Need help please :)

socky2011

I've been studying old exam papers for a few weeks now and this one has really gotten me stuck any help would really be appreciated
the question is:

A 0.024 sample of N2O4(g) is allowed to come to equilibrium with NO2(g) in a 0.372 L reaction vessel at 25 deg celsius. this is represented by
N2O4(g) <--> 2NO2(g) K= 4.61 * 10^-3 at 25 C

(i) write the expression used to determine the equilibrium constant in this reaction

(ii) using the expression you obtained in part (i) and the equilibrium constant value calculate the concentration of N2O4(g) and NO2(g) at equilibrium

ressem

Part (i) is

The equilibrium constant for the reaction
K = (num of Moles per Litre of NO2) Squared / (Num of moles of N2O4)

It's squared because you get two moles of NO2 for each mole of N2O4

So for (ii)
You've got the numbers to calculate the initial number of moles of each.
At equilibrium, the amount of N2O4 will have dropped by x, the amount of NO2, will have increased by 2x.

Plug these into your equation and calculate x.

socky2011

thanks mate i think i can do them now