# At some temperature, a 100-L reaction vessel contains a mixture that is initially 1.0 mol Co and 2.0 mol H2. The vessel also contains a catalyst so that the following equilibrium is attained: CO(g) + 2H2(g) <--> CH3OH(g) At equilibrium, the mixture contains 0.100 mol CH3OH. In a later experiment in the same vessel, you start with 1.0 mol CH3OH. How much methanol is there at equilibrium?

Dec 23, 2014

Methanol's concentration at equilibrium is $0.001 M$.

Before solving for the concentration, try and intuitively imagine what will happen. Since the second experiment is done under the same conditions as the first one (same 100-L vessel, same catalyst), we could estimate that the equilibrium concentration for this reaction will equal the equilibrium concentration of the first experiment.

Let's approach the problem by using the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart); for the first experiment, the concentrations of the gases are $0.01 M$ for $C O$ and $0.02 M$ for ${H}_{2}$ (because they're placed in a 100-L vessel)

....$C {O}_{\left(g\right)} + 2 {H}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} O {H}_{\left(g\right)}$
I:....0.01........0.02.......................0
C....(-x)..........(-2x).....................(+x)
E:...(0.01-x)...(0.02-2x)...............x

Since we know that the equilibrium concentration of $C {H}_{3} O H$ is $0.001 M$, we can determine the equilibrium constant for this reaction, knowing that the equilibrium concentrations of $C O$ and ${H}_{2}$ are 0.01 - 0.001 = 0.009 and 0.02 - 2 * (0.001) = 0.018, respectively,

${K}_{1} = \frac{C {H}_{3} O H}{\left[C O\right] \cdot {\left[{H}_{2}\right]}^{2}} = \frac{0.001}{0.009 \cdot {0.018}^{2}} = 343$

Let's set up the second experiment. The starting concentration for $C {H}_{3} O H$ will now be $0.01 M$.

....$C {H}_{3} O {H}_{\left(g\right)} r i g h t \le f t h a r p \infty n s C {O}_{\left(g\right)} + 2 {H}_{2 \left(g\right)}$
I:...0.01M......................0..............0
C.....(-x).......................(+x)...........(+2x)
E...(0.01-x)...................x...............2x

We know that for the reverse reaction, ${K}_{2} = \frac{1}{K} _ 1 = \frac{1}{343} = 0.003$, which means that we can determine the equilibrium concentrations of all the gases by finding $x$.

${K}_{2} = 0.003 = \frac{x \cdot {\left(2 x\right)}^{2}}{0.01 - x} = \frac{4 {x}^{3}}{0.01 - x}$. This equation produces 3 values for $x$, 2 negative, which will be discarded since concentrations cannot be negative, and one positive, $x = 0.009$.

Therefore, the concentrations at equilibrium are

$\left[C {H}_{3} O H\right] = 0.01 - 0.009 = 0.001 M$
$\left[C O\right] = 0.009 M$
$\left[{H}_{2}\right] = 2 \cdot 0.009 = 0.018 M$