Question

The equilibrium constant for the reaction ${{\text{H}}_{\text{2}}}{\text{O + CO}} \to {{\text{H}}_{\text{2}}}{\text{ + C}}{{\text{O}}_{\text{2}}}$ is $0.44$ at 1260K. What will be the equilibrium constant value for the reaction ${{\text{H}}_{\text{2}}}{\text{ + C}}{{\text{O}}_{\text{2}}} \to {{\text{H}}_{\text{2}}}{\text{O + CO}}$?

Answer 1

Apply the law of equilibrium on both the equation and relate their equilibrium constants. The given reactions are reversible to each other.

Formula used:
If a reaction is given as ${\text{aA}} + {\text{bB}} \to {\text{cC}} + {\text{dD}}$ then,
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{{\text{[C]}}}^{\text{c}}}{{{\text{[D]}}}^{\text{d}}}}}{{{{{\text{[A]}}}^{\text{a}}}{{{\text{[B]}}}^{\text{b}}}}}$
where ${\text{aA}}$ and ${\text{bB}}$ are the reactants and ${\text{cC}}$ and ${\text{dD}}$ are the products,${{\text{K}}_{\text{c}}}$ is the equilibrium constant.

Complete step by step solution:
In the question, the value of equilibrium constant is given which is $0.44$. The equation is ${{\text{H}}_{\text{2}}}{\text{O + CO}} \to {{\text{H}}_{\text{2}}}{\text{ + C}}{{\text{O}}_{\text{2}}}$ …… (1)
Then, we are asked to find the equilibrium constant value for the reaction
${{\text{H}}_{\text{2}}}{\text{ + C}}{{\text{O}}_{\text{2}}} \to {{\text{H}}_{\text{2}}}{\text{O + CO}}$ …….(2)
To find the equilibrium constant, apply the law of equilibrium to this given equation and calculate ${{\text{K}}_{\text{c}}}$.
But first we need to know what law of equilibrium is and how we can calculate ${{\text{K}}_{\text{c}}}$.
It is defined as the product of molar concentrations of the products, each raised to the power of their stoichiometric coefficients divided by the product of the concentration of reactant raised to the power of their stoichiometric coefficients is constant at constant temperature and is called equilibrium constant.
Now, applying this concept in equation (1) we get,
${{\text{K}}_{\text{c}}} = \dfrac{{[{{\text{H}}_{\text{2}}}][{\text{C}}{{\text{O}}_{\text{2}}}]}}{{[{{\text{H}}_{\text{2}}}{\text{O}}][{\text{CO}}]}} = 0.44$
We have to calculate the value of equilibrium constant for the reaction for equation (2). Let it be ${\text{K}}{{\text{'}}_{\text{c}}}$.
${\text{K}}{{\text{'}}_{\text{c}}} = \dfrac{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{O][CO]}}}}{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{][C}}{{\text{O}}_{\text{2}}}{\text{]}}}}$
Now we can see that equation (2) is the reversible reaction of equation (1). We all know that if the reaction is reversed, the value of equilibrium constant is also inverted. This is one of the characteristic properties. We can relate both the equations as:
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{K}}{{\text{'}}_{\text{c}}}}}$
$\Rightarrow {\text{K}}{{\text{'}}_{\text{c}}}{\text{ = }}\dfrac{{\text{1}}}{{{{\text{K}}_{\text{c}}}}}$
Now, putting the values in the above equation we get,
${\text{K}}{{\text{'}}_{\text{c}}} = \dfrac{1}{{0.44}}$
$\Rightarrow {\text{K}}{{\text{'}}_{\text{c}}} = 2.27$

Hence, the value of equilibrium constant ${\text{K}}{{\text{'}}_{\text{c}}}$ comes out to be $2.27$.

Note:
While solving such questions one should always take care of the stoichiometric coefficients involved in the reaction. Remember the characteristics properties to relate the equations e.g. if the equation is reversed, equilibrium constant will become $\dfrac{{\text{1}}}{{{{\text{K}}_{\text{c}}}}}$, if divided by 2, then $\sqrt {{{\text{K}}_{\text{c}}}}$ and if multiplied by 2, then it will be${{\text{K}}^{\text{2}}}$.