Answer:
The concentration of CH₃OH in equilibrium is [CH₃OH] = 2,8x10⁻¹ M
Step-by-step explanation:
For the equilibrium:
CO (g) + 2H₂(g) ⇄ CH₃OH(g) keq= 14,5
Thus:
14,5 =
![([CH_(3)OH])/([CO][H_(2)]^2)](https://img.qammunity.org/2020/formulas/chemistry/college/m4iihte2m41lyw6dnvwjti9hipegx0qtf0.png)
In equilibrium, as [CO] is 0,15M and [H₂] is 0,36M:
14,5 =
![([CH_(3)OH])/([0,15][0,36]^2)](https://img.qammunity.org/2020/formulas/chemistry/college/s99gtgdv2z395l7b0ubu10xkf3vaza6azk.png)
Solving, the concentration of CH₃OH in equilibrium is:
[CH₃OH] = 0,28M ≡ 2,8x10⁻¹ M
I hope it helps!