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

`Keq=11.5`

Step-by-step 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=([CH_3OH])/([CO][H_2]^2)`

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

`Keq=(x)/(([CO]_0-x)([H_2]_0-2x)^2)`

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

`[CO]_0=(0.45mol)/(1.00L)=0.45M\\`

`[H_2]_0=(0.57mol)/(1.00L)=0.57M\\`

`[CO]_(eq)=(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=(0.17M)/((0.45M-0.17M)(0.57M-2*0.17M)^2)\\\\Keq=11.5`

Best regards.