An equilibrium mixture contains 0.350 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00 L container.

Question

CO(g)+H2O(g)↽−−⇀CO2(g)+H2(g)

How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol once equilibrium has been reestablished?

Answer 1

CO_(g) + H₂O _(g) <======>. CO_2(g). +. H_2(g).

Initial Eq. Conc. (M): 0.200 0.200 0.350 0.350

(since the volume is 1.00 L, number of moles and concentration in molarity are numerically the same)

Equilibrium constant, K_c =. (0.350)(0.350) / (0.200)(0.200) = 3.0625

Let assume the increase in conc. of CO₂ = X molar.

When the conc. of CO₂ is increased and the reaction is allowed to re-equilibrate, the conc. of CO increased to 0.300M, and conc. of H₂O also has to increase to 0.300 M, since CO and H₂O are formed with 1:1 ratio in the reverse reaction. The conc. of H₂ has to decrease by 0.100 M (there is 1:1:1 ratio between H₂, CO and H₂O, so the decrease in the conc. of H₂ has to be equal to increase in conc. of CO and H₂O).

CO_(g) + H₂O _(g) <======>. CO_2(g). +. H_2(g).

The new equilibrium conc.(M): 0.300, 0.300, (0.350 + X), 0.250

If we substitute the new equilibrium concentrations in the K_c expression:

K_c. =. [CO₂][H₂] / [CO][H₂O]. =. (0.350+ X) (0.250) / (0.300)(0.300) = 3.0625 -------(i)

(K_c is a constant for a given reaction and changes only with temperature)

Solve eqn. (i) for X:

3.0625 = (0.0875 +. 0.250X) / 0.0900

(3.0625)(0.0900). = 0.0875. +. 0.250X

0.2756 = 0.0875 + 0.250X

X = 0.7525 M

Answer: the number of moles of CO₂ added = 0.753 mole.