Question

Consider the following gaseous equilibria given below:

1. ${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3$; Eqm. Constant = ${\mathrm{K}}_1$
2. ${\mathrm{N}}_2+{\mathrm{O}}_2\rightleftharpoons 2\mathrm{NO}$; Eqm. constant = ${\mathrm{K}}_2$
3. ${\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2\rightleftharpoons {\mathrm{H}}_2\mathrm{O}$; Eqm. constant= ${\mathrm{K}}_3$

The Equilibrium constant for the reaction: $2{\mathrm{NH}}_3+\frac{5}{2}{\mathrm{O}}_2\rightleftharpoons 2\mathrm{NO}+3{\mathrm{H}}_2\mathrm{O}$, In Terms of ${\mathrm{K}}_1$, ${\mathrm{K}}_2$ and ${\mathrm{K}}_3$will be _________.

Answer 1

Step-1: Calculate the Equilibrium constant

• In a chemical equilibrium, the equilibrium constant is written as the ratio of the concentration of product to the concentration of reactant raised to the power of their respective stoichiometric coefficient.

${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3$

${\mathrm{K}}_1=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$---------------------------- (i)

${\mathrm{N}}_2+{\mathrm{O}}_2\rightleftharpoons 2\mathrm{NO}$

${\mathrm{K}}_2=\frac{{\left[\mathrm{NO}\right]}^2}{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]}$--------------------------------- (ii)

${\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2\rightleftharpoons {\mathrm{H}}_2\mathrm{O}$

${\mathrm{K}}_3=\frac{\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[{\mathrm{H}}_2\right]{\left[{\mathrm{O}}_2\right]}^{{\displaystyle \frac{1}{2}}}}$ --------------------------- (iii)

Step-2: Equilibrium constant of $2{\mathrm{NH}}_3+\frac{5}{2}{\mathrm{O}}_2\rightleftharpoons 2\mathrm{NO}+3{\mathrm{H}}_2\mathrm{O}$

$\mathrm{K}=\frac{{\left[{\mathrm{NO}}_2\right]}^2{\left[{\mathrm{H}}_2\mathrm{O}\right]}^3}{{\left[{\mathrm{NH}}_3\right]}^2{\left[{\mathrm{O}}_2\right]}^{{\displaystyle \frac{5}{2}}}}$----------------------(iv)

Step-3: Calculate $\mathrm{K}$ in terms of ${\mathrm{K}}_1,{\mathrm{K}}_2,\mathrm{and}{\mathrm{K}}_3$

Cubing equation (iii)

${\mathrm{K}}_3^3=\frac{{\left[{\mathrm{H}}_2\mathrm{O}\right]}^3}{{\left[{\mathrm{H}}_2\right]}^3{\left[{\mathrm{O}}_2\right]}^{{\displaystyle \frac{3}{2}}}}$----------- (v)

multiplying eq(v) and eq(ii)

${\mathrm{K}}_3^3\times {\mathrm{K}}_2=\frac{{\left[\mathrm{NO}\right]}^2{\left[{\mathrm{H}}_2\mathrm{O}\right]}^3}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3{\left[{\mathrm{O}}_2\right]}^{{\displaystyle \frac{5}{2}}}}$----------- (vi)

Divide eq(vi) by eq(I)

$\frac{{\mathrm{K}}_3^3\times {\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{{\left[\mathrm{NO}\right]}^2{\left[{\mathrm{H}}_2\mathrm{O}\right]}^3\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3{\left[{\mathrm{O}}_2\right]}^{{\displaystyle \frac{5}{2}}}{\left[{\mathrm{NH}}_3\right]}^2}=\frac{{\left[{\mathrm{NO}}_2\right]}^2{\left[{\mathrm{H}}_2\mathrm{O}\right]}^3}{{\left[{\mathrm{NH}}_3\right]}^2{\left[{\mathrm{O}}_2\right]}^{{\displaystyle \frac{5}{2}}}}=\mathrm{K}$

Hence, The Equilibrium constant $\mathrm{K}$ in terms of ${\mathrm{K}}_1$, ${\mathrm{K}}_2$ and ${\mathrm{K}}_3$ is $\mathrm{K}=\frac{{\mathrm{K}}_3^3\times {\mathrm{K}}_2}{{\mathrm{K}}_1}$