Question

Be sure to answer all parts. Consider the following equilibrium process at 686°C: CO2(g) + H2(g) ⇌ CO(g) + H2O(g) The equilibrium concentrations of the reacting species are [CO] = 0.0500 M, [H2] = 0.0410 M, [CO2] = 0.0880 M, and [H2O] = 0.0380 M. (a) Calculate Kc for the reaction at 686°C.(b) If we add CO2 to increase its concentrationto 0.30 mol/L, what will theconcentrations of all the gases be when equilibrium isreestablished?

Answer 1

Answer:

a)The equilibrium constant at 686°C is 0.527.

b) Concentrations of all the gases be when equilibrium is reestablished:

$[CO_2]=0.2834 M$

$[H_2]=0.02440 M$

$[CO]=0.06660 M$

$[H_2O]=0.05460 M$

Explanation:

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

Equilibrium concentrations:

$[CO]=0.0500 M$

$[H_2]=0.0410 M$

$[CO_2]=0.0880 M$

$[H_2O]=0.0380 M$

The value of an equilibrium constant will be given as:

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

$K_c=\frac{0.0500 M\times 0.0380 M}{0.0880 M\times 0.0410 M}=0.527$

The equilibrium constant at 686°C is 0.527.

b)

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

Initially

0.30 M    0.0410 M              0.0500 M  0.0380 M

At reestablishment of an equilibrium;

(0.30-x) M    (0.0410-x) M             (0.0500+x) M  (0.0380+x) M

The value of an equilibrium constant will be given as:

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

$0.527=\frac{(0.0500+x)M\times (0.0380+x)M}{(0.30-x)M\times(0.0410-x)M}$

Solving for x;

x = 0.0166

Concentrations of all the gases be when equilibrium is reestablished:

$[CO_2]=(0.30-x)=(0.30-0.0166) M=0.2834 M$

$[H_2]=(0.0410-x)=(0.0410-0.0166) M=0.02440 M$

$[CO]=(0.0500+x)=(0.0500+0.0166) M=0.06660 M$

$[H_2O]=(0.0380+x)=(0.0380+0.0166) M=0.05460 M$