Question

The reaction $ZnO(s) + CO(g) \rightleftharpoons Zn(g) + CO_2(g)$ has an equilibrium constant of 1 atm at 1500 K. The equilibrium partial pressure of zinc vapour in a reaction vessel if an equimolar mixture of CO and $CO_2$ is brought into contact with solid ZnO at 1500 K and the equilibrium is achieved at 1 atm is

• A. 0.68 atm
• B. 0.76 atm
• C. 0.24 atm
• D. 0.5 atm

Answer 1

Answer: (C)

0.24 atm

Answer 2

The reaction is $ZnO(s) + CO(g) \rightleftharpoons Zn(g) + CO_2(g)$. The equilibrium constant in terms of partial pressures is $K_p = \frac{P_{Zn} \cdot P_{CO_2}}{P_{CO}}$. Given $K_p = 1$ atm at 1500 K.

Let the initial partial pressures of CO and $CO_2$ be $x$ atm each. $P_{Zn,i} = 0$. At equilibrium, let $P_{Zn,eq} = p$. Then, $P_{CO_2,eq} = x + p$ and $P_{CO,eq} = x - p$.

Total pressure at equilibrium is 1 atm: $P_{total} = P_{CO} + P_{Zn} + P_{CO_2} = (x-p) + p + (x+p) = 2x + p = 1$. So, $x = \frac{1-p}{2}$.

Equilibrium partial pressures in terms of $p$: $P_{Zn,eq} = p$ $P_{CO_2,eq} = \frac{1-p}{2} + p = \frac{1+p}{2}$ $P_{CO,eq} = \frac{1-p}{2} - p = \frac{1-3p}{2}$

Substitute into $K_p$: $K_p = \frac{p \cdot \frac{1+p}{2}}{\frac{1-3p}{2}} = 1$ $\frac{p(1+p)}{1-3p} = 1$ $p + p^2 = 1 - 3p$ $p^2 + 4p - 1 = 0$

Using the quadratic formula, $p = \frac{-4 \pm \sqrt{4^2 - 4(1)(-1)}}{2(1)} = \frac{-4 \pm \sqrt{20}}{2} = -2 \pm \sqrt{5}$. Since pressure must be positive, $p = -2 + \sqrt{5} \approx 0.236$ atm.