Question

${{5}}{{.1g}}$ of ${{N}}{{{H}}_4}{{SH}}$ is introduced in $3.0{{L}}$ evacuated flask at ${{32}}{{{7}}^ \circ }{{C}}$. $30\%$ of the solid ${{N}}{{{H}}_4}{{SH}}$ is decomposed to ${{N}}{{{H}}_3}$ and ${{{H}}_2}{{S}}$ as gases. The ${{{K}}_{{p}}}$ of the reaction at ${{32}}{{{7}}^ \circ }{{C}}$ is:
(${{R = 0}}{{.082atm}}{{.mo}}{{{l}}^{ - 1}}.{{{K}}^{ - 1}},$ molar mass of ${{S = 32g}}{{.mo}}{{{l}}^{ - 1}},$ molar mass of ${{N = 14g}}{{.mo}}{{{l}}^{ - 1}}$ ).
A. $0.242 \times {10^{ - 4}}{{at}}{{{m}}^2}$
B. $0.242{{at}}{{{m}}^2}$
C. $4.9 \times {10^{ - 3}}{{at}}{{{m}}^2}$
D. $1 \times {10^{ - 4}}{{at}}{{{m}}^2}$

Answer 1

The forward and reverse reactions proceed at the same rate at equilibrium. Also, when equilibrium is achieved, both the reactants and products remain constant. Equilibrium constant expression is applicable only when the reactant concentrations and product concentrations are constant at equilibrium.

Complete step by step solution:
It is given that $\alpha = 30\% = 0.3$
Mass of ${{N}}{{{H}}_4}{{SH}}$, ${{{m}}_{{{N}}{{{H}}_4}{{SH}}}} = 5.1{{g}}$
Volume, ${{V = }}3.0{{L}}$
Temperature, ${{T = 32}}{{{7}}^ \circ }{{C = 327 + 273 = 600K}}$
Molar mass of ${{S}}$, ${{{M}}_{{S}}} = 32{{g}}.{{mo}}{{{l}}^{ - 1}}$, molar mass of ${{N,}}{{{M}}_{{N}}}{{ = 14g}}{{.mo}}{{{l}}^{ - 1}}$.
Molecular mass of ${{N}}{{{H}}_4}{{SH}}$, ${{{M}}_{{{N}}{{{H}}_4}{{SH}}}} = 14 + \left( {5 \times 1} \right) + 32 = 51{{g}}.{{mo}}{{{l}}^{ - 1}}$
Now consider an equilibrium reaction,
${{aA}} \rightleftharpoons {{bB}} + {{cC}}$
The equilibrium constant, ${{{K}}_{{c}}} = \dfrac{{{{\left[ {{B}} \right]}^{{b}}}{{\left[ {{C}} \right]}^{{c}}}}}{{{{\left[ {{A}} \right]}^{{a}}}}}$, where $\left[ {{A}} \right],\left[ {{B}} \right],\left[ {{C}} \right]$ are the concentrations of ${{A,B,C}}$ and ${{a,b,c}}$ are the stoichiometric coefficients.
Also we know that the ideal gas equation is ${{PV}} = {{nRT}}$, where ${{P}}$ is the pressure, ${{V}}$ is the volume, ${{n}}$ is the number of moles of gas, ${{R}}$ is the gas constant and ${{T}}$ is the temperature.
${{P}} = \dfrac{{{n}}}{{{V}}}{{RT}} \Leftrightarrow {{P}} = {{CRT}}$, where ${{C}}$ is the concentration expressed in ${{mol}}{{{L}}^{ - 1}}$ and ${{R = 0}}{{.082Latm}}{{{K}}^{ - 1}}{{mo}}{{{l}}^{ - 1}}$
Now let’s write the reaction at equilibrium.
${{N}}{{{H}}_4}{{SH}} \rightleftharpoons {{N}}{{{H}}_3} + {{{H}}_2}{{S}}$
Molecular mass of ${{N}}{{{H}}_4}{{SH}}$, ${{{M}}_{{{N}}{{{H}}_{{4}}}{{SH}}}} = 51{{g}}.{{mo}}{{{l}}^{ - 1}}$
Thus the number of moles of ${{N}}{{{H}}_4}{{SH}}$, ${{{n}}_{{{N}}{{{H}}_4}{{SH}}}} = \dfrac{{5.1}}{{51}} = 0.1{{mol}}$
${{N}}{{{H}}_4}{{SH}} \rightleftharpoons {{N}}{{{H}}_3} + {{{H}}_2}{{S}}$

| ${{N}}{{{H}}_4}{{SH}}$| ${{N}}{{{H}}_3}$ | ${{{H}}_2}{{S}}$
---|---|---|---
Initial:| $0.1$| $0$ | $0$
After time t| $0.1\left( { - 1 - \alpha } \right)$| $0.1\alpha$| $0.1\alpha$

Thus the number of moles at equilibrium can be represented as:

$0.1\left( { - 1 - 0.3} \right)$	$0.1 \times 0.3$	$0.1 \times 0.3$
$0.07$	$0.03$	$0.03$

Thus ideal gas equation will be:
${{P}} \times {{3}}{{.0L}} = \left( {0.03 + 0.03} \right){{0}}{{.082}} \times {{600}}$
On simplification, we get
${{P = }}\dfrac{{2.952}}{3} = 0.984{{atm}}$
Thus ${{{P}}_{{{N}}{{{H}}_3}}} = {{{P}}_{{{{H}}_2}{{S}}}} = \dfrac{{{P}}}{2} = \dfrac{{0.984}}{2} = 0.492{{atm}}$
Equilibrium constant with respect to the partial pressure is represented by ${{{K}}_{{p}}}$.
Thus ${{{K}}_{{p}}} = {{{P}}_{{{N}}{{{H}}_3}}} \times {{{P}}_{{{{H}}_2}{{S}}}} \rightleftharpoons {{{K}}_{{p}}} = 0.492 \times 0.492 = 0.242{{at}}{{{m}}^2}$
Thus the equilibrium constant ${{{K}}_{{p}}} = 0.242{{at}}{{{m}}^2}$

Hence, the correct option is B.

Additional information:
Equilibrium constant does not depend on the initial concentrations of reactants and products. It is also temperature dependent. If the equilibrium constant has a very large value, then the reaction is approaching completion.

Note: Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant in the forward reaction. If it is greater than one, then it is product favored. If it is less than one, then it is reactants favored. Here, it is less than one. So it is reactant favored.