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 .

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

We can write the law of mass action as:

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

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

Finally, we can compute the equilibrium constant:

Best regards.

the answer for sodium is 0.186

the answer for rubidium is 0.251