Question

An equilibrium mixture contains 0.6500.650 mol of each of the products (carbon dioxide and hydrogen gas) and 0.2000.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00 L container.

CO(g)+H2O(g)−⇀↽−CO2(g)+H2(g)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 0.3000.300 mol once equilibrium has been reestablished?

Answer 1

Sol :-

Equilibrium concentration of each products CO₂ (g) and H₂ (g) = Moles/Volume in L

= 0.650 mol / 1.00 L

= 0.650 M

Similarly,

Equilibrium concentration of each reactants CO (g) and H₂O (g) = Moles/Volume in L

= 0.200 mol / 1.00 L

= 0.200 M

Given reaction is :

CO (g) + H₂O (g) <----------> CO₂ (g) + H₂ (g)

Expression of Equilibrium constant i.e. K_c(Equal to the product of the molar concentration of products to the product of the molar concentration of reactants raise to power of their stoichiometric coefficient at equilibrium stage of the reaction).

K_c = [CO₂].[H₂]/[CO].[H₂O]

= (0.650)²/(0.200)²

= 10.5625

Let concentration of CO₂ = C mol/L

So,ICE table is :

.............................CO (g)................... +.................. H₂O (g) <----------> CO₂ (g)................ +.................. H₂ (g)

Initial (I).................0.200 M........................................0.200 M..................C mol/L.....................................0.650 M

Change (C)............+0.100 M......................................+0.100 M...............-0.100 M.....................................-0.100 M

Equilibrium (E).......0.300 M........................................0.300 M....................(C-0.100) M..............................0.550 M

again, Expression of Equilibrium constant i.e. K_c

K_c = [CO₂].[H₂]/[CO].[H₂O]

10.5625 = (C-0.100).(0.550)/ (0.300)²

10.5625 x (0.300)² / (0.550) = C-0.100

C-0.100 = 1.728

C = 1.728 + 0.100

C = 1.828 mol/L

Also, Moles of carbon dioxide = 1.828 mol/L x 1.00 L = 1.828 mol

Moles of carbon dioxide added must be = 1.828 mol - 0.300 mol

Hence, moles of carbon dioxide added = 1.528 mol