[cliverlong]

I was asked a question to which I only have a partial answer

The question is essentially to explain the changes to position of equilibrium predicted by Le Chatelier but using kinetic arguments.

Removal of a reactant or product seems explicable to me. One says if one "side" of the reaction suddenly loses some material, e.g. removing ammonia from Born Haber creation of ammonia there will be less ammonia (obviously). Now the N₂ and H₂ combination reaction and the NH₃ decomposition still proceed at the same rate. However, there is an excess of H₂ and N₂ therefore this will produce more NH₃ until equilibrium is restored

Now if we think about the effect of pressure, by Le Chatelier

N₂ + 3H₂ <> 2NH₃

will shift to the right when pressure is applied (fewer moles on right therefore occupies smaller volume)

However, increasing pressure will bring ALL the molecules closer together, increasing the frequency of collisions, hence surely the rate of forward and back reactions both increase and the eqm position does NOT move?

I guess the rates of the two reaction directions don't increase by the same "amount" but I can't quantify that to support the Le Chat prediction

Similarly considering raising temperature. Le Chat will predict the equm move in the endothermic direction but kinetically all the reactions speed up because there are more collisions with sufficient energy therefore eqm position will NOT shift.

What am I getting wrong?


Thanks

Clive

[Yggdrasil]

Your intuitions are correct that both forward and backward rates will increase when you increase the pressure or increase the temperature, but you need to look more closely at the magnitudes of the increases in order to explain Le'Chatalier's principle.  A full explanation requires looking at the changes quantitatively to see if one rate increases more than the other or whether the rates increase by the same amount.  As it turns out, in both cases, one of the rates will increase more than the other rate, leading to the observed change in the position of the equilibrium.

In the case of pressure, you need to look at how the rates depend on the concentrations.  For example, assume we have a reaction X + Y  Z where the forward rate is given by v_f = k₁[X][Y] and the reverse rate is given by v_r = k_-1[Z].  You can see that increasing the pressure (i.e. increasing the concentration of every reactant) will increase the forward rate more than the reverse rate.

With temperature, you need to think of the Arrhenius expression for the rate constant, k = Ae^-E_a/k_BT.  The activation energy (E_a) for the endothermic direction will be greater than the activation energy for the exothermic direction.  If you calculate the effect of increasing temperature, you will see that the increase in temperature accelerates the endothermic reaction more than the exothermic reaction.

[cliverlong]

Excellent that makes perfect sense.

One clarification

In the case of pressure, you need to look at how the rates depend on the concentrations.  For example, assume we have a reaction X + Y  Z where the forward rate is given by v_f = k₁[X][Y] and the reverse rate is given by v_r = k_-1[Z].  You can see that increasing the pressure (i.e. increasing the concentration of every reactant) will increase the forward rate more than the reverse rate.

Say we have your example

X + Y  Z

but the rate equation for the forward equation is

v_f = k₁[X]

due to a change in X being the rate limiting step

then the order of the forward and reverse reactions will now be the same

v_f = k₁[X]
v_r = k_-1[Z]

so changing pressure won't affect this? In fact thinking about this if the rate limiting step is down to

v_f = k₁[X]

can't be affected by pressure since it isn't due to the interaction of two molecule so increasing collisions won't have any impact.



Thanks

Clive

[Yggdrasil]

That's an excellent question.  I'll now show why this will not happen.  Chemical equilibrium and chemical kinetics are connected by the fact that, at equilibrium, the forward rate and backward rate of a reaction are equal.  In other words:

v_f = v_r

Now, let us assume that the forward and backward rates follow the expression that you described.  Substituting these into the equation (using the equilibrium concentrations) give us the following relation:

k₁[X]_eq = k_-1[Z]_eq

Which can be rearranged to show give:



This implies a fundamental relation at equilbrium: [Z]/[X] is constant.  However, from the definition of the equilibrium constant, we know that this is NOT the relationship we'd expect.  The relationship above suggests that [Y] can take on any value as long as [Z]/[X] = k₁/k_-1.

Instead, we know that the equilibrium constant K is given by:



Clearly, [Y] cannot vary independently of [X] and [Z].  Rather, the quantity [Z]/[X][Y] stays constant at equilibrium.  If you think about the math involved, you can prove to yourself that if this is true, then no other combination of [X], [Y], and [Z] can be a constant value.  Therefore, if we know that if the backward rate is given by v_f = k₁[X], then the backward rate must have the form v_r = k_-1[Z]/[Y].

The principle above is known the principle of detailed balance and it comes from the assumption that all chemical reactions are reversible.  It's very useful for checking kinetic schemes and mechanisms to make sure that they make physical sense.

[cliverlong]

<< snip >>
With temperature, you need to think of the Arrhenius expression for the rate constant, k = Ae^-E_a/k_BT.  The activation energy (E_a) for the endothermic direction will be greater than the activation energy for the exothermic direction.  If you calculate the effect of increasing temperature, you will see that the increase in temperature accelerates the endothermic reaction more than the exothermic reaction.

Yep, made my way through a simplified example and I think I get it. The key seems to be that e^-E_a/k_BT term and the effect of a change in temperature and that the endothermic direction has the higher activation energy than the exothermic direction.

Excellent. Ta. Clive