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].
[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].
