you first -- ignoring the mathematics -- explain to yourself what you're trying to accomplish in every step and make sure you understand the relevant concepts (proportions and unit conversion).

Your first step is incorrect. Think about it this way:
$5$ people have $10$ eyes. This group of people has $14$ eyes. How many people in the group?
The correct solution is:
This group has $14/10 = 1.4$ times more eyes than the known group. Thus, they have $1.4$ times more people. $5$ people $\cdot$($14$ eyes/$10$ eyes) $ = 5$ people $ \times 1.4$ (unitless ratio) $= 7$ people. Here, the units make sense. This is correct.
In your calculation you do ($5$ people)/($10$ eyes $\times 14$ eyes)$= 1$ people/($28$ eyes$^2$). This unit people/eyes$^2$ makes no sense. This is incorrect.
First you find the amount of $\ce{Ni}$ in the final sample.
Your sample absorbs $0.410$ which is slightly less than $0.522$ ($\pu{6 ppm }$sample which). So your sample has $(0.410/0.522) \times \pu{6 ppm} = \pu{4.713 ppm}.$
Water solution has density of $\pu{1 kg/L}$, so $\pu{500 mL} \to \pu{500 g}$. $\pu{4.713 ppm}$ in $\frac{\pu{500 g}}{1\,000\,000} = \pu{0.5 mg}$. $\pu{0.5 mg} \cdot 4.713 = \pu{2.356 mg}$.
The total weight of the sample was $\pu{0.35 g} = \pu{350 mg}$
$$100\ \% \cdot \frac{\pu{2.356 mg}}{\pu{350mg}} = 1.003\ \%$$
You should round it down to $2$ digits, because that was the accuracy of other data. The answer is  $1.00\ \%$.

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text:
Whether copper can combine with oxygen in equal proportion?
Copper sulphide contains 66.6% copper and sulphur trioxide contains 40% sulphur. According to law of reciprocal proportions copper is 80% in copper oxide. The calculation: In the first, for 33.4% sulphur, there is 66.6% copper. In the second, there is 40% sulphur. Therefore, in the resultant there should be 80% copper. My question is why cannot we calculate the other way round:In the second, for 40% sulphur there is 60% oxygen; therefore, for 33.4% there should be 50% oxygen. Hence, the resultant combination should have equal proportion of copper and oxygen. Is there any such substance?

Do you mean $\ce{CuO}$? This is just normal copper (II) oxide.

No. CuO is from the first calculation. I want to know the substance according to the second calculation and it ought to be CuO4.

The Law of Reciprocal Proportions states: "If two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine."
Wikipedia has a different phrasing: "If element A combines with element B and also with C, then, if B and C combine together, the proportion by weight in which they do so will be simply related to the weights of B and C which separately combine with a constant weight of A."
Note this is a statement about the weights of substances, not about the molar fractions or volumes.
For example,
Copper sulfide ($\ce{Cu2S}$) contains 66% copper by moles. It does not, however, contain 66% copper by weight. Instead it contains $(2\cdot63.55)/( 2\cdot63.55+32.06) = 79.9\%$ copper by weight, or 20.1% sulfur by weight.
Likewise, while sulfur trioxide ($\ce{SO3}$) contains 25% sulfur on a molar basis, it contains $(32.06)/( 32.06+3\cdot16.00) = 40.0\%$ sulfur on a mass basis, or 60.0% oxygen by weight.
So when we apply the law of reciprocal proportions to copper oxide, we need to make use of the 20.1% and 40.0% numbers for sulfur. For example, if we had two masses of $\ce{Cu2S}$ and $\ce{SO3}$ each containing 1 g of sulfur, then we would have $1\rm{g}\cdot(79.9/20.1) = 3.98$ grams of copper and $1\rm{g}\cdot(60.0/40.0) = 1.50$ grams of oxygen. The ratio here is $3.98/1.50 = 2.65$.
Now, when we examine copper (II) oxide ($\ce{CuO2}$), for example we see that the ratio of masses for copper and oxygen are $63.55/(2\cdot16.00) = 1.99$ ... and 2.65 is very close to a simple multiple of 1.99 (specifically, 4/3 - the difference is due to rounding.)

The fact that this is always true is trivial, once you know atomic theory. So while the Law of Reciprocal Proportions is important in a historic context - as it helped lead to the development of modern atomic theory - basically no one uses it as such these days.
Say we have the compounds $A_hB_j$, $A_kC_l$ and $B_mC_n$.
In $A_hB_j$ the masses occur in the ratio $(j\cdot M_b) / (h\cdot M_a)$. In $A_kC_l$ the masses are in the ratio $(l\cdot M_c)/(k\cdot M_a)$, so for a fixed mass (say $M_a$g) of A, the ratio of the masses is $(k\cdot j\cdot M_b)/(h\cdot l\cdot M_c)$
Now for the compound $B_mC_n$, the masses occur in the ratio $(m\cdot M_b) / (n\cdot M_c)$. Taking the proportion of this with the previous proportion, we see that the masses cancel, leading to $(k\cdot j\cdot n)/(h\cdot l\cdot m)$. Now because molecules form with small, whole number coefficients in their molecular formulas, this final quantity will typically be a simple, small fraction. (a.k.a. a "simple multiple" of each other.)
So all the Law of Reciprocal Proportions is doing is saying that 1) all atoms of a given element have about the same average weight 2) molecules form from (typically) small, integer numbers of atoms. -- That's it.
This is why you don't see many people discussing it. It's not all that interesting to people who are accustom to thinking atomically.

When did the mass of oxygen become 18.01??

@f'' Oops. You're right. I'll fix that. However, the fact it worked out in the end is just more evidence that the masses of the atoms doesn't really matter, as long as they're constant.

You are not finishing the calculations properly and it is a coincidence that your first one got the right answer.
When you calculate the proportions in copper oxide, you have to find both the copper and the oxygen. You have correctly found that the masses of copper and sulfur are in a ratio of 80 to 40. From sulfur trioxide, you can find that the masses of sulfur and oxygen are in a ratio of 40 to 20 (one-third of 60 because there are three oxygens). Combining these, you know that the ratio of copper and oxygen's masses is 80 to 20. Therefore the percentage of copper in CuO is 80/(80+20), which is 80%. Your calculation only worked because the sum of the numbers (80+20) was 100.
Doing the calculation the other way, you need to remember to divide the oxygen by 3 because there are three of them in sulfur trioxide. You should find that the ratio of oxygen's mass to sulfur's mass is 16.7 to 33.4 and the ratio of sulfur's mass to iron's mass is 33.4 to 66.6. This makes the percentage by mass of oxygen in CuO equal to 16.7/(16.7+66.6), which is 20% as expected.

This is the law of proportion.
For chemicals it works much the same way as for other objects. consider this proportion: If 3 people have 6 eyes, then how many eyes 5 people have?
3 ->6
5 ->x
x=5*6/3 = 10 eyes. This is correct. You suggest to do something like
x=3*6/5 = 3.6 eyes. Which is incorrect.

Even your first calculation is not convincing. You don't account for valency difference in sulfur. You compare CuS (2 valent sulfur and 2 valent Cu) with SO3 (2 valent oxygen and 6 valent sulfur). Now you want to compare CuO (2 valent O and 2 valent Cu). Your step from CuS to SO3 is not justified because you don't account for the change of valency in sulfur. Please expand your calculation to make it easier to follow.

You mentioned direct proportion and inverse proportion in mathematics. My logic is not about the above. In my first calculation I found volume of copper. In the second I tried to find the volume of oxygen. However, the two calculations produce different results. I don't think the law is violated.

@MathivananPalraj you can only use proportion if the oxidation of the atom is the same. You can go from  MgS to MgCl2 to SCl2. Likewise you can go H2O, CO2, CH4. This is because in first example they all have valence 2 and in the second example H, O, and C have valences 1, 2, 4 respectively. What you are trying to do is to go from CO to CO2. In this case proportion law is not applicable. Does it make more sense now?


text:
Reaction of salicylic acid with PCl5
What it is the product when salicylic acid (2-hydroxybenzoic acid) is reacted with $\ce{PCl5}$?

According to me the product should be:

But why the product is supposed to be:

See this answer. http://chemistry.stackexchange.com/questions/35359/can-oh-group-on-the-benzene-ring-of-4-hydroxybenzoic-acid-react-with-pcl5

Photo shows PCl3 on the photo - not PCl5.
You have two potentially reactive groups. Phenol (with OH) and carboxylic acid to consider. carboxylic acid turns into chloro anhydride just as both pictures show. Now alcohols react with PCl3, but phenols don't. YOu have a phenol, so it will not react, so the corect answer is (B).

Why phenol will not react

@koolman, look at the posts linked in the comments below. They answer that question in quite some detail.


text:
Acetic acid reaction sequence
Can anyone tell me where I am doing it wrong for the following reaction?
In the first step, Br should be attached to the alpha carbon.
In second step, the bromide is replaced by cyanide.
I am able to get the third product that is final compound.

Heating a carboxylic acid often leads to decarboxylation. So decarboxylate your product from step 2 (well actually the product formed following acidification in step 3!)

@Leeser can you tell me the products formed

Step 1: $\alpha$-bromination
$$\ce{CH3-COOH + Br2/P -> Br-CH2-COOH}$$
Step 2: Nucleophilic Substitution by Sodium Cyanide
$$\ce{Br-CH2-COOH + NaCN -> NaBr + NC-CH2-COOH}$$
Step 3: Acid-catalyzed Conversion of Nitrile to Amide or Acid
(1) First hydration... $$\ce{NC-CH2-COOH + H2O -> H2N-CO-CH2-COOH}$$
(2) Further heating... $$\ce{H2N-CO-CH2-COOH -> HOOC-CH2-COOH + NH4+}$$
Special Notes
Be VERY CAREFUL with Step 2 as you are potentially forming $\ce{HCN}$ which could get everyone in the lab killed if you're not careful.

But @sixtytrees has made different products

@user123733 No, he only went step further with decarboxylation, without any good reason.

There are two possibilities. You will have a two step reaction
NC-CH2COOH -> HOOC-CH2-COOH This is malonic acid. It can be further decarboxylated to
HOOC-CH2-COOH + H$^+$ + heat -> CO2 + CH3COOH
the final product is acetic acid.


text:
How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of CO2 and H2O will be formed?
How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of $\ce{CO2}$ and $\ce{H2O}$ will be formed?
After balancing the left-hand side produces 5 volume of oxygen and 2 volume of acetylene. However, the right-hand side produces 4 volume of carbondioxide and 2 volume of water. Why doesn't the volume on LHS and RHS match? Or how is it matched?
Is it correct that 100 ml of oxygen and 40 ml of acetylene produce 80 ml of $\ce{CO2}$ and 40 ml of $\ce{H2O}$? If so, what happened to 20 ml on the right hand side?

I think this shouldn't be on hold. The poster has clearly made an attempt by balancing the equation (albeit with a mistake - it should be 4CO2 on the right side) and is asking why the total number of moles is not conserved.

While I can see the effort in the question, and hence will reopen it, I strongly recommend writing out the whole equation, i.e. $\ce{2HCCH + 5O2 -> 4CO2 + 2H2O}$. Then you can see, that acetylene is decomposing when combustion occurs. Since you make a dramatic change in the molecular geometry of the compounds, a volume change is certainly expected to occur, too. Also make sure that you understand that these are molar ratios and only within the ideal gas approximation they are proportional to the occupying volumes.

@Martin Thanks for enlightening. Is that the right hand side occupies less space?

The volume can change in the course of the reaction. For example, if you add $\ce{Zn}$ to $\ce{HCl}$ the volume of the system changes. In the reaction $\ce{2H2 + O2 -> 2H2O}$ three volumes of gases turn into 1 (or even zero, if you allow water to condensate).
While mass remains constant, volume can change in chemical reactions.
Behavior of gases (under low pressure) is described by ideal gas theory. In gas a molecule is a small particle that has a large free space around it. The amount of free space is independent from of the molecular mass. lightest molecule $\ce{H2}$ occupies as much space as a heavy gas $\ce{SF6}$. As a result, density of $c\ce{H2}$ is much smaller than density of air (this is how helium baloons fly) which in turn is smaller than density of $\ce{SF6}$.
If two molecules form a dimer the density of the gas doubles, and the volume decreases by a factor of two (if the pressure is kept constant.
From 12 carbons and 24 hydrogens you can make 6 molecules of ethane ($\ce{C2H4}$) or 4 molecules of cyclopropane ($\ce{C3H8}$) or three molecules of cyclobutane ($\ce{C4H8}$). The amount of atoms is the same, the weight is conserved, but the volume and density are not.


text:
Is enthalpy of activation the same as activation energy? Is there a purely mathematical way to find that value?
I asked about the effect of pressure on a reaction, and since then I learned a lot more. One of the things I learned was the Eyring equation:
$$\kappa = \frac{\kappa_B T}{h} e^\frac{\Delta^\ddagger S°}{R} e^\frac{- \Delta^\ddagger H°}{R}$$
where:
$\begin{multline}
\begin{split}
\kappa &= \text{reaction rate} \\
\kappa_B &= \text{Boltzmann constant} \\
T &= \text{temperature} \\
h &= \text{Planck's constant} \\
R &= \text{gas constant} \\
\Delta^\ddagger S° &= \text{entropy of activation} \\
\Delta^\ddagger H° &= \text{enthalpy of activation} \\
\end{split}
\end{multline}$
My questions:

Is the enthalpy of activation the same thing as the activation
energy?
Is there a purely mathematical way to determine both $\Delta^\ddagger S°$ and $\Delta^\ddagger H°$? (without having a laboratory to use to observe the reaction)

(1) Enthalpy of activation is a component of activation energy. The other component is temperature * entropy of activation. Enthropy controls "heat difference" while entropy controls "measure of disorder difference".
$\Delta$G = $\Delta$H - T*$\Delta$S. This applies to activation energy too.
(2) DFT functionals rely on 4 parameters per atom make predictions. So, if you give 4*5=20 parameters, then you can in theory predict behaviour of any compound, build with only C, H, N, O, S atoms. This covers most of biochemistry, excluding nucleic acids (you would need additional 4 parameters to characterize P).
These 4 parameters are known for most of the atoms (excluding some elements, going after uranium). But those four parameters were experimentally measured by fitting behavior of modeled molecules to real molecules.
Should we count it as "purely theoretically predict?"
Also there are purely ab initio methods like Hartree-Fock with even less input required. You would still need some input. In math you can build many different plausible worlds. For example purely "newtonian" vs "newtonian + quantum mechanics" words. experiment is the only way to tell, which model is correct.

Wow, I downloaded JDFTx and it is way beyond my understanding. Do you have an idea of something easier to use?

In my PhD QM 30% of work was relevant to QM. I am now working as a coder.  JDFTx is above my head. It is a "plugin" (one of the DFT potentials) to work with a very specific case "calculations of electronic systems in contact with liquid environments". It is not a mainstream approach so you might not find tutorials.

Feasible steps for you would be (1) ACD/ChemSketch to draw molecules -> (2)  convert to 3D (in ChemSketch) -> (3) use Gaussian to compute properties. If you try to build Minecraft level from ab initio QM, it will not work. Currently professors in computational chemistry develop methods to model protein-protein interaction. A computational model of a whole cell is a dream in the field. The problem isn't that they can't find solutions to the equations. The problem is that using modern supercomputers you would use a billion years to model a nanosecond of life of a single cell.

Licensed Gaussian costs money, but it is a de-facto standard. You can check alternatives here https://www.reddit.com/r/chemistry/comments/s6m0w/opensource_gaussian_alternative/, but I've never worked with those.

All that just to predict entropy and enthalpy of activation of a reaction?

Try to find Gaussian software and see some tutorials on how to use it. This would be a good start.

@sixtytrees Are you claiming that by computational means alone you can accurately calculate the properties of transition states?

@porphyrin Yes, you can. TS for small molecules work well. CF3 substituent slowed a reaction I was working on by a factor of 500 (measured) 450-470 (computed at DFT(B3LYP)/6-31++ depending on solvent model). This agreement is good given simplicity of B3LYP.  I could have used a better solvation model, but this accuracy was sufficient for my purposes. It is indeed hard to predict KIE for enzymatic reaction when tunneling effects become significant. We can always tweak the functional to give us results we need for a small set of reactions, but we don't have a robust method for all cases.

I don't know the technical terms, all I want is to be able to know the values of enthalpy and entropy of activaton of a given reaction. Are those values properties of the reaction, meaning are they constant? If so, the rate of the reaction can be calculated as a function of T. Or, maybe enthalpy and entropy of activation are not constant for a given reaction, but vary as temperature and pressure vary? maybe enthalpy is a function of the reaction equation and temperature, and entropy a function of the reaction equation and pressure? I just guessing so you guys can see what I'm looking for.

I'm not talking about proteins or anything that complex, maybe homopolymers as polyethylene, I don't need enzimes or any of that, just simple reactions between simple compounds, like Ca(OH)2 + K2CO3 → CaCO3 + 2 KOH

By the way, Gaussian is paid aand extremely expensive (like $ 6,000 a license). I found this other one called Orca (license is free), I wanted to see if you ever heard of it, and, if yes, if it can do what I need.

"Ca(OH)2 + K2CO3 → CaCO3 + 2 KOH".  You would think this exchange is simple. It is actually very complex. You have to account for formation of bulk phase (precipitate). First of all it is bulk (so you need a lot of atoms in your model. Second, you need solvent molecules. Third, you need a decent model of a crystal.Trying to predict which crystal lattice will form in precipitation of CaCO3 is already a state of art computation. The easiest thing to model is the reaction of CH3Cl with CN-. Ideally it should have two molecules one of which is neutral and the other is charged.

Gaussian wants money. Someone in internet might share a "free copy" with you. For your purposes you should Gamess or Orca are a good start. I haven't use either of them, but heard good words about Gamess. You should open a separate question about QM modelling so that we stay on topic. There we can discuss models that you should start with, methods you can use and so on. Btw, even modelling liquid water is difficult. Water model (there are many things you can compare with experiment) is a first benchmark of the quality of your QM model.

I don't know what the title of the question should be (this is how lost I am at this point. I will get it all together eventually, but right now I'm totally lost). Any suggestions? I don't want to make a question with a bad title. Even though this forum is much friendlier and less strict (more chill) than stackoverflow, I don't want to show lack of effort.

@sixtytrees about what you said of the reaction between calcium hydroxide and potassium carbonate, is all that necessary to determine the enthalpy of activation and the entropy of activation. What I want is the following: given a reaction equation, a temperature and a pressure, what fraction of each reactant (in moles) will participate in the reaction and what will be the fraction of products (in moles) obtained. And then be able to tweak temperature and pressure to achieve a 100% reaction, where all reactants react to form the products.

@FinnTheHuman "What makes a reaction an easy or a difficult target for computational modeling?" start with a question like this.

@sixtytrees I have no idea

I am saying that is a good title for your new question.


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Why is the boiling point of alkyl iodide is more than alkyl fluoride?
As far as I know, the more the polar nature of the organic compound, the more will be the energy required to break the intermolecular bonds and thus a higher boiling point. Here in this case, $\ce{R-F}$ is more polar in nature than $\ce{R-I}$. Then, how does $\ce{R-I}$ have a higher boiling point than $\ce{R-F}$?

See this page: "The increase in boiling point as you go from a chloride to a bromide to an iodide (for a given number of carbon atoms) is also because of the increase in number of electrons leading to larger dispersion forces. There are lots more electrons in, for example, iodomethane than there are in chloromethane - count them!"

@JasonB, the increase in number of electrons is not significant, most of them are contracted into tight and inert inner shells. Now, increase in atoms size and polarizability is quite impressive.

@permeakra - I think that is the point made on that page (or it should be), that the number of electrons in $I$ that are not in closed inner shells is much higher than for fluorine.  You have the 10 4d electrons, which do contribute to the polarizability.  But you make a great point, that isn't made on that page, that it is the size of these orbitals that really impacts the polarizability, and therefore the magnitude of the dispersion forces.

Because dispersion forces are more powerful than we usually give them credit for, maybe?

The main factor affecting the boiling point is the molecular weight, as can be pictured from the noble gases boiling point in the periodic table (the noble gases are those which have the weakest interactions due to their complete outer shell). But note that if you want to consider all the effects altogether, you must use Kelvins instead of Celsius. So going from 44°C to 88°C is not twice as much but a mere 14% increase.
The other factor affecting boiling point is the nature and strength of intermolecular interactions, usually in the order ionic > hydrogen bond > dipole-dipole > Van der Waals.
In the comparison of RF vs RI boiling point, molecular weight is definitely in favor of higher boiling points for RI, at least for small molecules. It is true that the C-F bond is much more ionized than C-I so it will create stronger dipole-dipole interactions but not sufficiently to compensate for the effect of the mass increase.
HTH

Assuming they have the same alkyl groups i.e. methyl fluoride, methyl chloride, methyl bromide, and methyl iodide.
In this case I considered the molecular mass of these compounds. The one with the highest molecular mass, which was methyl iodide was supposed to have the highest boiling point (BP) since BP increases with increase in molecular weight.
The one with the highest molecular weight has more electrons which creates more temporary bonds with other molecules.  The more the bonds the more the energy needed to break them, hence higher boiling point.

Polarity consideration is important when dealing with interactions between molecules. A hydrogen bond (H-bond) is a good example. Water has polar $\ce{H-O}$ bonds, as a result partially positive $\ce{H}$ interacts with partially negative $\ce{O}$ from a dofferent molecule as a result water is liquid at $T_\mathrm r$ and solidifies at $0\ \mathrm{^\circ C}$, while methane of about the same molecular weight is a gas up until $-168\ \mathrm{^\circ C}$.
The reason for $\ce{HF}$ to be liquid and $\ce{HI}$ to be gas is different. $\ce{H}$ and $\ce{F}$ have similar sized orbitals and the molecule is polar. As a result intramolecular H-bonds are strong. In $\ce{HI}$ the overlap is very poor (minute $\ce{H}$ orbitals don't overlap well with huge diffuse orbitals of $\ce{I}$. As a result $\ce{HI}$ doesn't form strong intermolecular $\ce{H}$ bonds and the bonding between $\ce{H}$ and $\ce{I}$ in the molecule is much weaker than in $\ce{HF}$. Thus, $\ce{HI}$ is both gas and a very strong acid (much stronger than $\ce{HF}$).
I will assume that by $\ce{R}$ you mean an alkyl group. In this situation neither $\ce{R-F}$, nor $\ce{R-I}$ can form hydrogen bonds. For example, the polarization of $\ce{$\mathbf{H{-}C}$H2F}$ is much closer to that in methane ($\ce{CH4}$) than in water, not to mention $\ce{HF}$. Some polarization exists but its effects are small. Much greater effect is the molecular weight and density that translate into much stronger London forces in $\ce{R-I}$ compared to $\ce{R-F}$. Both weight and density favor $\ce{R-I}$ to be less volatile than $\ce{R-F}$. The effect of polarization is not large enough to compensate weight difference.
So, how large is this polarization effect in $\ce{R-F}$? Well, compare bp of ethane ($\ce{CH3CH3}$) and fluoromethane ($\ce{CH3F}$) and iodomethane ($\ce{CH3I}$). $T_\mathrm{bp}(\text{ethane})=-88\ \mathrm{^\circ C}$ and $T_\mathrm{bp}(\ce{CH3F})=-77\ \mathrm{^\circ C}$, $T_\mathrm{bp}(\ce{CH3I})=+44\ \mathrm{^\circ C}$. The effect is there but it is very small.

You should expand the third paragraph into description of dispersion forces and why they are increased with move from fluorine to iodine. As of now, your answer does describe nicely, why polarity does not play a role here, but what does ?

I would go with "molecular weight is the main contributor" as the rule of thumb. Sure, there are dispersion fores, but consider this. CH3I weigh 142 g/mol and boils at 44°C. C10H22 weighs 142 g/mol and boils at 174.1°C. If I go into dispersion forces I would have to explain why C10H22 has so much higher boiling point without dispersion forces.  I can, but that would be a convoluted off-topic story that TC didn't ask for.

I don't buy "molecular weight" as the main contributor.  It absolutely has to be about the forces between the molecules.  Weight only matters through gravity, and I assure you it isn't gravity that keeps a CH3I molecule in the condensed phase at the same temperature a CH3F molecule escapes into the gas.

MoF6 boils at 34 C while undeniably heavier .WF6 boils at 19 C. Consequetly, general reference to 'weight' is not an acceptable answer.

I think high boiling point of alkyl iodide in comparison to alkyl fluoride is because of the alkyl iodide's higher molecular mass which in turn results in increase of van der Waals' force.

@ShuvamShah As I pointed above, increase in mass may result in decrease of vdW forces. And strictly speaking, atomic mass has nothing to do with vdW forces, polarizability does.

@permeakra I read it from here: http://chemistry.stackexchange.com/questions/28499/relation-between-molar-mass-and-van-der-waals-force

@ShuvamShah - from the answer on that page "the main factor in work here is the volume occupied by electrons", which happens to correlate with mass.  But saying that the mass causes the increase in forces makes it seem like it's a gravity issue

@JasonB Yes, it is NOT a gravity issue. But accurate accounting for the factors is not suitable for this format. A question like" why London dispersion forces are stronger in CD4 compared to CH4?" doesn't have a short and simple answer.

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


text:
Aren't hydrogens on less substituted groups more electrophilic?
I am having some trouble understanding the behaviors of some Alkene reactions that I have been studying. I took two semesters of O-chem a couple of years ago and am trying to refresh before an advanced class, but I have found two reactions in my studies that aren't making sense.
At the top of page 13 of the e-book found here: http://crab.rutgers.edu/~alroche/Ch08.pdf ... I see that the water molecule is attacking the carbon with two methyl groups on it... but wouldn't the carbon with only one methyl group be more electrophyllic? Because of induction, I would think that the carbon with two methyl groups would be more stable than the other, and so the carbon with less substituents would be more electron-poor.... So why does the water attack the more substituted carbon?
Secondly, in the image below (sorry it's turned), where the second proton transfer is taking place, you see the water molecule pulling the hydrogen off of the more substituted carbon. But wouldn't the hydrogens on the methyl group adjacent to the positive charge be way more electrophyllic, thus creating a double bond on the end of the chain?
I appreciate any help on this. Thanks for your time and effort!

These are both fairly standard questions which will be answered by any good organic chemistry text in detail. If you have access, Warren & Clayden- Organic Chemistry is excellent.

I have been searching in my textbook and online. My textbook does not seem to explain it well, and it has been very difficult to find online due to the very specific questions I am asking. That was why I came here.

I have a copy of the book I mentioned in front of me, the questions are answered in any undergraduate text (i.e. not very specific at all). But I dont mind trying to answer if that would help, I think my point was that a decent book would probably do it more succinctly

Water can attack both carbocation and hydrogen atom. Sometimes it attacks C, sometimes it attacks H.
Whenever it attacks C alcohol forms so the reaction goes back to reactants. When it attacks H the product forms. We are interested in product, so we show only this path.


text:
Derivation of the form of the time correlation function in NMR spectroscopy
For a given protein, I know that the NMR spectrometer magnet generates a field $B_0$ and that the interactions with the spins in the local environment generates a much smaller field $B_\mathrm{loc}$ (not necessary aligned with $B_0$).
Owing to Brownian collisions between solvent molecules and the protein, the atoms associated with these spins move, hence making $B_\mathrm{loc}$ a function of time ($B_0$ is constant).
We define a time correlation function$$G(t,\tau) = \overline{B_\mathrm{loc}(t)B_\mathrm{loc}(t+\tau)},$$ where $G(t,\tau)$ is a stationary random function (see Time evolution of correlation functions (specifically Onsager's hypothesis) in time correlation link)
Hence $G(t,\tau)$ only depends on $\tau$, the delay in measuring $B_\mathrm{loc}$.
So, for simplicity we set $t = 0$ and note that
$$G(t,\tau) = \overline{B_\mathrm{loc}(0)^2} \mathrm{e}^{-\tau/\tau_\mathrm{c}},$$
where $\tau_\mathrm{c}$ is the correlation time, the time it takes for the whole molecule to rotate by 1 radian in a process called rotational diffusion.
How do I derive the second equation from the first?

You can't prove this, simply because it is not generally true. But it still makes sense to assume this functional form because it seems to fit to many systems.

Your function $G(\tau)= \langle B_x(t)B_x(t+\tau) \rangle$ is an autocorrelation function. If it is only a function of $\tau$, the time delay, and not t, this is called the stationary assumption.
The fluctuating field has  zero average $\langle B_x(t) \rangle =0$. The bracket implies averaging over a long time for a single spin or average over many spins at a particular time, which is the ergodic hypothesis.
The magnitude of the fluctuating fields is the mean square fluctuation $\langle B_x^2(t)\rangle \ne 0$ and is positive thus the mean square is not zero.
It is also necessary to know how rapidly the fields fluctuate and this is where the autocorrelation $G(\tau)= \langle B_x(t)B_x(t+\tau) \rangle \ne 0$ comes in. We compare the field at one point t with another point $t+\tau$ some time later. If $\tau$ is small then the value of $B_x(t)$ and $B_x (t+ \tau )$ are similar and $B_x(t)B_x(t+\tau)$ is large and positive, by comparison if $\tau$ is large the product $B_x(t)B_x(t+\tau)$ is small approaching zero as $\tau$ increases.
Thus the general shape is that $G(\tau)$ is large at small $\tau$ and decays away to zero with increase in $\tau$. Often we assume that the decay is exponential thus $$G(\tau)= \langle B_x^2 \rangle \mathrm{e}^{-|\tau|/\tau_\mathrm{c}}$$ where $\tau_\mathrm{c}$ is a correlation time. This has the correct qualitative form but is difficult to justify on the basis of proper theory.
You state that $\tau_\mathrm{c}$ is caused by rotational diffusion thus it will, in general, be small in fluid solution and large in viscous ones. Of course $\tau_\mathrm{c}$ also depends on the temperature (smaller at larger $T$) and size of the whole molecule's rotation (as a prolate or oblate ellipsoid) or of a  mobile group therein, depending upon which property you are observing. Groups in a protein can undergo 'wobbling in a cone' type motion in addition to rotation of the whole protein. Note also that if the molecule is not spherical then it will have more than one rotational relaxation time so $G(\tau)$ may decay with more than one $\tau_\mathrm{c}$. Hope that this helps!

Thanks, so no way to prove that assumption? I am assuming spheric molecule and there is an average $\tau_c$ for whole protein.

And if you could explain the real significance of the spectral density function I get when I Fourier Transform $G(\tau)$ that would be great. It is of the form $J(\omega) = \frac{2\tau_c}{1+\omega^2\tau_c^2}$

I am trying to prove the exponential quality of $G(\tau)$, so all help would be appreciated.

@HighEnergy The Fourier transform of an exponential is the Lorentzian function $J(\omega)$ so its the same thing as $G$ but in the frequency domain. It may make it easier to interpret to look at it this way. You should also probably look at the Weiner-Kninchin relations as these allow you to find the power spectrum (or spectral density) of an autocorrelation.

You gave me information that I needed (and knew somewhat since I had looked at the Weiner-Kninchin relations). I am curious for a given frequency what $J(\omega)$, the spectral density function, means. Also, could I use the fluctuation-dissipation theorem to show that because NMR relaxation (dissipation) is exponentially decaying the time correlation function of $\mathrm{B_{loc}}$ (fluctuation owing to Brownian collisions) should as well? That is my ultimate goal.

$J(\omega )$  is the power (or average energy) in the frequency range $\omega $ to $\omega  +-\delta \omega $, i.e. if the signal is $f$ the average energy is $<f^2>$ and this is because the autocorrelation signal is related to the energy or power in a given signal via. Wiener-Kninchin eqns. Note also that autocorrelation produces the square modulus of its fourier transform, thus as all phase info is lost and so the fourier transform cannot be inverted to get to the original function.  Trying to use the fluctuation-dissipation theorem is a good idea. Look at rotational motion.


text:
Spontaneous reaction between chromium(III) and aluminum

I'm trying to rationalize this problem. It's been a while since I've done redox problems. So here goes, and any input would be great. Basically - am I right?
Here is what happens when you store chromium ion in an aluminum container:
$\ce{Cr^3+ + Al -> Cr + Al^3+ ~~~~~~E^{o}_{cell}=}$ ${+ 0.94}$$\ce{V}$
$\ce{Cr^3+}$: oxidizing agent
$\ce{Al}$: reducing agent
$\ce{Cr}$: conjugate reducing agent
$\ce{Al^3+}$: conjugate oxidizing agent
This reaction is spontaneous because we are going from strong to weak (just like in a spontaneous acid-base reaction).
So if we are going from strong to weak, all these statements are true:
$\ce{Cr^3+}$ is a stronger oxidizing agent than $\ce{Al^3+}$.
$\ce{Al}$ is a stronger reducing agent than $\ce{Cr}$.

And what is your question?

@Shadock - am I right?

For me the reaction you are considering is not correct. Because the reduction potiential of Chromium is -0.74V and the one of the Aluminium is -1.66V.

@Shadock - it is still a spontaneous reaction. The standard cell potential is positive.

Oops I haven't read the problem correctly. Let me think

Yes, your rationalization here is correct. Chromium ions will oxidize aluminum container and deplete chromium while contaminating the solution with aluminum ions. It's exactly the same process, as for instance, copper nitrate with an iron paper clip experiment that we have all seen in a chemistry class at some point. But here you have an aluminum container instead of a paper clip :)


text:
What is incorrect about this depiction of the hydrolysis of an ester?
First of all, I'd like to apologize for again asking a question solely featuring the deficiencies of my textbook.
I have a few specific questions, and a general one about the image at the bottom of this question:

Is the product really an acid? It doesn't even have an $H^+.$
It's stated that hydrolysis can happen when esters are treated with an acid as well as a base. But acids only donate $H^+$ so where would the extra $O$ come from to produce an alcohol as a product?
It's stated that a molecule of water is produced. Is this statement incorrect, or is the diagram incorrect?
In general, how would we change the diagram/information below to make it correct?

If you can answer any/all of these questions, I would greatly appreciate it.

Thanks again for this professional community.

In Figure 8, the alcohol oxygen should not be red.

Can I ask what the book is as it seems poorly written?

@Beerhunter It's Nelson Chemistry 12, a terrible, terrible high school textbook from Ontario, Canada.

There are indeed two mechanisms for ester hydrolysis, one using base catalysis and one using acid catalysis. The mechanisms of both are provided below (as an aside, you mention that your textbook isn't great- you should consider getting a different one, even if your teacher gives assignments out of the other book, it'll give you a second opinion and something to cross check your book against).

Base catalysed ester hydrolysis

Source: Organic Chemistry (Wothers and Clayden)
Acid catalysed ester hydrolysis

Source: Organic Chemistry (Wothers and Clayden)

Is the product really an acid? It doesn't even have an H+.
Yes, it is an acid. The acid is actually initially formed however because of the basic conditions of the base catalysed mechanism it gets deprotonated to give the carboxylate. Upon workup the carboxylate salt (in your example the sodium salt of the carboxylate) would protonate to give the carboxylic acid (COOH).
It's stated that hydrolysis can happen when esters are treated with an acid as well as a base. But acids only donate H+ so where would the extra O come from to produce an alcohol as a product?
The product is a gain a carboxylic acid, and as you can see from the mechanism above, water is involved, which is where the additional oxygen comes from.  Its worth noting that with the acid catalysed mechanism the whole thing is reversible.
It's stated that a molecule of water is produced (in the base catalysed mechanism). Is this statement incorrect, or is the diagram incorrect?
It could be water or an alcohol depending on whether hydroxide or the alkoxide does the deprotonation. Water is most likely however, as shown above. In the acid catalysed mechanism it is the alcohol that is formed.
In general, how would we change the diagram/information below to make it correct?
See the mechanisms above, taken from Clayden and Wothers. In general the diagram in your textbook looks essentially 'fine', though I dont entirely like the way the structures are drawn (its a very american textbook way of doing it.)

Good answer!
I'd reinforce that the book doesn't seem very didatic, when it doesn't show the mechanisms and write "acid" under the conjugate base of the acid. $\ce{ROO-}$ isn't an acid (actually it's a base), but it's a result from deprotonation of the carboxylic acid.

I think the alcoxides are more nucleophilic than the base hydroxide though. At least when we conduct a base catalysed esterification reaction, we first react the base with alcohol, to form alcoxides, and the latter make the nucleophilic attack on the acylglycerides. If hydroxide was a stronger nucleophile, it would make more sense to use it directly and then react the formed carboxylic acid with the alcohol.

@IanC, in what step are you referring to, and for the acid/the base?    I don't quite understand

Sorry, I was a little confusing in my comment. For the step of deprotonation of the base catalysed hydrolysis, you said that water was more likely to deprotonate the carboxylic acid instead of the alcoxide. If the concentration of hydroxide is much greater than the concentration of alcoxide this makes sense (which I think is the case in the reaction discussed), but it's good to point out that alcoxides are stronger bases than hydroxides and if present in the same ammounts will more readily attack the carboxylic acid.

@NotEvans. Wouldn't be better to say that the first mechanism is base-induced and not base-catalyzed as the $\text{OH}^{-}$ isn't produced at the final step?


text:
Which base pair is more stable, U:G or I:C?
As a result of damage (usually deamination), DNA will sometimes contain uracil and inosine bases. I am trying to determine which unnatural base pairing combo would be more stable: uracil paired with guanine or inosine paired with cytosine. See the images below.

Tinoco has a series of papers on the subject, usually in NAR.   http://nar.oxfordjournals.org/content/10/1/341.full.pdf

One can discuss the strength of H-bonds, but that would be half of the picture. It is more important how well (or poorly) this pair fits in the whole helical structure. Stacking interactions (arene-arene interactions) are playing important role in stabilization of DNA. If the pair disturbs backbone shape the binding will be poor. Google scholar search for relevant papers might be the most productive way to go.


text:
Lemon juice and secret messages
The experiment is described in the following link: http://www.sciencekids.co.nz/experiments/invisibleink.html
I carried out a small research but I was unable to find what exactly happens. In other words which particular acids, present in lemon juice, are oxidized (e.g. citric acid, malic acid etc.)? And what causes the brownish colour?
Any kind of help would be appreciated.

Possible answer, http://www.scientificamerican.com/article/bring-science-home-invisible-ink/

Actually my question is older than the one presented by Mockingbird.

Lemon juice contains sugars (carbohydrates) and acid (lemon acid). Sugars turn into carbon+water when heated in presence of acid. This is a standard thin layer chromatography analysis for sugars (spray plate with acid and heat up - black spots are sugars). The reaction is
C12H22O11 (sucrose) + H$^+$ (catalytic) -> 12C (charcoal) + 11 H2O (steam).
Paper also has polysugars (cellulose), but cellulose is more robust. Eventually even paper will turn black by the same reaction.
You can use sugar + vinegar as the ink. Result will be the same. It proves that decomposition of sugars in presence of acid is the reaction responsible for revealing this ink.

authors: Αντώνιος Κελεσίδης


text:
Iron/Iron(III) Concentration Cell
From what I understand about concentration cells - which I never learned about when I took general chemistry - they rely on the fact that if we have differing concentrations of some solute separated by a permeable membrane, they will try to exchange solute and/or solvent to achieve the same concentration or as close to it as possible. Correct?
Now, in a concentration cell, there is no physical exchange of ions. Instead, a salt bridge facilitates the flow of electrons, which either:
a) helps convert solids into ions
or
b) helps convert ions into solids

So in the above problem, we obviously have unequal concentrations of iron(III) ions. We can describe the problem using cell notation:
${|Fe|Fe^{3+} (4.8*10^-7 M)||Fe^3+ (1.3*10^-3 M)|Fe|}$
The flow of electrons will be from half-cell A to half-cell B. These are the reactions:
Half-cell A: $\ce{Fe -> Fe^3+ + 3e^-}$
Half-cell B: $\ce{Fe^3+ + 3e^- ->Fe}$
This increases the $\ce{Fe^3+}$ concentration in half-cell A while decreasing the  $\ce{Fe^3+}$ concentration in half-cell B. This will continue until both half-cells have the same $\ce{[Fe^3+]}$.
So, we can also describe half-cell A as the anode, because it is where oxidation occurs, and half-cell B as the cathode, as it is where reduction occurs.
So far so good?
And if we want to calculate the electrochemical potential of this particular half-cell (wired up to make a spontaneous reaction), we can apply the Nernst equation as follows:

So wait, what's the question?  Are you looking for clarification on what's occurring in a redox reaction?  Are you trying to determine if this is a valid cell?

@Checkmate - is what I posted correct?

This all looks correct.  It's worth noting that the voltage generated by this cell occurs through the flow of electrons from half cell A to half cell B as the system reaches equilibrium.  It's also worth noting that the final concentrations of ions should eventually become nearly identical given enough time and free interaction between the two half cells (at which point the voltage becomes 0).
For further reading, this site has some good examples:
http://chemwiki.ucdavis.edu/Core/Analytical_Chemistry/Electrochemistry/Voltaic_Cells/Electrochemical_Cells_under_Nonstandard_Conditions/Concentration_Cell
While wikipedia has a (brief) overview of the theory:
https://en.wikipedia.org/wiki/Concentration_cell


text:
What would the pH change be if I drop one gram of sodium hydroxide into a pH 8.1 solution
I have an experiment where I drop 1 gram of lab-concentration Sodium Hydroxide into a litre sea water (pH 8.1). What should be the expected change in pH? Would this change be constant? i.e if I was to drop 2 grams of NaOH into the sea water, would the change double?

As I understand, you're looking for a more or less precise calculation? If so, we need to know the volume of said "little sea water" since the pH would depend on the resulting concentration of the sodium hydroxide. Also, what do you mean by "lab-concentration" sodium hydroxide??

@ChemistryHelpCenter thanks a lot for replying. The volume is a litre of sea water. Lab concentration means about pH 13-13.4

If we know that the pH is 13 (let's take it as 13 for the simplicity), then we can back-calculate the concentration of sodium hydroxide in this solution from $\ce{14=pH + pOH}$. This gives us pOH=1, thus the concentration of sodium hydroxide of 0.1M. Since 0.1M sodium hydroxide is hardly a very concentrated solution, we can safely assume that 1 gram of the solution is going to be roughly 1 mL. Thus, with each 1 gram of the lab solution you're adding about 0.0001 moles of $\ce{NaOH}$ into your 1 liter of sea water giving you $1*10^{-4}$M solution with the pH = 10.
Using the same trick to calculate the concentration of $\ce{OH-}$ in your sea water as we did for the stock sodium hydroxide solution, we get about $1.25*10^{-6}$M, which is negligibly small compared to $1*10^{-4}$M solution that you get after adding sodium hydroxide to it. Thus, after 1 gram of sodium hydroxide in a liter of sea water your pH should go from 8.1 to about 10. If you add 2 grams, you'll get a $2*10^{-4}$M solution, which yields pH of about 10.3.


text:
how to mathematically predict reaction results

I've been trying to computationally achieve the following (I have no lab, I'm 14 years old, and I want to make a Minecraft mod to teach real Chemistry):

I want to mathematically predict the range of temperature and pressure (if the reaction is between gases) or temperature and concentration (if the reaction is in a solution) in which a given chemical reaction can happen, and the rate at which it happens, with the variation of those two arguments.

EDIT 1:

Regarding the first part of @sixtytrees answer I suppose I must calculate $K$.

I assume $\Delta G$ would be:
$$\begin{align}\Delta G &= \Delta G_\mathrm f^°{_\text{methanol}} - \Delta G_\mathrm f^°{_\text{carbon monoxide}} - 2 \Delta G_\mathrm f^°{_\text{hydrogen}}\\[6pt]
&= -162.5 - 137.3 - 0\\[6pt]
&= -299.8\ \mathrm{\frac{kJ}{mol}}\end{align}$$
Now to solve for $K$:
(I believe $R$ must be $0.0083144598\ \mathrm{\frac{kJ}{mol\ K}}$, because $\Delta G$ is in $\mathrm{\frac{kJ}{mol}}$
Also, I think we have to take $25\ \mathrm{^\circ C}$ as the temperature which is $298\ \mathrm K$)
$$\begin{align}\ln K &= \frac{\Delta G}{-RT}\\[6pt]
&= \frac{-299.8}{-0.0083144598 \cdot 298}\\[6pt]
&= \frac{-299.8}{-2.4777090204}\\[6pt]
&= 120.99887336713996006405304847878\\[6pt]
K &= \mathrm e^{120.99887336713996006405304847878}\end{align}$$
And now I realize I must have made some mistake because this number will be too big.
☹

EDIT 2:

I just woke up and I noticed I miscalculated $\Delta G$, because I added them together when I should be subtracting. Let's see how it goes now.

$$\begin{align}\Delta G &= \Delta G_\mathrm f^°{_\text{methanol}} - \Delta G_\mathrm f^°{_\text{carbon monoxide}} - 2 \Delta G_\mathrm f^°{_\text{hydrogen}}\\[6pt]
&= -162.5 - (-137.3 - 0)\\[6pt]
&= -25.2\ \mathrm{\frac{kJ}{mol}}\end{align}$$

Now $K$ will be:

$$\begin{align}\ln K &= \frac{\Delta G}{-RT}\\[6pt]
&= \frac{-25.2}{-0.0083144598 \cdot 298}\\[6pt]
&= \frac{-25.2}{-2.4777090204}\\[6pt]
&= 10.170685820053125395644218884808\\[6pt]
K &= \mathrm e^{10.170685820053125395644218884808}\\[6pt]
&= 26126\end{align}$$

Still not as small as 10 that @sixtytrees predicted.

$$26126 = \frac{[\ce{CH3OH}]^1}{[\ce{H}]^2[\ce{CO}]^1}$$

What now?

EDIT 3:

I've been thinking, and decided Chemical Equilibrium is not something I need to bother with, because as the reaction progresses, it is assumed that the final products are drained out as they are formed, and additional pressure is added to account for the reduction of total molarity. This way I don't need to distinguish between reversible reactions, irreversible reactions, and virtually irreversible reactions (when there are trace amounts of the reactants in the final products).

This only works if the reactants are gases, I don't know how to sustain the rate of the reaction if I take away the products from a solution.

The original question remains, now that equilibrium is irrelevant, is it possible to predict the temperature and pressure/concentration range of the reactants in which the reaction can occur?

I downloaded Orca, by the way. No idea how to use it.

EDIT 4:

Ok, so let´s say this is all happening in a sealed container with 1L of volume, so

$\begin{align}
26126 &= \frac{[CH_3OH]}{2^2 \cdot 1^1}\\
[CH_3OH] &= 26126 \cdot 4 = 104,504
\end{align}$

What does this mean? It can't mean I end up with more than one hundred moles in 1 liter. It makes no sense, I don't have enough reactants to get so much product. So, I am clearly doing it all wrong, can someone explain it to me?

Have a look at wikipedia pages on rate equations. There are many solved equations for simple reactions but complex equations have to be solved numerically using standard numerical methods, Euler to Runge-Kutta etc.

@FinnTheHuman, describe, what do you want to include in your engine and we might suggest what is doable and what is not. Keep in mind that scientists cannot model a single cell from the first principles. Currently they are modelling a small section of mouse brain in real time, but they model individual cells. Computational study is the future of science, so you are looking in the right direction.

Ignore downvoters - you are here to find answers to your valid questions, not to win the popularity contest.

@sixtytrees I tried to follow your answer to find K so I could calculate the rest, but K ended up being an extremely huge number, so something must be wrong with my calculations

This is a really cool and inventive mod idea - good job.

@ChildishJack Thank you very much. The idea started as just a new energy system based in real physics and chemistry, but now I want to have a "Chemistry Workbench" in my mod, where people can make chemical reactions happen. I'm still not sure about how it will work. I know there will be input and output valves for gases/solutions (the input for the reactants, the output for the products), there will be slots for catalysts, but I won't think about that for now, and there will be a way to fine tune temperature and pressure (at the expense of the energy your engines will be generating)

Regarding Edit 4, your answer is incorrect because you are not considering the "extent of reaction".  You need to solve for the final amount of MeOH as a function of the starting amounts of reactants.  This is generally handled by declaring a variable that represents the extent.  See this PDF, particularly the bottom of page 4 where the concept of extent is introduced mathematically.

To provide a little context, entire books, degrees and careers are focused on answering the question of "how do chemicals react under specific conditions".  I think your goal to introduce some of these calculations into a Minecraft mod is interesting, but I suspect you will benefit greatly from limiting the scope of the chemicals and reactions you want to consider.  One suggestion: starting with combustion reactions would give a range of chemicals and variables that are still interesting.  Plus it is a practical process for providing motive force.

There are two parts to quantification of reactions:
(1) How far it goes? (Thermodynamics, equilibrium). This all is dictated by ground state energy.
(2) How fast it goes? (Kinetics, reaction rates). This is dictated by  transition state energy vs ground state energy.
Equilibrium is much easier to understand. Let [Y] be concentration of reagent Y. You say that for 2A + B = C+3D +4E you have:
K = [C]$^1$[D]$^3$[E]$^4$/([A]$^2$* [B]$^1$)
Note how powers are equal to coefficients in the equation.
$\Delta$G= -RT*ln(K)
$\Delta$G is Gibbs energy (measured and tabulated for all of inorganic chemicals from school textbook).
R is a constant (Universal gas constant)
T is temperature.
How to use it? Well, imagine that you have
2*H2+ CO <-> MeOH
Lets say that K=10 under that conditions. I am sure I am off. Correcr way to find it is to use $\Delta$G= -RT*ln(K). You know T, R is constant. $\Delta$G is known for MeOH, H2 and CO. Calculate the difference to find $\Delta$G. You will need to find $\Delta$G for methanol vapor or take it for liquid MeOH and add $\Delta$G of evaporation.
You start with 1M of CO ( atm at 273C) and 1.5M of H2 (67.2 atm at 273C). Total pressure 112 atm. (Mol to pressure conversion using ideal gas law). What will be the equilibrium amount of MeOH if ?  You say that
Let x be the amount of product at equilibrium. H2 = 1.5 M - 2*x and CO = 1-x.
x/(1.5-2*x)$^2$*(1-x)=10=K
Now you have a cubic equation to solve (1.5-2x)$^2$*(1-x)*10 - x = 0;
With more reagents you have algebraic equation of higher power, but there are methods to solve those numerically and the idea is simple enough to explain.
This gets a bit more complicated when you have multiple reactions going. But This is the basis. When system gets complex you start ignoring some small members to simplify it. (1.0001)$^5$ ~ 1.0005 (reasonably accurate). YOu do the same in chemistry.

Kinetics is trickier. For A+ 2B -> C the first approximation is that
rate = k*[A]*[B]$^2$.
Also "k often increases by a factor of 2-4 per 10C of increase in temperature".
Now to treat it accurately you need to solve a system of equations. each equation is either
r$_i$ = $\delta$[A$_i$]/$\delta$t
or
r$_i$ = ($\delta$[A$_i$]*$\delta$[B$_i$])/$\delta$t
Rate if i$^{th}$ reaction = change of concentration of i$^{th}$ reagent per time period as the period goes to zero).
You end up with a system of ordinary differential equations. But don't get tricked by the word "ordinary". Most of them don't even have solutions in "elementary functions" (functions that you study in school or university as a chemist.
To understand what is a differential equation you would need to know what is an integral. We study differential equations in second year of university and that probably the most difficult subject in university.

Predicting reaction results does not necessarily have to be so complicated. Consider a simple system of reactants - for example if you made esters from alcohols and acids (example: ethyl alcohol plus acetic acid -> ethyl acetate) you could predict the products (and their relative amounts) from a given set of reactants in different quantities, because formation of an ester bond is pretty much the same energetically, regardless of which acids and alcohols you use. If you started with three alcohols and three acids, there are 9 possible esters. It gets more interesting when you add in difunctional acids and alcohols - then you start to get polymer chains. This kind of system can be simulated using a simple computer program.

You can start with a rough model and then gradually improve it. Doctors start with frogs in their carrier, then remove appendix, then do hart surgery. You should do the same. Choose 30 chemicals. Record properties of each chemical (melting/boiling point, density, molecular weight, color, solubility in (or reaction with) water. Now for each pair of chemicals for (30*15 combinations) record if they react with each other in dry state, in water, in acidified water and in basic water, the rate of the reaction, the products (gas/precipitate/color change). 1000 conditions is a lot of typing, but it is worth it. Besides, most of the cases will be "no visible reaction". You need a good reference book to get most of the data. For the rest of the reactions we can help. You can see what reaction looks like on youtube to know what to show in your mod. You will have your virtual lab where you can set stuff on fire or blow up things. Include this in your mod and see how it goes. Do people play it, do they ask "give us more explosions, we don't want precipitates? Expand the mod as needed.

But is my Edit 2 correct so far?

Judging by your attempts it looks like you haven't learnt about thermodynamics  yet nor calculus nor advanced probability. The fundamental  problem in your attempts is that you don't take in account the energy transfers, activation energy, etc. These energy changes that occur are scientifically represented in curves, you are trying to solve a certain part of that curve. Therefore, you would use a combination of differentiation  and integration to do that. Secondly, you see if reactions  were mathematically deterministic then there would be no use for a different language for chemistry (I assume you know basic chem). Finally, chemistry is not structured as maths, in other words it is just a hodgepodge of substance that react under the right circumstances.


text:
What would happen if I mixed a single water molecule in a beaker of hexane?
What would happen if I mixed a single water molecule in a beaker of hexane? Would it sink or float?
How about 2 water molecules? 3? Etc?
In other words, how many water molecules does it take to be hydrogen bonding together to become more dense than hexane?

This is a clever question because the molecular weight of water is only 18.01528 g/mol while the molecular weight of hexane is 86.17536 g/mol. Naturally, you would expect the lower mass to rest on top, but water has a higher density... and we all know that water will sink! But there are other forces at work.
So, supposing that you neglect the fact that one molecule of water would be in a vapor-state, Google says the surface tension of Hexane @ 20 °C is 18.43 mN/m... so placing the water molecule ontop of the hexane would probably cause the molecule to float (like a feather on water).
However, if you suberged the molecule it should dissolve (Google says that the solubility of water in hexane is 0.01% at 20°C). Now, once the saturation point of hexane has been met, such that one more single molecule of water would cause a change to occur, that change would probably result in a condensation and precipitation of the dissolved water, because the reality is, that the hexane will become super-saturated, such that the extra water molecule will cause many water molecules to fall out of solution.
Understanding that you are transferring energy is key to understanding this. Consider pushing a ball over a hill, once the ball reaches the top and begins to fall, it falls into the valley. Once you push a boat to the tipping point, it falls to the bottom.
So the number of molecules necessary for hydrogen bonding together to become more dense than hexane would be the number of water molecules dissolved at the maximum supersaturation point (maybe 0.0105%, +1 molecule) minus the number of molecules at the saturation point (0.01%)
Now having a saturated solution, with water on the bottom of the container, adding another molecule of water would generally cause one of the other dissolved molecules to fall out solution (because there is an equilibrium... some of the water molecules at the bottom are dissolving, while others are falling out of solution).

Need we consider Brownian motion here?

I think supersaturation is a moot point because it is usually achieved by raising the temperature of a solvent to allow more solute to dissolve before allowing the solvent to cool slowly. For hexane at 20 °C, it would be as you pointed out 0.1%. Then  if you add a single water molecule above that limit it would probably float to the top. However, once enough molecules are added (around 6 I think) to create a solvation shell around a single water molecule, then the density of that 7H2O complex would be higher than hexane and sink to the bottom.

@nova Supersaturation is practically done like in the way that you described, but the solution would also supersaturate by delicately adding one molecule at a time. And again, they would all float due to surface tension... they would need to be physically pushed to submerge. Then they would be dissolved; then the hexane would force a bunch of the other molecules out of solution... like rain.

Ok, good point. But my argument still stands for what happens once the supersaturated point is reached. Any addition molecules would begin to float up until enough amalgamate to create a sphere of hydration that would then sink.

@nova They would not float. They are dissolved throughout the solution. At supersaturation, they would continue to be dissolved in solution. Adding one more would cause a cascade of many water molecules to condense and precipitate.

It seems like you assume that water is not soluble in hexane at all. This is incorrect. It is not miscible, but solubility of water in hexane 0.01% at 20°C. That is 100 mg/L or 5.5 mM which is actually a decent amount of molecules. Distribution in liquid in a beaker will be the same at all heights.
A good question would be "what is the distribution of water molecules on the walls of the beaker and in the volume?" Beaker walls can accomodate small amount of molecules. The extent of absorption is following Langmuir adsorption model. It can be ignored at high concentrations (mM of dissolved material), but starts playing a significant role when you go into nanomolar or picomolar range. At that concentrations the actual amount of material in solution is significantly smaller than calculated  because of absorption.

Suppose that you have a sealed beaker with some hexane liquid in it at room temperature. Now add one water molecule. The answer is obviously that the single water molecule will spend some time in the vapour phase and some dissolved in the hexane. Once the water molecule hits the liquid surface there is a chance that it will enter the liquid. If there is a strong interaction between the water and the liquid then the tendency will be to spend longer in the liquid than vapour. Hexane and water have a poor interaction, the hexane cannot effectively solvate a water molecule's dipole as it has no appreciable dipole itself and hence a small dielectric constant (relative permittivity)$^*$. The low dielectric solvent means that the electric filed of the water's dipole can spread out over many hexane molecules causing a positive interaction energy. Exactly how big is difficult to say.
The water molecule can occasionally overcome any intermolecular interactions in the liquid phase as there is an exponential distribution of energies in all molecules at a finite temperature. This is given by the Boltzmann distribution. Thus a few molecules have far greater energy than the average ($3RT/2$) and eventually these molecules will impart some of their excess energy to the water molecule and it will be ejected into the vapour phase. These processes will continually repeat themselves.
Now if we consider this from a thermodynamic viewpoint, we will have to assume numerous water molecules in the presence of the hexane.  We need to calculate the free energy which consists of the enthalpy change $\Delta H$ and entropy change $\Delta S$ as in the formula $\Delta G=\Delta H -T\Delta S$.
The enthalpy of water entering the hexane will probably be small and positive for reasons outlined above, but there is also an entropic factor. When two liquids are mixed the (ideal) entropy of mixing$^{**}$ is $\Delta S= RT(n_1ln(x_1)+n_2ln(x_2) )$ where $x_1$ is the mole fraction of water and $x_2$ that of hexane and $n_1$ and $n_2$ are the respective number of moles. As the solubility of water in hexane is low approx $10^{-2}$ molar, if not smaller, then the mole fraction of hexane can be approximated as 1 and  $\Delta S= RTn_1ln(x_1)$ which evaluates to approx -1 kJ/mol.
The enthalpic (heat) term can be shown to be $(n_1+n_2)x_1x_2w$ where $w$ is an energy term allowing for the interaction between solute-solute $\epsilon_{11}$, solvent-solvent and solute-solvent molecules, $w \approx 2\epsilon_{1,2}-\epsilon_{1,1}-\epsilon_{2,2} $ where $\epsilon$ represents an interaction energy. As the interaction between water and hexane is smaller than hexane-hexane interaction($\epsilon_{1,1}$) and water-water interactions, the interaction energy is positive. This means that the free energy for dissolving $\Delta G$ is going to be small and probably negative overall, and this means that the solubility is going to be small also.
As the amount of water increases approaching the solubility limit there will be microscopic regions of pure water and pure hexane until the water separates and being the most dense of the two liquid forms the lower layer.In this situation water is dissolved in the hexane, to its solubility limit, and hexane also in water. The extent of solubility is given by Henry's law. The vapour pressure P of a solute above a solution is given by $P= k_Hx$ where x is the solute mole fraction and $k_H$ the Henry Law constant which is only determined by experimental and depends on solute/solvent and temperature. Values of $k_H$ (in units of Bar) are several thousands for relatively insoluble solutes such as water in hexane or vice versa.
$^*$ A low dielectric constant $\epsilon _0$ means that an electric field spreads a considerable distance r from its source compared to a solvent with a high constant.  The interaction energy scales as $\approx 1/(\epsilon _0 r)$, low dielectric (alkane) solvents have $\epsilon _0$ in the range 2-4, whereas water is 78 so the effect is considerable.
$**$ Entropy is defined from statistical thermodynamics as $S=k_Bln(\Omega)$ , where $k_B$ is the Boltzmann constant and $\Omega$ the number of configurations available to the system. The calculation of the entropy starts by working out the number of ways of placing molecules into sites. Suppose that there are N1 and N2  molecules and that they can occupy in total N1+N2 sites. There is a chance of putting the first molecule into N1+N2 sites , the second in N1+N2-1 sites etc. The total number of possibilities is (N1+N2)!. However, molecules of one type cannot  be distinguished from one another thus we must divide by N1! and similarly for N2 by N2!. This produces $\Omega=\frac{(N1+N2)!}{N1!N2!}$. Using Sterling's approximation for factorials the equation for $\Delta S$ can be obtained.
This is the answer to the previous version of the question with ethane instead of hexane.
Ethane is a gas, did you mean ethane? Assume that your beaker is sealed but in thermal contact with the outside. i.e. sitting on the bench. On adding a single water molecule the thermal energy of the ethane will impart its energy to the water and it will be found in the gas phase. If you keep adding water the same thing happens but eventually liquid water will form at the bottom/sides of the container in equilibrium with water vapour; rate of evaporation equal to rate of condensation. Adding more water will allow more ethane to dissolve as the water volume and gas pressure increases in the container (as it is sealed). Some small amount of the ethane will also dissolve in the water. The amount is determined by Henry's law.

Changed ethane to hexane. Typo.


text:
Why is there a discrepancy among sources on the atomic radius of some elements?
My book says the atomic radius of gallium is less than that of aluminium, but I found out different atomic sizes on different sites.
For example, this says gallium is 136 pm and aluminium is 118 pm, but Wikipedia says Ga is 130 pm, Al is 125 pm.
Both of them give different atomic radii but are consistent in ordering: both sites say Ga is larger than Al. What is correct?
My book says atomic size of Cr, Mn, Fe is about 117 pm and Zn is 125 pm; these again differ from the values give in the above websites. Which source provides correct values of the atomic radii?

You need to check is what value is being quoted and how it was obtained. Usually the radii are called 'hard sphere radii'  or 'van der Waals packing radii' and 'bare ion radii' for ionic crystals. Results are obtained from x-ray and neutron diffraction studies on solids, (especially crystals) and from gas solubility , viscosity and self -diffusion data on liquids. The values often differ by only a few percentage points. The data you have may also be obtained from theory. Values of most atoms are between 0.1 and 0.2 nm. If ions are hydrated the radius is different from the bare ion.

to expand on porphyrin's answer just a bit, such a atomic size is "averaged" also. So Na is a different size in in solid crystals of NaF, NaCl, NaBr and NaI.

http://chemistry.stackexchange.com/questions/42831/why-ionic-radii-of-cu2-is-less-than-zn2

Atomic radii are not really well-defined. There is a great deal of uncertainty associated with them, mainly due to the somewhat philosophical question ‘where does an atom end?’ Orbitals, in theory, extend into infinity for all (mathematical) intents and purposes, yet the probability of locating an electron in Far, Far Away is zero by all sensible rounding methods. However, the decrease is exponential, so you need to define a cutoff value if you want to know where an orbital ends to draw the beautiful pictures you probably have seen before.
This only gets worse when considering entire atoms rather than orbitals. We typically consider atoms as spherical (hence radius). But for every element except hydrogen, helium, lithium and beryllium (possibly including boron, too) there is at least one additional non-spherical orbital occupied. Can we still truly define an atom as a sphere? And even if we can, consider the case of transition metals; should we be talking about some cutoff value for the outermost s orbital or about one for the lower shell’s d subshell, whose electrons are considered valence electrons?
Now in theory, you could ignore all those philosophical discussions and just take the solid-state structure of an element, measure the shortest distance between two nuclei, divide it by 2 and call that the atomic radius. Well, that is what is done. However, many elements have different structures that they adopt or can adopt at room temperature and standard pressure. For other, especially non-metals, the shortest interatomic distance is not an atom-atom contact but an interatom bond. In a select few, the interatomic bond can have different bond order values; sixtytrees noted carbon whose most stable phase is graphite (bond order 1.5) but which also features a diamond allotrope with a bond order of 1.
The covalence problem can be overcome with a set of correction factors that introduce further uncertainties and are added differently by different sources.
All these issues mean that there is not one ‘true’ way to determining atomic radii; and since the methods differ, so will the results. It is only natural for some elements to be larger than others in one scale but smaller in another.

The whole problem gets infinitely worse with ionic radii. Ionic compounds exist as crystals with anions and cations — but where to the cations end and the anions begin? How do we deal with the fact that no bond is truly ionic and no bond between two different elements truly covalent?
Even if you determine e.g. the ionic radii of $\ce{Na+}$ and $\ce{Cl-}$ by various methods, (e.g. from comparing sodium fluoride, sodium carbonate, magnesium chloride, ammonium chloride …) and then add the two values together to arrive at an interionic distance in the $\ce{NaCl}$ crystal, it is very likely that your result is off.

Atomic radius is harder to work with than ionic radius. Ionic radius can be measured from crystal structure of an appropriate salt. Ionic spheres are hard and not really compressible.
For atomic radius consider carbon. Should we assign radius based on diamond (bond order = 1), graphite (bond order = 1.5) or polyacetulene (bond order = 2)?


text:
Cutting a Solid with a sharp edged tool

The third point of difference suggests that amorphous solids can't be cut into two parts of regular surfaces
But glass being an amorphous solid is cut into regular pieces often.
What am I getting wrong?

Your textbook is correct. Glass surface may appear smooth, but in reality it is really rough and irregular if viewed under a microscope. Here you will find some pictures. Have you ever seen an insect walking on a window? It is possible because of those irregularities on the glass surface. Insect legs have tiny hooks which graple to all these lumps, cracks and holes on the glass.

everything is rough when viewed under a microscope. I don't think anyone can possibly attain a smooth surface. The problem is that how can I judge if a solid is uniform  throughout(crstalline) or not (amorphous) on the basis of whether it can be cut smoothly thorugh just one of its layers


text:
Thermodynamics in Solution (solvation) from quantum chemistry
The gist of the question is "how do I get a heat of formation in solution" - but let me give it some context to be more specific.
I am interested in calculating the thermodynamic values of molecules in solution. Due to the number of jobs, time etc. I have to use DFT methods. These are not as accurate as for example CCSD related methods but still provide good answers in reasonable timeframes (plus I'm checking up on estimated values, so I don't immediately need the highest acutacy).
As atomisations are hard to calculate with DFT methods (see a book by Cramer & comment on the ORCA forum), isodesmic reactions offer a better option.
So far so good. This allows me to use literature values for reference compounds to calculate the heat of formation of a species.
Now I need the heat of formation of the species in the liquid phase.
I can repeat my quantum chemistry calculations with a solvation model and again obtain values - however I now lack the reference compounds.
Based on my understanding of chemistry (see textbook by Cramer) I can relate the Gibbs free energy of the gas phase molecule with the solvated molecule.
So using the following:

$G_{liq} = G_{gas} + G_{sol}$
$\delta G = \delta H_f - \delta S ~T_{LO}$
I can work out the liquid phase heat of formation of the molecule.
(Please correct me if I am wrong.)

Now here is a little problem:
For organic molecules, there can be multiple local minima - but only one absolute total minimum.
The total minimum can correspond to a different structure in the gas and liquid phase, so do I compare structures and consider them or do I use the total minimum in both cases, I.e the lowest energy structure to calculate my thermodynamic data?
Reference:
Essentials of Computational Chemistry: Theories and Models, 2nd
Edition Christopher J. Cramer
http://eu.wiley.com/WileyCDA/WileyTitle/productCd-0470091827.html

The best way is to take all the structures into account and use Maxwell-Boltzmann distribution to weigh them accordingly. Here is what I would do
For each chemical structures/isomers (here I used F to denote Gibbs free energy),
$ F_{liq} =  F_{gas} +  F_{solvation}$
Now, $ F_{gas} =  E_{scf} - k_BTlnQ$
Where, $Q=q_{translational} \times q_{vibrational} \times q_{rotational} $
Using appropriate method (implicit solvation or QM/MM) you can find $F_{solvation}$. Now you have $F_{liq}$ for each isomers. Then use the following method to weigh them according to Maxwell-Boltzmann distribution.
In the above image, I use $\epsilon_i$ to denote $F_i$.


text:
How can I best achieve a stable solution of Pt-Nanowires in water
I am synthesizing platinum nanowires ($50nm×20\mu m$) in aqueous solution. Unfortunately they start to aggregate rapidly, which is not desired.
How can I keep them dispersed?

I think there are several things that could work for you, but I suspect the driving issue in your choice will be whether the additive interferes with your process.  Could you state in a few sentences the gist of your process?


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Cis/trans stereochemistry dictating pKa
Why is trans-4-phenylcyclohexane-1-carboxylic acid a stronger acid than the cis isomer?
In the cis-isomer the carboxyl group is axial; and, given the 1,4 distance from the phenyl, why would solvation be any more difficult?

Welcome to Chemistry.SE. This seems like it might be a homework type question. What are your thoughts on the matter?

Hi Ben, thanks for replying. In the cis-isomer the carboxyl group is axial; and, given the 1,4 distance from the phenyl,  why would solvation be any more difficult? These Qs can be maddening! Tony

@tony, do you have values for the pH?   What kind of order of magnitude difference is observeD?

The bulky phenyl group will preferentially occupy an equatorial position. Therefore, the trans acid has the carboxyl group equatorial and the cis acid has it axial, as you have rightly pointed out.
The difference in acidity is to do with the resulting solvation of the carboxylate group. The negatively charged carboxylate group is more strongly solvated by water molecules, and has a larger effective steric bulk, than the neutral, protonated form.
So, there is an increase in the steric bulk of the group upon deprotonation. Such a process would be more favoured if the group is equatorial, i.e. the trans isomer is more acidic.
Of course, in the absence of pKa values, such an explanation is only theoretical in nature. Solvation is a very difficult concept to quantify objectively.

A possible reason has to do with the need for the ionized carboxylate group to be solvated by water molecules--lowering energy by delocalizing its negative charge. More water molecules can contact/hydrogen bond with/solvate the carboxylate group of the trans isomer than is possible for the cis isomer, due to the phenyl group being placed closer to the carboxylate in the cis isomer.