Answer:
The concentration of CH₃OH in equilibrium is [CH₃OH] = 2,8x10⁻¹ M
Explanation:
For the equilibrium:
CO (g) + 2H₂(g) ⇄ CH₃OH(g) keq= 14,5
Thus:
14,5 = [tex]\frac{[CH_{3}OH]}{[CO][H_{2}]^2}[/tex]
In equilibrium, as [CO] is 0,15M and [H₂] is 0,36M:
14,5 = [tex]\frac{[CH_{3}OH]}{[0,15][0,36]^2}[/tex]
Solving, the concentration of CH₃OH in equilibrium is:
[CH₃OH] = 0,28M ≡ 2,8x10⁻¹ M
I hope it helps!