Answer:
[CH₃OH] = 1.1 M
Explanation:
Let's consider the synthesis of methanol at high temperature.
CO(g) + 2 H₂(g) ⇄ CH₃OH(g)
The equilibrium constant Kc is:
[tex]Kc=\frac{[CH_{3}OH]}{[CO].[H_{2}]^{2} }[/tex]
We know the concentrations at equilibrium, [CO] = 0.03M and [H₂] = 0.04 M. So, we can find [CH₃OH].
[tex][CH_{3}OH]=Kc.[CO].[H_{2}]^{2} =2.3 \times 10^{4} \times 0.03 \times 0.04^{2} =1.1M[/tex]
