Calculate the value of [tex]K_{p}[/tex] for the equation

[tex]\[C (s) + CO_2(g) \rightleftharpoons 2 CO (g) \quad K_{p} = \, ?\][/tex]

given that at a certain temperature:

[tex]\[
\begin{array}{ll}
C (s) + 2 H_2 O (g) \rightleftharpoons CO_2(g) + 2 H_2(g) & K_{p1} = 3.35 \\
H_2(g) + CO_2(g) \rightleftharpoons H_2 O (g) + CO (g) & K_{p2} = 0.745
\end{array}
\][/tex]



Answer :

To determine the equilibrium constant [tex]\( K_p \)[/tex] for the reaction:

[tex]\[ C (s) + CO_2 (g) \rightleftharpoons 2 CO (g) \][/tex]

we are given the equilibrium constants for two related reactions:

1. [tex]\( C (s) + 2 H_2O (g) \rightleftharpoons CO_2 (g) + 2 H_2 (g) \)[/tex], with [tex]\( K_{p1} = 3.35 \)[/tex]
2. [tex]\( H_2 (g) + CO_2 (g) \rightleftharpoons H_2O (g) + CO (g) \)[/tex], with [tex]\( K_{p2} = 0.745 \)[/tex]

We need to manipulate these reactions to arrive at the target reaction and determine its equilibrium constant.

Step-by-Step Solution:

1. Reverse the second reaction:

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

Reversing it gives:

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

When a reaction is reversed, its equilibrium constant becomes the reciprocal of the original equilibrium constant. Therefore:

[tex]\[ K'_{p2} = \frac{1}{K_{p2}} = \frac{1}{0.745} \][/tex]

2. Adjust the first reaction:

[tex]\[ C (s) + 2 H_2O (g) \rightleftharpoons CO_2 (g) + 2 H_2 (g) \][/tex]

We need to multiply the stoichiometric coefficients by [tex]\( \frac{1}{2} \)[/tex]:

[tex]\[ \frac{1}{2} C (s) + H_2O (g) \rightleftharpoons \frac{1}{2} CO_2 (g) + H_2 (g) \][/tex]

When we multiply the reaction coefficients by a factor, the equilibrium constant is raised to the power of that factor. Therefore:

[tex]\[ K'_{p1} = \sqrt{K_{p1}} = (3.35)^{1/2} \][/tex]

3. Combine the modified reactions:

[tex]\[ \begin{array}{rll} (\frac{1}{2}) C (s) + H_2O (g) & \rightleftharpoons & \frac{1}{2} CO_2 (g) + H_2 (g) , & \sqrt{3.35} \\ CO (g) + H_2O (g) & \rightleftharpoons & H_2 (g) + CO_2 (g) , & \frac{1}{0.745} \end{array} \][/tex]

Adding these two reactions, we get the desired reaction:

[tex]\[ (\frac{1}{2}) C (s) + H_2O (g) + CO (g) + H_2O (g) \rightleftharpoons \frac{1}{2} CO_2 (g) + H_2 (g) + H_2 (g) + CO_2 (g) \][/tex]

Simplifying the sums of the reactants and products:

[tex]\[ C (s) + CO_2 (g) \rightleftharpoons 2 CO (g) \][/tex]

The equilibrium constant for the combined reaction is the product of the equilibrium constants of the individual reactions:

[tex]\[ K_p = \sqrt{3.35} \times \frac{1}{0.745} = \frac{\sqrt{3.35}}{0.745} \][/tex]

4. Calculate the value:

[tex]\[ \sqrt{3.35} \approx 1.83 \quad \text{(rounded to two decimal places)} \][/tex]

Therefore:

[tex]\[ K_p = \frac{1.83}{0.745} \approx 2.46 \quad \text{(rounded to two decimal places)} \][/tex]

So, the equilibrium constant [tex]\( K_p \)[/tex] for the reaction [tex]\( C (s) + CO_2 (g) \rightleftharpoons 2 CO (g) \)[/tex] is approximately [tex]\( 2.46 \)[/tex].