One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation, H2O (g) + CO (g) ⇌ H2 (g) + CO2 (g). Calculate the equilibrium constant for the reaction.
Solution 1
The given reaction is:
`H_2O_(g) + CO_(g) ↔ H_(2(g)) + CO_(2(g))`
Initial conc 1/10M 1/10M 0 0
At equilibrium (1-0.4)/10 M (1-0.4)/10 M 0.4/10 M 0.4/10 m
=0.06M =0.06M =0.04M =0.04M
Therefore, the equilibrium constant for the reaction,
`K_c = ([H_2][CO_2])/([H_2O][CO])`
`= (0.04 xx 0.04)/(0.06 xx 0.06)`
= 0.444 (approximately)
Solution 2
Number of moles of water originally present = 1 mol
Percentage of water reacted =40%
Number of moles of water reacted = 1 x 40/100 = 0.4 mol
Number of moles of water left = (1 – 0.4) = 0.6 mole According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hydrogen and 0.4 mole of carbon dioxide.
Thus, the molar cone, per litre of the reactants and products before the reaction and at the equilibrium point are as follows:
`H_2O(g) + CO(g) ⇌ H_2(g) + CO_2(g)`
Initial moles/litre 1/10 1/10 0 0
Mole/litre at the equilibrium point
(1-0.4)/10 = 0.6/10 (1-0.4)/10 = 0.6/10 0.4/10 0.4/10
Applying law of chemical equilibrium point,
Equilibrium constant (`K_c`) = `([H_2(g)][CO_2(g)])/([H_2O(g)][CO(g)]) = ((0.4/10 "mol L"^(-1)) xx (0.4/10 "mol L"^(-1)))/((0.6/10 "mol L"^(-1))xx(0.6/10 "mol L"^(-1)))`
`= 0.16/0.36 = 0.44`
