I've been stuck with this problem:
The following reaction has Kc = 2.8 · 10
2 at T=1000 K:
C
2H
4 + H
2 C
2H
6A reactor with volume V = 10 L initially contains 0.32 mol C
2H
4, 0.16 mol H
2 and 0.68 mol C
2H
6.
Calculate the number of C
2H
6 moles after:
a) the addition of 1.00 mol of C
2H
6 in the reactor.
b) the volume of the reactor is increased by 20 L.
I know the answer to a) must be 1.515 moles, while b) must be 0.645 moles.
I can see these numbers make sense by LeChatelier principle. By adding 1 mole to 0.68 mol you get 1.68 mol of C
2H
6, but a part of it gets converted back to its reactants; by adding 20 L to the volume of the recipient, you shift the equilibrium position towards the reactants (because they are gaseous, so they are entropically favoured)
I tried to solve this exercise with ICE tables, resulting in
Numerator: 1.68 + x
Denominator: (0.32 - x) × (0.16 -x)
But when I solve for Numerator/Denominator = Kc, I do not get the answers above. I also do not know how to include volume in my calculation. I've tried using the ideal gas law and converting into partial pressures, but I still do not get the correct answer.