Questions: The following reaction has an equilibrium constant of Kp=2.26 x 10^4 at 298 K CO(g)+2 H2(g) ⇌ CH3OH(g) Calculate K'p for the following reaction: 3 CH3OH(g) ⇌ 3 CO(g)+6 H2(g)

Solution Steps

The given reaction is: \[ \mathrm{CO}(\mathrm{g}) + 2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3}\mathrm{OH}(\mathrm{g}) \] with an equilibrium constant \( K_{\mathrm{p}} = 2.26 \times 10^{4} \).

We need to find the equilibrium constant \( K_{p}^{\prime} \) for the reaction: \[ 3 \mathrm{CH}_{3}\mathrm{OH}(\mathrm{g}) \rightleftharpoons 3 \mathrm{CO}(\mathrm{g}) + 6 \mathrm{H}_{2}(\mathrm{g}) \]

Notice that the new reaction is the reverse of the given reaction, but multiplied by 3.

For the reverse reaction: \[ \mathrm{CH}_{3}\mathrm{OH}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g}) + 2 \mathrm{H}_{2}(\mathrm{g}) \] the equilibrium constant \( K_{\mathrm{p, reverse}} \) is the reciprocal of \( K_{\mathrm{p}} \): \[ K_{\mathrm{p, reverse}} = \frac{1}{K_{\mathrm{p}}} = \frac{1}{2.26 \times 10^{4}} = 4.4248 \times 10^{-5} \]

Since the new reaction is the reverse reaction multiplied by 3, the equilibrium constant for the new reaction \( K_{p}^{\prime} \) is: \[ K_{p}^{\prime} = (K_{\mathrm{p, reverse}})^{3} = \left(4.4248 \times 10^{-5}\right)^{3} \]

\[ K_{p}^{\prime} = (4.4248 \times 10^{-5})^{3} = 8.6696 \times 10^{-14} \]

Final Answer