Question

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?

Answer 1

Answer:

a) The value of the $K_c$ is 0.52.

b)Concentration of all species when equilibrium reestablishes:

$[CO_2] = 0.4748 M$

$[H_2] = 0.0198 M$

$[CO] = 0.0752 M$

$[H_2O] =0.0652 M$

Explanation:

a)  $CO_2(g) + H_2(g)\rightleftharpoons CO(g) + H_2O(g)$

Equilibrium concentration of species :

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

The expression of equilibrium constant is given as :

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

$=\frac{0.050 M\times 0.040 M}{0.086 M\times 0.045 M}=0.517\approx 0.52 M$

b)

$CO_2(g) + H_2(g)\rightleftharpoons CO(g) + H_2O(g)$

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

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

$0.52=\frac{(0.050+x) M\times (0.040+x) M}{(0.50-x) M\times (0.045-x) M}$

Solving fro x;

x = 0.0252

Concentration of all species when equilibrium reestablishes:

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

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

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

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