Consider the following equilibrium process at 686°C. CO2(g) + H2(g) equilibrium reaction arrow CO(g) + H2O(g) The equilibrium concentrations of the reacting species are [CO] = 0.050 M, [H2] = 0.045 M, [CO2] = 0.086 M, and [H2O] = 0.040 M. (a) Calculate Kc for the reaction at 686°C. WebAssign will check your answer for the correct number of significant figures. (b) If we add CO2 to increase its concentration to 0.50 mol/L, what will the concentrations of all the gases be when equilibrium is reestablished?

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Tringa0 Tringa0 · Answer 1 · 2020-02-21T07:49:59+01:00

Answer:

a) The value of the [tex]K_c[/tex] is 0.52.

b)Concentration of all species when equilibrium reestablishes:

[tex][CO_2] = 0.4748 M[/tex]

[tex][H_2] = 0.0198 M[/tex]

[tex] [CO] = 0.0752 M[/tex]

[tex][H_2O] =0.0652 M[/tex]

Explanation:

a) [tex]CO_2(g) + H_2(g)\rightleftharpoons CO(g) + H_2O(g)[/tex]

Equilibrium concentration of species :

[tex][CO] = 0.050 M, [H_2] = 0.045 M, [CO_2] = 0.086 M, and [H_2O] = 0.040 M[/tex]

The expression of equilibrium constant is given as :

[tex]\K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}[/tex]

[tex]=\frac{0.050 M\times 0.040 M}{0.086 M\times 0.045 M}=0.517\approx 0.52 M[/tex]

b)

[tex]CO_2(g) + H_2(g)\rightleftharpoons CO(g) + H_2O(g)[/tex]

Initially:

0.50 M 0.045 M 0.050 M 0.040 M

At equilibrium ;

(0.50-x) M (0.045-x) M (0.050+x) M (0.040+x) M

[tex]K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}[/tex]

[tex]0.52=\frac{(0.050+x) M\times (0.040+x) M}{(0.50-x) M\times (0.045-x) M}[/tex]

Solving fro x;

x = 0.0252

Concentration of all species when equilibrium reestablishes:

[tex][CO_2] = (0.50-x) M=(0.50-0.0252)M = 0.4748 M[/tex]

[tex][H_2] = (0.045-x) M= (0.045-0.0252) M=0.0198 M[/tex]

[tex] [CO] = (0.050+x) M=(0.050+0.0252)M = 0.0752 M[/tex]

[tex][H_2O] = (0.040+x) M=(0.040+0.0252) M=0.0652 M[/tex]