Consider the reaction: 2 SO2 (g) + O2 (g) ⇋ 2 SO3 (g)

Question

4.0 mol of SO2 and 2.2 mol of O2 were placed in a 10 L flask and allowed to reach equilibrium. The flask was found to contain 2.6 mol SO2. Calculate the equilibrium constant K for this reaction.

Answer 1

(SO2) = mols/L = 4.0 mols/10 L = 0.40 M
(O2) = 2.2/10 = 0.22 M
........................ 2SO2 (g) + O2 (g) ⇋ 2SO3 (g)
I.......................0.40M..........0.22 M...........0
C.........................-2x...........-x...........+2x
E......................0.40 - 2x.....0.22-x........2x
The problem tells you that mols SO2 at equilibrium was 2.6 mols. That in 10 L is 2.6/10 = 0.26 M so you know that 0.40-2x = 0.26. Solve for x, evaluate 0.40-2x = (SO2) at equilibrium, 0.22-x = (O2) at equilibrium, and 2x = (SO3) at equilibrium. Substitute those evaluations into Keq expression and solve for K. Post your work if you have any questions.

Answer 2

Okay thanks that makes sense but I am confused for this next one

Consider the reaction: 2 IF5 (g) + I4F2 (g) ⇋ 3 I2 (g) + 6 F2 (g)
6.0 mol of IF5 and 8.0 mol of I4F2 are placed in a 5.0 L container. At equilibrium, 6.0 mol of I4F2 are present. Calculate K for this reaction.