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

Kc= 144. If .4 mol of A and B are placed in a 2 Liter container, what will be the concentration of C at equilibrium?

I said the concentration of A= .2 because (.4/2=.2)
and because 1 mol of A = 1 mol of C,

the concentration of C also = .2
But the answer was .185. What did I do wrong?

2 answers

You ignored the fact that the reaction reaches equilibrium subject to the Kc = 144; i.e., your numbers are based on what the reaction STARTED with and not what they are at equilibrium.
...........A + B ==> C + D
initial.0.2M 0.2M ..0...0
change....-x....-x...+x..+x
equil....0.2-x 0.2-x..x...x

Kc = 144 = (x)(x)/(0.2-x)(0.2-x)
Solve for x.
thanks!