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#### Mao

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##### Selected Tutorials for VCE Chemistry U4
« on: September 10, 2010, 09:54:22 pm »
+5
These tutorials do not aim to give a summary of the course nor to include every detail like in textbooks. These tutorials are made to complement what you have learnt in class, enhance your understanding, and possibly boost exam performance. These tutorials are not vital to your exam success, and may benefit some more than others. Don't stress if there are materials that go straight over your head, they are most probably beyond what is required.

NOTE: because including pictures in these posts will be fairly difficult (and mainly, I can't be bothered), if you want to get the most out of the advanced (read: beyond the course) materials, I suggest you have a stack of paper and a graphics calc next to you, and work your own way through each step. Also, a friend/significant other to calm you would also be useful in the case you rage balls.

This is a response to this thread. I promised you guys, now I will slowly deliver in dribs and drabs.

This thread will remain locked until all tutorials have been delivered, I aim to maintain coherence within this thread. In the meanwhile, someone please create a 'Q&A' thread and I will answer any questions there.

It's been a while since I looked at the study design, so I may have given an outcome the wrong number or something (or miss an outcome altogether! let's hope that does't happen). PM me and let me know if I've made a mistake.
« Last Edit: September 26, 2010, 08:12:50 pm by Mao »
Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015

#### Mao

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##### Re: Selected Tutorials for VCE Chemistry U4
« Reply #1 on: September 10, 2010, 11:07:39 pm »
+2
Rate of Reaction

As you all should know, the kinetic theory states a successful reaction requires:
• A collision
• Collision must have enough energy
• Collision must occur at the right place in the right orientation
The first two conditions are well covered in your classes. The last one is rarely touched upon. Detailed knowledge of this is not required, but here is an example:
- for the oxidation of 1-hexanol to hexanoic acid, an oxygen is added on the the terminal carbon (going from CO to COO)
- this oxygen is from a water molecule, thus the reaction involves a collision between hexanol and water
- If this water collides with 1-hexanol at the terminal carbon with the hydroxy, a reaction may take place.
- If this water collides with the carbon at the other end or in the middle of the chain, you won't be so lucky.
- The water must also collide into the terminal carbon with the O pointing towards the C. If it decides to 'reverse' into the carbon with the H directed at the C, reaction won't take place either.

You must acknowledge that a successful reaction requires a collision in the correct orientation. You don't need to know how this applies to any of the below methods for changing the rate of reaction.

You should all remember from class that rate of reaction can be increased by:
• Increasing concentration (for solutions)
• Increasing surface area (for solids)
• Using a catalyst
• Increasing temperature
• Increasing pressure (for gases)
• stirring (for solutions)
The first four things should be fairly automatic. The latter two aren't mentioned as much.
The first two points are very easily explained, increasing concentration will increase the number of particles, thus collisions.
- Increasing surface area will increase the rate of collision as more atoms/molecules are exposed.
The last two points aren't usually mentioned, but can also be easily explained.
- For stirring, it mixes certain unreacting 'regions', thus increasing 'contact'.
- For increasing pressure, think back to PV = nRT, I can rearrange this and say $P \propto \frac{n}{V} T$. When I increase pressure, either $\frac{n}{V}$ must increase (density of particles increase, thus number of collisions increase), or T must increase (which then increase rate of reaction).

Catalysts are more tricky. Most know it to the extent that it lowers activation energy, thus increasing the proportion of successful collisions. This is correct, and enough for VCE standards, but for those who always wondered how it actually work, here it is:

To define what a catalyst is, it is a chemical or substance that participates in a reaction, but is not consumed by it. It generally works by offering an alternative pathway that has a lower activation energy, thus allow reaction to proceed faster. This 'alternative pathway' doesn't usually involve creating completely different chemicals.

Usually, a slow reaction is either because of
- an unstable intermediate or transition state
--- thus the catalyst reacts with this intermediate to stabilize it, or reacts with the reactants to form a more stable transition state.
--- With the intermediate steps more stabilized, more successful reactions are observed
- or a particular reactant has a very low concentration (here the catalyst has similar chemical properties to the reactant, and thus reacts in place and causes the rate to increase, a very good example is the depletion of ozone with chlorine radicals, you don't really see this kind of reactions in VCE).

There are two ways catalysts interact with reactants.

Homogeneous catalysis refer to catalysts that are in the same phase as the reactants (i.e. liquid in liquid, gas in gas, [solid in solid is a little silly]). You don't see much homogeneous catalysis in industry because you want to reuse the catalyst, but since the catalyst is in the same phase, you'll have to somehow extract it from the solution or gas mixture(adding operation time and cost). Homogeneous catalysis occurs mostly in nature (such as enzymes, depletion of ozone layer, etc), and is very efficient.

Heterogeneous catalysis is different in the sense that the phases are different. This brings a problem of contact. For a catalyst to be useful, it must be part of the reaction, thus it must come into contact with the reactant (or in other words, there must be a collision between catalyst and reactant). However, due to the reactant being in a different phase, you cannot maximize the contact by mixing (in the case of homogeneous catalysis). Thus, to increase the effectiveness of heterogeneous catalysts, you will need to maximize the surface area, such as using small granules (most metal catalysts) or bubbling reactant gas through a large layer of liquid catalyst (fluidized catalyst beds in sulfuric acid production).

For temperature, some physicists here may know that it is not a 'real' quantity, it is rather an averaged statistical measure of kinetic energy on the molecular scale.
- Let $\tilde{v_i}$ be the velocity of a particle, then the temperature is $T \propto \displaystyle \sum_{i} \frac{1}{2}m_i |\tilde{v_i}|^2$.
- Thus increasing temperature always increases the average kinetic energy, thus the average velocity.
- This means there will be more collisions, and a greater proportion is successful.
Many of you would have seen the Maxwell-Boltzmann distribution and see how increasing temperature shifts the distribution to the right, and thus more atoms will have sufficient energy to react.

Most reaction rates increase with temperature, but there are some reactions that have rate decreasing with temperature (very rare, typically involve creating crazy radicals without activation energy barriers). Don't worry about this, everything you deal with at VCE level will have increasing rate with increasing temperature.

Here is a couple of first year level concepts that can aid in understanding how rate of reaction changes:

1. The rate of reaction is proportional to concentration.
- Thus for aqueous reactions (since all we deal with is the (aq) phase in VCE, and the more common solvents we actually do reactions in are (dmso), (thf), (cryo), (dcm), etc..),
- say A + B --> C,
--- then $RATE \propto [A]$ and $RATE \propto [B]$,
--- we combine these together to give $RATE_{forward} \propto [A][B]$.
--- Given the same kind of conditions (solvent, temperature, background light, taste in music, how much you had for breakfast, whether Australia has a competent government or not), we can say that there is a constant k such that $RATE_{forward} = k [A][B]$, where k varies with temperature and other conditions.
--- With this, we say that doubling the concentration of A will double the rate of reaction, doubling both A and B will make it 4x as fast, etc.

2. The rate constant k is then narrowed down by the Arrhenius equation $k=\alpha e^{-E_a/RT}$, where Ea is the activation energy, R and T you are familiar with, and $\alpha$ is a constant dependent on the conditions (such as type of solvent, light, etc).
In this case, we are saying $k \propto e^{-1/T}$, graph this with your calculator and you will see how reaction rate increases with temperature.

We will be drawing upon these two concepts in the next tutorial for equilibrium reactions. I will show you why K is defined the way it is, and show you how a mathematical proof for how K changes with temperature for endothermic and exothermic reactions.
« Last Edit: September 26, 2010, 06:36:57 pm by Mao »
Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015

#### Mao

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##### Re: Selected Tutorials for VCE Chemistry U4
« Reply #2 on: September 26, 2010, 06:35:23 pm »
+3
Equilibria - Part 1

What you should know
1. Dynamic equilibrium is defined as the rate of forward reaction equaling the rate of backward reaction. In particular:
- All reactions are equilibrium reactions (backwards reaction is always possible)
- All systems will tend towards equilibrium
- System appears to not change with time at equilibrium

2. For systems at equilibrium, we can change volume/pressure and add/remove reactants/products. Here, I will reinvent the wheel and clearly lay out the logical steps involved with Le Chatelier's Principle, that:
- Some change is made to the system. This has an immediate effect (e.g. increase in amount of products).
- The system will try to oppose this change. This is a subsequent change (e.g. decrease the amount of products)
- The system facilitates this change (move backwards, consuming products)
- Equilibrium is re-established.
Note that LCP implies this subsequent change will only be partial. It will never fully counteract the initial change. This can be proved mathematically. One such example is given here

3. Changing the temperature is the only way to change the equilibrium constant.
- It is harder to release heat to a hot environment, thus exothermic reactions favour cold temperatures
- It is harder to absorb heat from a cold environment, thus endothermic reactions favour hot temperatures
- A given system will always have an endothermic reaction and an exothermic reaction (forward/back). If temperature is raised, the endothermic reaction will be favoured, and the system will move in that direction, and vice versa.

4. You should be able to describe everything using Q (concentration quotient) and K (equilibrium constant).
- At equilibrium, Q = K
- For a 'normal' change in the system, Q changes, and subsequently moves towards K
- If temperature is changed, K changes, Q is unchanged, and subsequently moves towards K

A few relevant notes
A. What exactly is 'pressure'?
- Pressure is the amount of force exerted by the gas (or liquid, but we don't worry about that at VCE) on an unit surface. This has little meaning to you. If you must associate it with some physical meaning: everyone should have done the sealed syringe/NO2/N2O4 experiment. As you push the plunger in, the gases inside become compressed and push back harder. This is the increased pressure.
- Pressure can be increased by both a decrease in volume, and an increase in temperature. Pressure is also increased by adding more gas into the same volume (pumping more into a fixed space).
- To state these more generally, we look at the ideal gas equation $PV = nRT$
-- Increase in V decreases pressure
-- Increase in n increases pressure. Since we're dealing with ideal gases where all gases are the same, this n is ALL gases present, even if they're not reacting.
-- Increase in T increases pressure.
- Thus, you must realise that pressure applies to the system as a whole. Whilst the 'concentration' of an ideal gas in a gaseous mixture is important (you will learn later on that this is called the 'partial pressure'), the system changes the overall pressure. Remember also that LCP implies partial change, if you increase the pressure by pumping in some gas, and the system subsequently changes to reduce pressure, the final pressure will still be higher than original.

B. LCP applied to pressure and volume changes
- Most of you would have been taught to apply LCP to gaseous systems using the word 'pressure'. That is if pressure is increased (such as by a decrease in volume), the system wants to decrease pressure and thus move to the side with less particles (note the ideal gas equation, a decrease in n is a decrease in P). The final pressure will still be higher than before the volume was decreased. This works in most of the cases, including cases where both sides have the same number of particles, so increasing pressure doesn't change anything.
- However, this has two problems
i) When presented with an aqueous system, where the volume has been doubled. We know that the system will tend towards the side with more particles. But why?
-- We can logically deduce this for gaseous systems using the ideal gas equation, where we link pressure with the total number of particles in the system.
-- There is no 'ideal aqueous solution' equation.
-- Most people explain this with the use of 'total concentration of the system has decreased, thus the system will move to the side with more particles to increase the total concentration'
-- What is this 'total concentration'? How is it applied? Does it really have a physical significance at all?
ii) When presented with a gaseous system, where some inert gas such as Ar is added. In this case, the number of particles in the system has increased, everything else is the same, thus pressure must increase.
-- The system 'should' move to the side with less particles to decrease pressure. But it doesn't change position at all.
-- Why is this change in pressure 'invalid'?
-- If you answer 'because it is an inert gas', explain why pressure caused by inert gas is not as 'worthy' as pressure caused by other gases. [In fact, it doesn't have to be inert, it just cannot react with anything in the system, so pumping in CO2/N2 generally wouldn't change anything either.]

To explain these two problems, we must dive deeper into what the equilibrium constant really is, and relate everything back to the very basic definition of the rate of forward reaction equal the rate of backward reaction. Materials after this point is beyond VCE. You can still use the explanations you've learnt in class (and I've debunked), even if you read and understood everything below, you're still going to have to use these wrong explanations because you won't have the space/time to express these ideas in an exam. But if you are a true scientist..

Rate of reaction and Equilibrium Constant (using the Arrhenius equation)

Firstly, we draw on the previous tutorial that rate of reaction for $A + B \leftrightharpoons C + D$ is

\begin{align*}
rate_{forward} & = k_1 [A][B] \\
rate_{backward} & = k_2 [C][D] \\ \end{align*}

Where k1 and k2 are some constants dependent on the environment (temperature, solvent, background lighting, winner of the 2010 AFL grand final [or lack of], etc). We can show these two equations are true empirically by doubling/tripling/quadrupling/halving concentrations and measure the rate of reaction. These equations are intuitively true by the kinetic theory (making the assumption that A and B collide to react, and C and D collide to react).

Then, at equilibrium, the two rates must equal to each other.

\begin{align*}
k_1 [A][B] & = k_2 [C][D] \\
\frac{k_1}{k_2} & = \frac{[C][D]}{[A][B]} \\
\end{align*}

Here, since the k1 and k2 are constants (we are not changing the reaction conditions, mainly that we're not changing the temperature [yet]), the fraction k1/k2 is a constant, and is our equilibrium constant K.

As an example, we can explain how the system changes if we add some A:
- higher concentration of A, thus more collisions between A and B, thus rate of forward reaction increases
- [C] and [D] are unchanged, thus rate of backward reaction decreases
- Since rate of forward is now greater than backwards, the system moves forward
- As the reaction proceeds, some A and B is consumed, thus rate of forward decreases. Similarly, C and D are produced, thus rate of backward increases
- This continues until the two rates equal each other, and equilibrium is re-established.
Again, we can mathematically prove LCP to show that the subsequent decrease in A is partial.

Applying this to pressure, we note that everything is now in terms of collision. Increasing pressure (by decreasing volume) means there are more particles inside a smaller box, thus more collisions, and faster reaction. This effect is greater for reactions with more particles (A + 2B benefits far more than A + B, this can be shown with some probability calculations), and thus the reaction going from more particles --> less particles becomes much faster, thus a shift in equilibrium position towards the less particles side.

This explains our inert gas problem, that even though pressure is increased, we have just added some inert gas that doesn't change the rate of collision (since A colliding into the inert gas won't do anything, and A still has the same chance of colliding into B), thus the rate has not changed, and as far as the equilibrium is concerned, nothing has changed.

This also explains the dilution problem, where A + 2B is affected more by dilution than A + B, thus A + B end up being much faster after dilution, and the equilibrium position shifts to the side with more particles. [Many of you probably know this by calculating the new Q by halving each concentration, and the new Q is a fraction of (or many times) the original K.]

Temperature dependence
By applying the Arrhenius equation, we can express k1 and k2 as $k = \alpha e^{-E_a/RT}$, where Ea is the activation energy, and $\alpha$ is a constant dependent on the conditions (but independent of temperature). In this case, we take the logarithm on both sides to yield

$\log_e k = -\frac{E_a}{R}\cdot \frac{1}{T} + \log_e \alpha$

Thus for a small increase in temperature, the new k is

$\log_e k* = -\frac{E_a}{R}\cdot \frac{1}{T + dT} + \log_e \alpha$

The ratio of k* and k can then be found as (for small change in temperatures)

\begin{align*}
\log_e k* - \log_e k & = -\frac{E_a}{R} \left( \frac{1}{T+dt} - \frac{1}{T} \right) \\
\log_e \frac{k*}{k} & = -\frac{E_a}{R} \cdot \frac{-dt}{T(T+dt)} \\
\log_e \frac{k*}{k} & \approx \frac{E_a}{R} \cdot \frac{dt}{T^2} \\ \end{align*}

It can be seen here that the ratio of k*/k is larger if E_a is larger. That is, for a small change in temperature, a reaction with a larger activation energy benefits more. [We can generalise this to any change in temperature by performing an integration over some temperature range.] To reiterate, reactions with larger Ea benefits more from temperature rise.

We apply this to an endothermic reaction, where the activation energy of forward is greater than activation energy of backward. In this sense, temperature rise favours the forward (higher Ea), thus the system moves forward. The subsequent increase/decrease in concentration of products/reactants then bring the system to equilibrium. The same thing applies to exothermic reactions, where temperature rise favours the backward reaction (higher Ea).

We can also express the equilibrium constant K in terms of the Arrhenius equation:

$K = \frac{k_1}{k_2} = \frac{\alpha_f e^{-E_{a,f}/RT}}{\alpha_{b}e^{-E_{a,b}/RT}} = \frac{\alpha_f}{\alpha_b} e^{(E_{a,b} - E_{a,f})/RT} = \frac{\alpha_f}{\alpha_b} e^{-\Delta H/RT}$ (note $E_{a,f} - E_{a,b} = \Delta H$ )

From this, we can see that $K \propto e^{-1/T}$ for endothermic reactions, and $K \propto e^{1/T}$ for exothermic reactions. Plot these on your graphics calculator and you'll see trends you are familiar with.

[On a personal note, the Arrhenius equation is one of the best in physical chemistry, it has a striking resemblance to the Boltzmann distribution, but I haven't worked out how they are related just yet.]

This is all for now, equilibrium part 2 will deal with acid/base equilibria and the major assumptions we use and where they are applicable.
« Last Edit: September 26, 2010, 08:02:30 pm by Mao »
Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015

#### Mao

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##### Re: Selected Tutorials for VCE Chemistry U4
« Reply #3 on: October 04, 2010, 08:59:12 pm »
+2
Equilibria - Part 1 (Supplementary)

So, I have two research reports due in 2 weeks, but the weather has made me feel happy. So what the hell, here's the next bit.

Following on from previous, we've covered the main topics of equilibria. This partial tutorial will be much more applied.

How to systematically deal with all equilibria
For any given system, reactions can only occur in stoichiometrical ratios. That is, the change in any species over the course of the reaction is stoichiometrically related to every other species. Thus, with this in mind, it is very easy to work out how a system changes and find the final concentrations in a system given enough information.
For example, if my system is $A + 2B \to 3C + 4D$, I initially have 1 mole of A, 1.6 mol of B, 0.3 mol of C and 0.8 mol of D. After the system equilibrates, there is now 0.4 mol of B. What is K?
Code: [Select]
     A    B     C     Di    1   1.6   0.3   0.8Δ       -1.2Δ  -0.6       +1.8  +2.4f   0.4  0.4   2.1   3.2The rest is trivial.

Heterogeneous reactants
When reactants are in different phases, you run into the problem of 'contact'. The equilibrium constant isn't actually defined in terms of concentrations, it is defined in terms of 'activity', a separate measurement you'll learn about later. You should know that this 'activity' can modify the equilibrium constant, but it is a constant and thus will not change the behaviour of the equilibrium constant (in the sense that K will still be a constant).

For gases, 'activity' is proportional to the 'partial pressure', which is proportional to the gaseous concentration given conditions that allow for ideal gas behaviour (i.e. most sub-critical situations, as opposed to supercritial situations [google this if you must, but there is a point where the pressure/temperature becomes too high that there is no longer a distinction between solid/liquid/gas]). Thus, you can express an equilibrium constant in terms of pressure. For heterogeneous systems (i.e. a gas/liquid system), K can be expressed as a constant in terms of both pressure and concentration. A past VCAA question did this (on the solubility of CO2, this question is also in checkpoints), where $K = \frac{pressure}{concentration}$, and the units are kPa.M-1. This is possible, and in fact quite plausible. So long as the units are the same (i.e. pressure in kPa, concentration in M), it is equally as valid as a 'normal' K, and you should use it the same way.

For solids, assume 'activity' or 'concentration' is 1. In other words, it does not affect the equilibrium constant, and simply does not appear in the expression for K.

And I am a bit too happy right now to continue this, thus, acid/base equilibria tute will come later. =]
Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015