The equilibrium constant, K c, is equal to 1.4 at 1200° K for the reaction: CO2(

Question

The equilibrium constant, K c, is equal to 1.4 at 1200° K for the reaction:
CO2(g) + H2(g) CO(g) + H2O(g)
If 0.65 moles of CO2 and 0.65 moles of H2 are introduced into a 1.0-L flask, what will be the concentration of CO when equilibrium is reached?

The equilibrium constant, K c, is equal to 1.4 at 1200° K for the reaction:
CO2(g) + H2(g) CO(g) + H2O(g)
If 0.65 moles of CO2 and 0.65 moles of H2 are introduced into a 1.0-L flask, what will be the concentration of CO when equilibrium is reached?

Explanation / Answer

we know that

conc = moles / volume

given

volume = 1 L

so

conc = moles

so

initial conc of C02 = 0.65 M

initial conc of H2 = 0.65 M

the given reaction is

C02 + H2 ---> C0 + H20

the equilibrium constant is given by

Kc = [C0] [H20] / [C02] [H2]

now

consider the reaction

C02 + H2 ---> C0 + H20

using ICE table

initial conc of C02 , H2 , C0 , H20 are 0.65 ,0.65 , 0 , 0

change in conc of C02 , H2 , C0 , h20 are -x , -x ,+x , +x

equilibrium conc of C02 , H2 , CO , H20 are 0.65 -x , 0.65 - x , x ,x

now

Kc = [C0] [H20] / [C02] [H2]

1.4 = x2 / [0.65-x]^2

solving we get

x = 0.3523

now

[CO]eq = x

[C0]eq = 0.3523

so

the conc of CO at equilibrium is 0.35 M

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