Question

How do you find the equilibrium constant for the reverse reaction of this problem?
Given:
Temperature: 1500 degree Celcius
Equilibrium Constant: 0.1764
Balanced equation: ${\text{CO}} + {\text{3}}{{\text{H}}_{\text{2}}} \to {\text{C}}{{\text{H}}_{\text{4}}} + {{\text{H}}_{\text{2}}}{\text{O}}$
Need:
Equilibrium constant for ${\text{C}}{{\text{H}}_{\text{4}}} + {{\text{H}}_{\text{2}}}{\text{O}} \to {\text{CO}} + {\text{3}}{{\text{H}}_{\text{2}}}$

Answer 1

To solve this we have to first write the expression for the equilibrium constant of both the reactions. From the two expressions, find the relationship between the equilibrium constants of the two reactions.

Complete solution:
A chemical equilibrium is a state where the rates of the forward reaction and the reverse reaction are equal, so the net amounts of reactants and products do not change.
An expression that describes the state of equilibrium in terms of concentrations of reactant and product is known as the expression for the equilibrium constant.
The expression for the equilibrium constant is the ratio of product of molar concentrations of product species to the product of molar concentrations of reactant species.
We are given a reaction as follows:
${\text{CO}} + {\text{3}}{{\text{H}}_{\text{2}}} \to {\text{C}}{{\text{H}}_{\text{4}}} + {{\text{H}}_{\text{2}}}{\text{O}}$
The expression for the equilibrium constant for the reaction is as follows:
${{\text{K}}_1} = \dfrac{{[{\text{C}}{{\text{H}}_{\text{4}}}][{{\text{H}}_{\text{2}}}{\text{O}}]}}{{[{\text{CO}}]{{[{{\text{H}}_{\text{2}}}]}^3}}}$ …… (1)
Where ${{\text{K}}_1}$ is the equilibrium constant.
We are given a reaction as follows:
${\text{C}}{{\text{H}}_{\text{4}}} + {{\text{H}}_{\text{2}}}{\text{O}} \to {\text{CO}} + {\text{3}}{{\text{H}}_{\text{2}}}$
The expression for the equilibrium constant for the reaction is as follows:
${{\text{K}}_2} = \dfrac{{[{\text{CO}}]{{[{{\text{H}}_{\text{2}}}]}^3}}}{{[{\text{C}}{{\text{H}}_{\text{4}}}][{{\text{H}}_{\text{2}}}{\text{O}}]}}$ …… (2)
Where ${{\text{K}}_2}$ is the equilibrium constant.
From equation (1) and equation (2),
${{\text{K}}_2} = \dfrac{1}{{{{\text{K}}_1}}}$
We are given that the equilibrium Constant: 0.1764 i.e. ${{\text{K}}_1}$ is 0.1764. Thus,
${{\text{K}}_2} = \dfrac{1}{{0.1764}}$
${{\text{K}}_2} = 5.669$
Thus, the equilibrium constant for the reverse reaction i.e. ${\text{C}}{{\text{H}}_{\text{4}}} + {{\text{H}}_{\text{2}}}{\text{O}} \to {\text{CO}} + {\text{3}}{{\text{H}}_{\text{2}}}$ is 5.669.

Note: The magnitude of the equilibrium constant gives an idea about the rate of the reaction. Higher the value of equilibrium constant higher is the concentration of products and thus, higher is the rate of the reaction. Thus, from the given reactions, the reverse reaction has higher rate.