brightsky's Chem Thread

[lzxnl]

This might sound like a dumb question, but why can't ethanoic acid function as a buffer by itself? Is it simply because there ain't enough ethanoate ions to consume extra hydronium ions introduced into solution with the addition of an acid?

A buffer works because any acid that is introduced is absorbed by the buffer and becomes an equimolar amount of weak acid. Similarly for bases. If you have CH3COO- and some CH3COOH, if you introduce some NaOH, the amount of CH3COOH drops and you get more CH3COO-, but that's a weak base so the change in pH won't be as large. The conjugates must BOTH be present.

Q2
My ‘intuitive’ method for determining the change in K for chemical systems is based upon the fact that high temperatures favour endothermic reactions, whereas low temperatures favour exothermic reactions. I will explain more shortly. But first, for any chemical equilibrium there is forward and a reverse reaction.
Now, if the forward reaction is exothermic ΔH<0 , that means the reverse reaction must be endothermic ΔH>0. At increased temperature, the endothermic reaction is favoured, so the reverse reaction occurs faster (greater rate of reaction) than the forward exothermic reaction does. This means that reactants are favoured more than products initially were. Hence, the point of equilibrium shifts to the left, meaning a smaller value of K. You can use this limited logic to deduced the effect on K of different systems under different temperatures.

Factors Affecting Entropy
The entropy of a chemical system is affected by:
Change in temperature
1.   Entropy increases as temperature increases.
2.   This is because as temperature increases, the molecules or ions undergo greater vibration (solids) or rapid motion (liquids and gases) and this reduces overall orderliness
3.   Note: at absolute zero, a substance has maximum order and so has zero entropy

You haven't really addressed WHY low temperatures favour exothermic reactions though. We can find that fact in the textbook; it's a matter of explaining why this is the case that I think is the question. I like Mao's explanation of using the Arrhenius equation; I tried to use it myself in explaining, but it's a bit confusing for those who aren't familiar with it.

Also, the third law of thermodynamics states that a perfect crystal at 0 K has zero entropy. That means, a PERFECT crystal with the minimum number of microstates has entropy 0. General non-perfect crystals will have higher entropy even at 0 K.
Particles still move at 0 K. Quantum mechanics forbids the absolute cessation of motion.

[brightsky]

A buffer works because any acid that is introduced is absorbed by the buffer and becomes an equimolar amount of weak acid. Similarly for bases. If you have CH3COO- and some CH3COOH, if you introduce some NaOH, the amount of CH3COOH drops and you get more CH3COO-, but that's a weak base so the change in pH won't be as large. The conjugates must BOTH be present.

but how do you apply le chatelier's principle to buffer solutions. there is more than reaction taking place. take a buffer solution consisting of ethanoic acid and sodium ethanoate for instance. so you start off with water. then you add ethanoic acid. hydrolysis occurs. CH3COOH(aq) + H2O(l) <--> CH3COO-(aq) + H3O+(aq). but the position of equilibrium lies to the left, and so you have more CH3COOH(aq) floating around than CH3COO-(aq) and H3O+(aq). now you add sodium ethanoate. the sodium ethanoate dissolves in water and you are left with Na+(aq) and CH3COO-(aq). Na+(aq) has no acid-base properties so we'll ignore it for the time being. since you have introduced more CH3COO-(aq), thus increasing the concentration of CH3COO-(aq), a nett back reaction will occur as the system regains equilibrium. so even more CH3COOH(aq) is produced. so you have a solution consisting of lots of CH3COOH, lots of CH3COO-, and some H3O+ to make the solution acidic.

now what happens if you add acid or a base? my understanding is that if you add an acid, you introduce more H3O+(aq) ions. most of these ions will, at some point, bump into a CH3COO-(aq) and react to form CH3COOH(aq). so concentration of H3O+(aq) does not increase that much, meaning that the pH will not drop dramatically, as one might expect. but the textbook explains it in terms of equilibrium. if you add H3O+(aq), then position of equilibrium will tip to the left, thus removing most of the H3O+(aq). which is correct? or do both of these events occur?

i have a feeling that the textbook's explanation is incomplete. this is why i spurred me to ask the previous question. if you just had a solution of ethanoic acid, then technically, if you add more H3O+(aq) ions, the same thing would occur. the position of equilibrium will tip to the left, and some of the H3O+(aq) ions will be thus removed from solution. yay or nay?


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also, a question regarding vce, is it true that our schools only pick one chapter to study out of ch19-22 (for those using the heinemann textbook)? will we be examined at the end of the year on the other chemicals?

[lzxnl]

Remember, it says that the system only partially offsets any disturbance in equilibrium, which means that the net result of adding H+ to a solution is to make it more acidic. PARTIALLY OFFSET
Add H3O+, yes, the system will try to use some of it up, but it can't use all of it up. You will still have more H3O+ in the end than initially.

I've heard that too. One non-examinable chemical to study

[brightsky]

Remember, it says that the system only partially offsets any disturbance in equilibrium, which means that the net result of adding H+ to a solution is to make it more acidic. PARTIALLY OFFSET
Add H3O+, yes, the system will try to use some of it up, but it can't use all of it up. You will still have more H3O+ in the end than initially.

I've heard that too. One non-examinable chemical to study

yeah which is why when you add an acid to a buffer solution, the pH still decreases, although to a much lesser extent than expected. but i don't think that really answers my question...

another question: why is the melting point of ammonia lower than that of water? technically, both can form 4 hydrogen bonds, even though water has an extra lone pair...

[09Ti08]

@ brightsky

Firstly, I think heinemann's explanation only focuses on the ethanoic acid and its ionization. It is true that an addition of H3O+ causes a net back reaction. However, I think that they have forgotten about the conjugate base. I think the H3O+ ions really have no choice other than reacting with CH3COO- ions. Normally, the buffer solution would contain equal molar concentration of both the acid and the conjugate base. Therefore, the majority of CH3COO- ions is the conjugate base produced by the dissociation of the sodium ethanoate, not by the ionization of CH3COOH.

For your second question, notice that a water molecule has two lone pairs and two hydrogens, therefore every one of them can involve in hydrogen bonding. Now consider the structure of ammonia, each nitrogen has only one lone pair but three hydrogens. There are not enough lone pairs to satisfy all the hydrogens. The amount of hydrogen bonding is limited. Therefore, the boiling point of water of water is higher than that of ammonia.

[lzxnl]

The equilibrium does tip to the left in both cases. But Le Chatelier's principle does not say by how much.

Imagine you have 0.1 mol of CH3COOH and 0.1 mol of CH3COO- in equilibrium in 1 L water. Let's add 0.01 mol HCl. We're left with 0.09 mol CH3COO-, 0.11 mol CH3COOH and 0.01 mol H+. Suddenly, we have some H+ that can react with the CH3COO-. A bit of this will react to a small but significant extent, which is the reduction in concentration of H+ you mentioned.

Now replace the 0.1 mol CH3COO- with 0.1 mol CH3COOH. After adding 0.01 mol HCl, we're left with 0.2 mol CH3COOH and 0.01 mol H+. But this H+ has nowhere to go. Problem? There's no weak base to react with now; CH3COOH is a horrible base in water.
The H3O+ has only the minute amounts of OH- produced by water and the Cl- to react with. What'll happen is that a tiny, tiny, tiny, tiny bit of it will react with the shoddy base that is Cl- to form HCl(aq) and a tiny, tiny, tiny bit will react with the OH- in the water. This way, the system does act to reduce the concentration of H+, but there's no real path for it to go, so effectively the addition of H+ isn't mitigated at all.

This is why the conjugate base is needed.

[zvezda]

this is almost spot on. You shouldn't say "as its activation energy has been reached". Rather, a greater proportion of collisions now exceed this activation energy (according to how the Boltzmann distribution vary with temperature).

This increase in the proportion of successful collision goes both ways, it helps forwards and backwards reactions. However, its impact is greater for endothermic reactions (compared to their respective backwards, exothermic reactions), since its activation energy is larger.

could you expand on this if possible? wouldn't an increase in temperate have a greater impact on exothermic reactions as they would need less energy in order to reach activation (ie increase in proportion of successful collisions)?

[SocialRhubarb]

I have no idea what a Boltzmann distribution is and I don't know if I even understand what's happening, but this is how I think about it.

An exothermic reaction has a lower activation energy than an endothermic reaction. And at a certain temperature, particles are going to have a certain amount of energy. Say, for argument's sake, that most of the particles have enough energy to trigger the exothermic reaction but not the endothermic reaction.

Then let's increase the temperature. It does increase the rate of the exothermic reaction, but it increases the rate of the endothermic reaction more, because, yes, increasing the temperature means that the particles gain more energy and the success rate of collisions for the exothermic reaction goes up, but a lot of them already had enough energy for the exothermic reaction anyway, so giving them even more energy wouldn't have a huge effect. Whereas for the endothermic reaction, which has a larger activation energy and hence not as many particles with sufficient energy to trigger it, the increase in temperature has an effect on more particles, so its rate is increased more than the exothermic reaction.

[Lasercookie]

also, a question regarding vce, is it true that our schools only pick one chapter to study out of ch19-22 (for those using the heinemann textbook)? will we be examined at the end of the year on the other chemicals?

The exam question will give you the option of choosing which chemical you'd like to answer about. You could look at previous Unit 4 exams for examples of how they word it (usually fairly general questions that'll apply to all of them). For the study design dot point, it says "application of equilibrium and rate principles to the industrial production of one of ammonia, sulfuric acid, nitric acid"

[zvezda]

The exam question will give you the option of choosing which chemical you'd like to answer about. You could look at previous Unit 4 exams for examples of how they word it (usually fairly general questions that'll apply to all of them). For the study design dot point, it says "application of equilibrium and rate principles to the industrial production of one of ammonia, sulfuric acid, nitric acid"

You only study one chemical, and its only ever going to be asked of in a SAC

[Phy124]

All of the key knowledge that underpins the outcomes in Units 3 and 4, and the set of key skills listed on page 12 of the study design are examinable except
• specific details related to the study of a selected chemical (one of: ammonia, sulfuric acid or nitric acid).

Source: http://www.vcaa.vic.edu.au/Documents/vce/chemistry/chem-specs-samp-w.pdf

[zvezda]

I have no idea what a Boltzmann distribution is and I don't know if I even understand what's happening, but this is how I think about it.

An exothermic reaction has a lower activation energy than an endothermic reaction. And at a certain temperature, particles are going to have a certain amount of energy. Say, for argument's sake, that most of the particles have enough energy to trigger the exothermic reaction but not the endothermic reaction.

Then let's increase the temperature. It does increase the rate of the exothermic reaction, but it increases the rate of the endothermic reaction more, because, yes, increasing the temperature means that the particles gain more energy and the success rate of collisions for the exothermic reaction goes up, but a lot of them already had enough energy for the exothermic reaction anyway, so giving them even more energy wouldn't have a huge effect. Whereas for the endothermic reaction, which has a larger activation energy and hence not as many particles with sufficient energy to trigger it, the increase in temperature has an effect on more particles, so its rate is increased more than the exothermic reaction.

ahh thanks for that. clears things up

[brightsky]

i see...thanks guys!

also, does the nernst equation hold for reduction half-equations? the nernst equation shows that for a redox equation, the higher the standard cell potential, the higher the Kc value. is this true for reduction half-equations as well?

edit: actually i better be more specific. wikipedia gives two subtly different equations for half-cells and full cells (link: http://en.wikipedia.org/wiki/Nernst_equation), so i was wondering whether a_red/a_ox = Q for a reduction half-equation.

thanks!

[lzxnl]

Yes but your half cell must be standardized against the standard hydrogen electrode which is 1 bar H gas and 1M hygrogen ions. The electrode potential of this half cell is by definition 0V. You can see where this goes.

[brightsky]

What do you write as the state symbol when you have a substance dissolved in a solvent that is not water?