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chemistry Kc question

Hello, I'm a bit confused. So I've learnt that Kc is only affected by temperature and not concentration as a change in concentration means that the equilibrium position moves to counter this change so there is no effect on Kc. However the same thing also happens when temperature changes so why is it that temperature still affects Kc??

Reply 1

Pigster

Consider the equilibrium reaction: A <-> B
For easy numbers sake, if at equilibrium you have 1 mol of A and of B in a 1 dm3 container, then Kc = 1/1 = 1
If you added a further 2 mol of A, then there would be 3 and 1, so 1/3 =/= Kc, i.e. it is no longer at equilibrium.
The forward reaction is favoured until both are 2 mol and hence 2/2 = Kc, i.e. it is back at equilibrium.
In this case, the position of the equilibrium hasn't changed, the mixture still has a 50:50 mix. This happens when the number of particles is the same on the LHS as RHS - just like what happens with changing pressure on a gas equilibrium.

Now consider A <-> 2B, with 1 mol of both in 1dm3. Kc = 1²/1 = 1
Now add 2 mol of A -> 1²/3 =/= Kc
So A converts into B until (I can't be bothered to do the quadratics) B²/A returns to 1. This doesn't happen when A and B are both 2 (since 2²/2 =/= 1) hence you don't have a 50:50 mix anymore. In order for the maths to make Kc = 1 again, you need more A than B, i.e. the equilibrium position has shifted to the LHS.

Now, consider A <-> B again, and we'll assume it is exothermic. For exothermic reactions, Kc decrease with increasing temperature. The reasons for that are complex and involve how the rate constants, k, for the forward and reverse reactions change with changing T.
So... at T₁, Kc = 1/1 = 1 and a a higher T Kc <1 and hence A must increase and B must increase, i.e. the equilibrium goes to the LHS.

If you have further Qs, please reply and I'll have another go.

Reply 2

Nitrotoluene

Original post by Frizzle16523

The rate constants in the equilibrium reaction are usually given as k1, k-1, k2, and k-2, representing the forward and backward rates of the reaction, and the equilibrium constant, Kc, is related to these rate constants by the equation Kc = k1/k-1. As the temperature changes, the rate constants k1 and k-1 also change.
The rate constants of chemical reactions are influenced by temperature, and their values change in response to temperature changes.
Specifically, the Arrhenius equation describes how rate constants increase or decrease as temperature increases or decreases. This means that as temperature rises, the rate constant typically increases, while as temperature falls, the rate constant typically decreases: k = Ae^(-Ea/RT)**, where A is the rate coefficient, Ea is the activation energy, R is the molar gas constant, and T is the temperature in Kelvin. **For a better understanding of what I have written, you should do the one-variable function study, where the independent variable is T.
As the temperature increases, the rate constants k1 and k-1 increase exponentially, which means that the forward and backward velocities of the reaction also increase. However, the equilibrium constant, Kc, is a ratio of these rate constants, so it also changes.
If the activation energy of the direct reaction is greater than the activation energy of the reverse reaction, an increase in temperature will create favourable conditions for the direct reaction, increasing Kc.
Conversely, if the activation energy of the reverse reaction is greater, an increase in temperature will create favourable conditions for the reverse reaction, decreasing Kc. In summary, temperature affects Kc by changing the rate constant in the equilibrium expression, which in turn affects the equilibrium constant itself.
The direction and magnitude of the change in Kc depend on the relative activation energies of the forward and reverse reactions.