For $A (g) + B (g) C (g)$ at constant temperature, an equilibrium mixture in a 10.0 L vessel contains 1.0 mol A, 2.0 mol B and 3.0 mol C. The vessel is then compressed isothermally to 5.0 L and equilibrium is re-established. The amount of C (in mol) now is

Answer: B

K_{c} = (3.0/10) / ((1.0/10) (2.0/10)) = 15

. After compression

Q_{c} = 7.5 < K_{c}

, so the reaction goes forward by

x

mol:

15 = \frac{( 3 + x ) /5}{[( 1 - x ) /5 ] [( 2 - x ) /5 ]}

, i.e.

3 x^{2} - 10 x + 3 = 0

, giving

x = 0.333

. So

n (C) = 3 + x = 3.33

mol.