Answer:
[tex]Keq=11.5[/tex]
Explanation:
Hello,
In this case, for the given reaction at equilibrium:
[tex]CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)[/tex]
We can write the law of mass action as:
[tex]Keq=\frac{[CH_3OH]}{[CO][H_2]^2}[/tex]
That in terms of the change [tex]x[/tex] due to the reaction extent we can write:
[tex]Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}[/tex]
Nevertheless, for the carbon monoxide, we can directly compute [tex]x[/tex] as shown below:
[tex][CO]_0=\frac{0.45mol}{1.00L}=0.45M\\[/tex]
[tex][H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\[/tex]
[tex][CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\[/tex]
[tex]x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M[/tex]
Finally, we can compute the equilibrium constant:
[tex]Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5[/tex]
Best regards.
