Answer to Question #142171 in Physical Chemistry for Iqfa

Question

For the reaction CO2(g) + H2(g) = CO(g) + H2O(g)
the equilibrium constant at 1000 K is 0.53.
(a) If a mixture at equilibrium in a 1 dm3 vessel contains 0.25 mole of CO and 0.6 mole of H2, how
many moles of H2O are there in the vessel?
(b) 5 moles of inert gas are added to the equilibrium mixture containing 1 mole of H2 and 1 mole of CO2
in 1 dm3 vessel. Predict equilibrium concentration of CO2 and H2O.

Answer 1

CO2(g) + H2(g) = CO(g) + H2O(g)

the equilibrium constant at 1000 K is 0.53.

(a) Equilibrium constant K = p(H2O)*p(CO)/{p(CO2)*p(H2)}

p = n*RT/V according to gas law, therefore:

K = 0.53 = n(H2O)*n(CO)/{n(CO2)*n(H2)} = 0.25n(H2O)/0.6n(CO2)

The ratio of n(H2O)/n(CO2) = 0.53*0.6/0.25 = 1.272. To find the number of moles of H2O and CO2, we should have an additional equation, e.g. for the total number of moles of gases, or alternately, total pressure in the vessel, but it is not indicated in conditions.

Therefore, let's assume that initially only reactants CO2 and H2 were in the vessel. Hence, the products CO and H2O should be produced in equal amounts in equilibrium, i.e. the number of moles of H2O is 0.25 mole.

(b) in equilibrium there are 1 mole of H2 and 1 mole of CO2. Equilibrium constant is

K = 0.53 = n(H2O)*n(CO)/{n(CO2)*n(H2)} = n(H2O)*n(CO)/{1*1}

Again, since total number of moles of gases or total pressure is not given, we assume that H2O and CO are produced equally. Then,

n(H2O) = n(CO) = sqrt(0.53) = 0.728 mole.

Therefore, the total number of moles of gases including inert gas is:

n(gas) = 5 + 1 + 1 + 0.728 + 0.728 = 8.456 mole

The concentrations of CO2 and H2O will be:

C(CO2) = 1 M, volume fraction of CO2 = 1/8.456*100% = 11.83 vol.%

C(H2O) = 0.728 M, volume fraction of H2O = 0.728/8.456*100% = 8.61 vol.%