You are an industrial chemist working for a company that makes methanol (CH 3 OH

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 327oC

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?

Explanation / Answer

Kc= [CH3OH]/{[CO][H2]^2}
= 0.15 / (0.24 * 1.1^2)
=0.5165

Since some CO was added, H2 concentration will go down because adding reactant shift equilibrium towards product side

Let added CO be y M

CO (g) + 2 H2 (g) <--> CH3OH (g)

0.24 + y       1.1                   0.15       (initial)
0.24 + y -x   1.1-2x            0.15+x      (at equilibrium)

hydrogen concentration changed by 0.3 M
so,
2x = 0.3 M
x = 0.15 M

Kc = (0.15+x) / {(1.1-2x)^2 (0.24 + y - x)}
0.5165 = (0.15+0.15) / {(1.1 - 2*0.15)^2 (0.24 + y - 0.15)}
0.5165 = (0.30) / {(0.8)^2 * (0.09 + y)^2}
(0.09 + y)^2 = 0.9076
0.09 + y = 0.9527
y = 0.86 M

Answer: 0.86 M

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