Equilibria - Part 1What you should know1. 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
here3. 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 notesA. 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

-- 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

is

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.

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 dependenceBy applying the Arrhenius equation, we can express k1 and k2 as

, where Ea is the activation energy, and

is a constant dependent on the conditions (but independent of temperature). In this case, we take the logarithm on both sides to yield

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

The ratio of k* and k can then be found as (for small change in temperatures)
 \\<br />\log_e \frac{k*}{k} & = -\frac{E_a}{R} \cdot \frac{-dt}{T(T+dt)} \\<br />\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:
/RT} = \frac{\alpha_f}{\alpha_b} e^{-\Delta H/RT})
(note

)
From this, we can see that

for endothermic reactions, and

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.