Question

You are an industrial chemist working for a company that makes methanol (CH₃OH) according to the equilibrium reaction shown below. You love your job, except that your assistant does not have the best notebook keeping skills. One day you set up the reaction with the equilibrium concentrations shown below.

CO (g) + 2 H₂ (g) <--> CH₃OH (g)

0.24 M       1.1 M                0.15 M   at 327^oC

Later on your assistant mentions that he added some more CO, but he doesn't remember how much. But he also remembers that as the equilibrium was being reestablished, the hydrogen concentration changed by 0.3 M, although he doesn't remember if it went up by that much, or down by that much. How much carbon monoxide did he add, in M?

Answer 1

K = [CH3OH]/[CO][H2]^2

K = 0.15 / 0.24 x (1.1)^2

K = 0.516

Adding CO will drive the reaction to the right, increasing the CH3OH and

decreasing the H2: Let y = additional [CO]

CO + 2 H2 -----------------> [CH3OH]

0.24+y 1.1 0.15 ----------- initial

-x . -2x .+x ----------- change

(0.24 + y –x) 1.1-2x 0.15 + x ----------- equilibrium

The hydrogen changed by 0.30

2x = 0.30 ; then x = 0.15

[CO] = 0.24 + y - 0.15 = y + 0.09

[H2] = 1.1 – 0.30 = 0.80

[CH3OH] = 0.15 + 0.13 = 0.28

(0.28)/(y + 0.09)(0.80)^2 = 0.52

0.28 = 0.52(y + 0.09)(0.80)^2

0.28 = 0.52(0.64y + 0.0576)

0.28 = 0.33y + 0.030

0.33y = 0.25

y = 0.76

amount of [CO] added = 0.76