For $C (s) + CO X_{2} (g) 2 CO (g)$ , $K_{p} = 1.9$ atm at $1000 K$ . Excess graphite is in contact with a gas mixture containing $2.0$ mol $CO$ and $1.0$ mol $CO X_{2}$ at a total pressure of $3.0$ atm and $1000 K$ . The net change is

Aforward ( $C$ is consumed), since $Q_{p} = 4.0$ atm $> K_{p}$
Bforward, since $Q_{p} = 2.00$ atm $< K_{p}$ when graphite is counted as a reactant
Cbackward ( $CO$ is converted to $CO X_{2}$ and $C$ ), since $Q_{p} = 4.0$ atm $> K_{p}$
Dnone, since $Q_{p} = K_{p}$

Answer: C

Mole fractions are

2/3

and

1/3

, so

p_{CO} = 2.0

atm and

p_{CO X_{2}} = 1.0

atm. Graphite is a pure solid and is omitted from

Q_{p}

, so

Q_{p} = p_{CO}^{2} / p_{CO X_{2}} = 4.0

atm. This exceeds

K_{p} = 1.9

atm, so the net reaction is backward.