Question

$28$ g of ${{N}_{2}}$ and $6$g of ${{H}_{2}}$ were kept at ${{400}^{{\mathrm O}}}C$ in $1$ litre vessel, the equilibrium mixture contained $27.54$g of $N{{H}_{3}}$. The approximate value of ${{K}_{c}}$ the above equation can be: ( in $mol{{e}^{-2}}litr{{e}^{2}}$).
A. $75$
B. $50$
C. $25$
D. $100$

Answer 1

Hint: ${{K}_{c}}$ is an equilibrium constant of a reversible reaction at equilibrium with respect to concentration. It describes the relationship between the concentration of reactants and products at equilibrium at a specific temperature. Putting these values in the corresponding equilibrium constant equation we can calculate ${{K}_{c}}$.

Formula used: Let us, consider a general reversible chemical reaction:
$aA+bB\rightleftharpoons mM+dD$
${{K}_{c}}=\frac{{{[M]}^{m}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$ [In terms of concentration]
[M] = Concentration of product
[D] = Concentration of product
[A] = Concentration of reactant
[B] = Concentration of reactant
a,b and m,d are the stoichiometric coefficients of reactants and products respectively.

Complete step by step solution:
The equilibrium constant ${{K}_{c}}$ is defined by the ratio of molar concentrations of products to molar concentrations of reactants, each of them raised to the power equal to the stoichiometric coefficient at a specific temperature. ${{K}_{c}}$ has a specific value at a particular temperature and it is temperature dependent. If the temperature of the system changes, the value ${{K}_{c}}$ also changes.

Initially, we have $28g$${{N}_{2}}$ $1$ mole ${{N}_{2}}$ and $6g$${{H}_{2}}$ or $3$ mol ${{H}_{2}}$ in $1$ litre vessel.
${{N}_{2}} \ \ + \ \ 3{{H}_{2}} \ \ \rightleftharpoons \ \ 2N{{H}_{3}}$
Initial concentrations at time, $t=0$ $N_2 = \text{1mol},$ $ H_2 = \text{3mol}$ $NH_3 = \text{0mo}l$
Change in concentrations $\alpha $ $\alpha $ $0$
Time, $t={{t}_{eq}}$ $(1-\alpha )$ $3(1-\alpha )$ $2\alpha $

$\alpha $ is the change in concentration of reactants and products at equilibrium.
Given, the amount of $N{{H}_{3}}$ equilibrium $27.54g$
The number of moles $N{{H}_{3}}$ at equilibrium,$\text{n=}\frac{\text{m}}{M}=\frac{27.54}{17}=1.62$
Here m = mass of $N{{H}_{3}}$
M = Molar mass of$N{{H}_{3}}$
Given, $2\alpha =1.62$
Or, $\alpha =0.81$

According to the formula, ${{K}_{c}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}$
$[N{{H}_{3}}]=\frac{{{n}_{N{{H}_{3}}}}}{V}=\frac{(2\times \alpha )mole}{1litre}=(2\alpha )mole.lir{{e}^{-1}}$
$[{{H}_{2}}]=\frac{{{n}_{{{H}_{2}}}}}{V}=\frac{3(1-\alpha )mole}{1 litre}=3(1-\alpha )mole.lir{{e}^{-1}}$
$[{{H}_{2}}]=\frac{{{n}_{{{H}_{2}}}}}{V}=\frac{3(1-\alpha )mole}{1litre}=3(1-\alpha )mole.lir{{e}^{-1}}$
${{K}_{c}}=\frac{{{[(2\alpha )mole.litr{{e}^{-1}}]}^{2}}}{[(1-\alpha )mole.litr{{e}^{-1}}]\times {{[3(1-\alpha )mole.litr{{e}^{-1}}]}^{3}}}$
${{K}_{c}}=\frac{{{(2\times 0.81)}^{2}}}{(1-0.81)\times {{[3(1-0.81)]}^{3}}}mol{{e}^{-2}}litr{{e}^{2}}$
${{K}_{c}}=\frac{2.6244}{0.19\times 0.1852}mol{{e}^{-2}}litr{{e}^{2}}$
${{K}_{c}}=74.58\approx 75mol{{e}^{-2}}litr{{e}^{2}}$
Therefore the approximate value of the equilibrium constant${{K}_{c}}$ is$75mol{{e}^{-2}}litr{{e}^{2}}$.
Thus, option (A) is correct.

Note: To approach these types of numerical problems, one must focus on writing an appropriate equilibrium constant equation according to the given equilibrium equation and also take care of stoichiometric coefficients of reactants and products. These will help to calculate the appropriate equilibrium constant value without any mistake.