Question

Initially, we place 2 mol of A and 4 mol of B in a 0.5 L flask. At equilibrium, the flask contains 0.3 mol of C. Determine the value of K.

A(g) + B(g) <--> 3C(g) + 2D(g)

Answer 1

.........A + B ==> 3C + 2D
I........2...4.....0....0
C.......-x..-x...+3x...+2x
E.......2-x.4-x..0.3...+2x

Column 3C means x must be 0.3/3 = 0.1 (I didn't change C, it still is 0.3 since 0.1 x 3 = 0.3) That makes column D equilibrium = 0.1*2 = 0.2 mols
Column A = 2-0.1 and colum B = 4-0.1.
Convert mols A, B, C, and D at equilibrium to molarity (mols/L = M) and substitute into the K expression. Solve for K.

Answer 2

My chem teacher said that initial concentrations of A and B and the equilibrium concentration of C need to be doubled to put the experiment into a L from the 0.5L we started with in the question.