Question

The equilibrium constant for the reactions ${\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)$ and $\mathrm{NO}\left(\mathrm{g}\right)\rightleftharpoons \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)$ are $\mathrm{k}$ and ${\mathrm{k}}_1$ respectively, the relation between $\mathrm{k}$ and ${\mathrm{k}}_1$ is

• A. $k={\left(k_1\right)}^2$
• B. $k={\left(\frac{1}{k_1}\right)}^2$
• C. $k^2=k_1$
• D. $k_1={\left(\frac{1}{k}\right)}^2$

Answer 1

The correct option is B

$\mathrm{k}={\left(\frac{1}{{\mathrm{k}}_1}\right)}^2$

• Equilibrium constant (K_C): For a chemical reaction, the equilibrium constant can be defined as the ratio between the amount of reactant and the amount of product which is used to determine chemical behaviour.
• The equilibrium constant for a general reaction $\mathrm{aA}+\mathrm{bB}\rightleftharpoons \mathrm{cC}+\mathrm{dD}$ is

${\mathrm{K}}_{\mathrm{C}}=\frac{{\left[\mathrm{C}\right]}^{\mathrm{C}}{\left[\mathrm{D}\right]}^{\mathrm{d}}}{{\left[\mathrm{A}\right]}^{\mathrm{a}}{\left[\mathrm{B}\right]}^{\mathrm{b}}}$

• In homogeneous equilibrium, all the reactants and products are in one phase only. If all the reactants and products are in the gaseous phase, the molar concentrations of all the substances are taken into consideration for writing the equilibrium constant.

Example: ${\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{I}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{HI}\left(\mathrm{g}\right)$

${\mathrm{K}}_{\mathrm{C}}=\frac{{\left[\mathrm{HI}\right]}^2}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{I}}_2\right]}$

• In heterogeneous equilibrium, the reactants and the products exist in different phases, all the reactants and products are not taken into consideration for writing the equilibrium constant. The concentration of the reactants and the products existing in the solid-state and in the pure liquid state are taken in unity since their concentration cannot be changed.

Example : ${\mathrm{NH}}_4\mathrm{HS}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)$

${\mathrm{K}}_{\mathrm{C}}=\frac{\left[{\mathrm{NH}}_3\right]\left[{\mathrm{H}}_2\mathrm{S}\right]}{\left[{\mathrm{NH}}_4\mathrm{HS}\right]}$

Since, $\left[{\mathrm{NH}}_4\mathrm{HS}\right]=1$, ${\mathrm{K}}_{\mathrm{C}}=\left[{\mathrm{NH}}_3\right]\left[{\mathrm{H}}_2\mathrm{S}\right]$

• Both of the given reactions are homogeneous equilibrium.
• The equilibrium constant for the reaction ${\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)$ is $\mathrm{k}$

therefore, $\mathrm{k}=\frac{{\left[\mathrm{NO}\right]}^2}{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]}$

• The equilibrium constant for the reaction $\mathrm{NO}\left(\mathrm{g}\right)\rightleftharpoons \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)$ is ${\mathrm{k}}_1$

therefore, ${\mathrm{k}}_1=\frac{{\left[{\mathrm{N}}_2\right]}^{{\displaystyle \frac{1}{2}}}{\left[{\mathrm{O}}_2\right]}^{{\displaystyle \frac{1}{2}}}}{\left[\mathrm{NO}\right]}$

• squaring both sides in the above equation : ${\left({\mathrm{k}}_1\right)}^2={\left\{\frac{{\left[{\mathrm{N}}_2\right]}^{{\displaystyle \frac{1}{2}}}{\left[{\mathrm{O}}_2\right]}^{{\displaystyle \frac{1}{2}}}}{\left[\mathrm{NO}\right]}\right\}}^2$

${\left({\mathrm{k}}_1\right)}^2=\frac{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]}{{\left[\mathrm{NO}\right]}^2}$

now reciprocating both sides : $\frac{1}{{\left({\mathrm{k}}_1\right)}^2}=\frac{{\left[\mathrm{NO}\right]}^2}{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]}=\mathrm{k}$

it can also be written as : $\mathrm{k}={\left(\frac{1}{{\mathrm{k}}_1}\right)}^2$

• Hence, the correct option is (B).