For $C (s) + CO X_{2} (g) 2 CO (g)$ at a fixed temperature, the gas mixture at equilibrium under a total pressure of 1.0 atm contains 0.50 mole fraction of $CO$ . The total pressure is raised to 4.0 atm at the same temperature and equilibrium is re-established. Calculate the mole percent of $CO$ in the new gas mixture, to the nearest integer.

Numerical answer type, answer in %, tolerance 0.5.

Answer: 30 %

At 1.0 atm,

p_{CO} = 0.50

and

p_{CO X_{2}} = 0.50

atm, so

K_{p} = (0.50)^{2} /0.50 = 0.5000

atm. At 4.0 atm,

p^{2} / (4.0 - p) = 0.5000

gives

p^{2} + 0.5000 p - 2.000 = 0

, so

p_{CO} = 1.1861

atm. Mole percent

= 1.1861/4.0 \times 100 \approx 29.65%