Question

At 500K, NOCl is 90% decomposed for the reaction 2NOCl<->2NO+Cl2. If we assume that 1.00mol of NOCl in 1.00L before dissociation, what is the value of Kc for equilibrium at 500K?

Answer 1

`K_c~=36`....

Explanation:

For `2NOCl(g) rightleftharpoons2NO(g) + Cl_2(g)`

We write `K_c=([NO]^2[Cl_2])/([NOCl(g)]^2)`

Now initially there were `1.00*mol*L^-1` with respect to `NOCl`; at equilibrium `[NOCl]=0.10*mol*L^-1`, `[NO]=0.90*mol*L^-1`, and `[Cl_2]=0.45*mol*L^-1`...

And so we fill in the numbers....

`K_c=((0.90*mol*L^-1)^2xx0.45*mol*L^-1)/(0.10)^2=36.5`