# 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?

##### 1 Answer

Methanol's concentration at equilibrium is

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

....

**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 **0.01 - 0.001 = 0.009** and **0.02 - 2 * (0.001) = 0.018**, respectively,

Let's set up the second experiment. The starting concentration for

....

**I**:...0.01M......................0..............0

**C**.....(-x).......................(+x)...........(+2x)

**E**...(0.01-x)...................x...............2x

We know that for the reverse reaction,

Therefore, the concentrations at equilibrium are