Question

A definite amount of solid $N{H_4}HS$ is placed in a flask already containing ammonia gas at a certain temperature and $0.50atm$ pressure. $N{H_4}HS$ decomposes to give $N{H_3}$ and ${H_2}S$ at equilibrium total pressure in flask is $0.84atm$ . The equilibrium constant for the reaction is:
(A) $0.30$
(B) $0.18$
(C) $0.17$
(D) $0.11$

Answer 1

The equilibrium constant can be written in terms of pressure and is known as ${K_p}$ . Also equilibrium constant is given as the pressure of the products divided by the pressure of the reactant.

Complete step by step answer:
In the given case there are few things that needed to be taken care of, first is that we already have a container which has ammonia gas in it whose pressure was initially $0.50atm$ . The second thing is that the reactant added here $N{H_4}HS$ is in solid form which is dissociating into gas molecules and as it is dissociating the number of moles of the ammonia and hydrogen sulfide gas will increase increasing the total pressure of the system. Also we are being provided with the total pressure when the system is at equilibrium. Now if we assume that $x$ is the degree of dissociation then we can write the initial and final conditions as:
\begin{array}{*{20}{c}} {}&{N{H_4}S{H_{(s)}}}& \rightleftharpoons &{N{H_{3(g)}}}& \+ &{{H_2}{S_{(g)}}} \\\ {t = 0}&0&{}&{0.50atm}&{}&0 \\\ {{t_{eq}}}&{1 - x}&{}&{(0.50 + x)atm}&{}&{xatm} \end{array}
Now we know that at equilibrium the total pressure ${p_t} = 0.84 = 0.50 + x + x$
So, from here we can calculate the value of $x$ which is given as:
$2x + 0.50 = 0.84 \\\ 2x = 0.34 \\\ x = 0.17 \\\$
Now the equilibrium constant is given by the expression:
${K_p} = \dfrac{{{p_{N{H_3}}}{p_{{H_2}S}}}}{{{p_{N{H_4}HS}}}}$
But we know that $N{H_4}HS$ is a solid so its pressure will be negligible and we can remove this term from the equilibrium constant. So, our new equilibrium constant will be:
${K_p} = {p_{N{H_3}}}{p_{{H_2}S}} = (0.50 + x)x$
${K_p} = 0.17 \times 0.67 = 0.11$ .

Note:
${K_p}$ is a unit less quantity which means it does not have any units of pressure as it is constant. Also while writing the equilibrium constant the concentration and pressure of solids are assumed to be one because that does not change sufficiently with time.