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: ); for the first experiment, the 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
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