Question

9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.28 mol of CO remaining. Keq at the temperature of the experiment is ________.

Answer 1

Answer:

$Keq=11.5$

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

$CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)$

We can write the law of mass action as:

$Keq=\frac{[CH_3OH]}{[CO][H_2]^2}$

That in terms of the change $x$ due to the reaction extent we can write:

$Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}$

Nevertheless, for the carbon monoxide, we can directly compute $x$ as shown below:

$[CO]_0=\frac{0.45mol}{1.00L}=0.45M$

$[H_2]_0=\frac{0.57mol}{1.00L}=0.57M$

$[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M$

$x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M$

Finally, we can compute the equilibrium constant:

$Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5$

Best regards.