Question

\[5.1{\text{ g}}\] \[{\text{N}}{{\text{H}}_4}{\text{SH}}\] is introduced in 3 L evacuated flask at \[327^\circ {\text{C}}\]. \[30\% \] of the solid \[{\text{N}}{{\text{H}}_4}{\text{SH}}\] decomposed to \[{\text{N}}{{\text{H}}_3}\]​and \[{{\text{H}}_2}{\text{S}}\]as gases. The \[{{\text{K}}_{\text{P}}}\]​ of the reaction at \[327^\circ {\text{C}}\] Is.\[{\text{R}} = 0.082\;{\text{L}}\;{\text{atm}}\;{\text{mo}}{{\text{l}}^{ - 1}}\], molar mass of \[{\text{S}} = 32{\text{gmo}}{{\text{l}}^{ - 1}}\], molar mass of \[{\text{N}} = 14{\text{gmo}}{{\text{l}}^{ - 1}}\].
A.\[1 \times {10^{ - 4}}{\text{at}}{{\text{m}}^2}\]
B.\[4.9 \times {10^{ - 3}}{\text{at}}{{\text{m}}^2}\]
C.\[0.242{\text{ at}}{{\text{m}}^2}\]
D.\[0.242 \times {10^{ - 4}}{\text{at}}{{\text{m}}^2}\]

Answer 1

Hint: First of all using we will calculate the equilibrium amount of both the gases that are formed by decomposition of solid \[{\text{N}}{{\text{H}}_4}{\text{SH}}\]. Then using the ideal gas equation, total pressure at equilibrium can be calculated. Using a balanced chemical equation we will get the pressure of each gas and hence equilibrium constant can be calculated.
Formula used:
\[{\text{PV}} = {\text{nRT}}\]
Here P is pressure, V is volume, N is number of moles, R is universal gas constant and T is temperature.

Complete step by step answer:
The decomposition reaction of \[{\text{N}}{{\text{H}}_4}{\text{SH}}\] is as follow:
\[{\text{N}}{{\text{H}}_4}{\text{SH}}\left( {\text{s}} \right) \to {\text{N}}{{\text{H}}_3}{\text{ }}\left( {\text{g}} \right){\text{ }} + {{\text{H}}_2}{\text{S}}\left( {\text{g}} \right)\]
Using the mass and molar mass we will first calculate the amount of \[{\text{N}}{{\text{H}}_4}{\text{SH}}\] in terms of mole. The molar mass of \[{\text{N}}{{\text{H}}_4}{\text{SH}}\] is 51: \[\dfrac{{5.1}}{{51}} = 0.1{\text{ M}}\].
Given 30 percent of the initial amount have been dissociated so 60 percent of \[{\text{N}}{{\text{H}}_4}{\text{SH}}\] is left and 30 percent of ammonia and 30 percent of hydrogen sulphide will form.
Now we will write the reaction when equilibrium is attained:
\[{\text{N}}{{\text{H}}_4}{\text{SH}}\left( {\text{s}} \right) \rightleftharpoons {\text{N}}{{\text{H}}_3}{\text{ }}\left( {\text{g}} \right){\text{ }} + {{\text{H}}_2}{\text{S}}\left( {\text{g}} \right)\]
\[{\text{0}}{\text{.1 }} 0{\text{ }} + {\text{ }} 0 \]
Now we will use the 30 percent of initial concentration here:
\[{\text{0}}{\text{.1 }} - \dfrac{{30}}{{100}} \times 0.1{\text{ }}\dfrac{{30}}{{100}} \times 0.1{\text{ }} + {\text{ }}\dfrac{{30}}{{100}} \times 0.1\]
The number of moles at equilibrium will be:
\[{\text{0}}{\text{.07 }} 0.03{\text{ }} + {\text{ }} 0.03\]
The total number of moles will be \[0.03 + 0.03 = 0.06\]. We do not consider the number of moles of \[{\text{N}}{{\text{H}}_4}{\text{SH}}\] here because we need to calculate the equilibrium constant in terms of pressure and hence solids do not have any pressure. Using the total number of moles we will calculate the total pressure at equilibrium using the ideal gas equation. We need to convert the temperature into Kelvin by adding 273 to the temperature in degree Celsius.
\[{{\text{P}}_{{\text{total}}}} \times 3 = 0.06 \times 0.082 \times 600\]
Solving we will get:
\[{{\text{P}}_{{\text{total}}}} = 0.984{\text{ atm}}\]
Now since the number of moles of two gases at equilibrium is same then their partial pressure is also same. So the partial pressure of each gas is \[\dfrac{{0.984{\text{ atm}}}}{2} = 0.492{\text{atm}}\].
\[{{\text{K}}_{\text{P}}} = {\text{ }}{{\text{P}}_{{\text{N}}{{\text{H}}_3}}} \times {\text{ }}{{\text{P}}_{{{\text{H}}_2}{\text{S}}}}\]
\[ \Rightarrow {{\text{K}}_{\text{P}}} = {\left( {0.492} \right)^2} = 0.2420{\text{ at}}{{\text{m}}^2}\]

Hence, the correct option is C.

Note:
The equilibrium constant for the reaction can be defined in various ways. For example it can be defined in terms of temperature, pressure, concentration, mole fraction, and number of moles. Equilibrium constants are only defined for the equilibrium process. We cannot apply equilibrium constant at the initial stages of the reaction; we only use equilibrium constant at the equilibrium state.