[guybrush]

Hi,

We have just gone through the equilibrium constant, however i can't get my head round the enthalpy change.

I have three values for the equilibrium of sulphur trioxide:

1000k at 2.55 x 10 -3.
500k at 2.5 x 10 power of 10
1100k at 0.13

I have been asked to interpret what these tell me about the enthalpy reaction and if there is any conflict in obtaining a fast reaction and a good yield of sulphur trioxide.

Does this mean at 500k that the reaction is at its highest and the bigger the temperature gets, the less equilibrium gets?

[onn]

Hi,

We have just gone through the equilibrium constant, however i can't get my head round the enthalpy change.

I have three values for the equilibrium of sulphur trioxide:

1000k at 2.55 x 10 -3.
500k at 2.5 x 10 power of 10
1100k at 0.13

I have been asked to interpret what these tell me about the enthalpy reaction and if there is any conflict in obtaining a fast reaction and a good yield of sulphur trioxide.

Does this mean at 500k that the reaction is at its highest and the bigger the temperature gets, the less equilibrium gets?

This is the contact process, SO2 + 1/2 O2 ---> SO3; all in gaseous states

forming SO3, this results in decrease in entropy(could be small). But that is not really vital at the moment. Using le chateliers principle to see how the eqm changes with temperature, you would see that you get higher yield at lower temperature, ie less heat favours forward reaction, so the forward reaction should be exothermic.

Now the catch is to differentiate kinetics and thermodynamics. What might seem thermodynamically feasible(ie eqm constant > o => delta G < 0 does not imply that rate is fast, they are independent of one another).
rate is determined by arrhenius eqn(in general), and you can see high temperature is required to overcome the activation energy barrier for reactions to happen, and that is where there is a conflict as to what is the best temperature to use.