Question

A flask containing 0.50 atm of ammonia contains some solid $N{H}_{4}HS$ which undergoes dissociation according to $N{H}_{4}HS\left(s\right)\rightleftharpoons N{H}_3\left(g\right)+H_{2}S\left(g\right).$ Calculate the pressure of $N{H}_3$ and $H_{2}S$ at equilibrium $K_P=0.11).$ Also calculate the total pressure.

• A. $P_{N{H}_3}$ = 0.6653 atm
  $P_{H_{2}S}$ = 0.1653 atm
  $P_{Total}$ = 0.8306 atm
• B. $P_{N{H}_3}$ = 0.2 atm
  $P_{H_{2}S}$ = 0.1 atm
  $P_{Total}$ = 0.3 atm
• C. $P_{N{H}_3}$ = 0.1521 atm
  $P_{H_{2}S}$ = 0.1653 atm
  $P_{Total}$ = 0.3174 atm
• D. $P_{N{H}_3}$ = 0.5521 atm
  $P_{H_{2}S}$ = 0.0158 atm
  $P_{Total}$ = 0.5679 atm

Answer 1

The correct option is A $P_{N{H}_3}$ = 0.6653 atm

$P_{H_{2}S}$ = 0.1653 atm
$P_{Total}$ = 0.8306 atm
$\begin{array}{cccc}& N{H}_{4}HS\left(s\right)& \rightleftharpoons N{H}_3\left(g\right)& +H_{2}S\left(g\right)\\ \text{Pressure before dissociation}& x& 0.5& 0\\ \text{Pressure after dissociation}& (x-P)& (0.5+P)& P\end{array}$
Let pressure equivalent to P atm is developed by $N{H}_3$ and $H_{2}S$ on dissociation of $N{H}_{4}HS,$ when 0.5 atm of $N{H}_3$ is already present.
$\therefore K_p=P_{N{H}_3}\times P_{H_{2}S}=(0.5+P)\left(P\right)$
or $0.11=(0.5+P)\left(P\right)$
$\therefore P=0.1653$
$\therefore P_{N{H}_3}=0.5+0.1653=0.6653atm$
$\therefore P_{H_{2}S}=0.1653atm$
$P_{Total}$ =$P_{N{H}_3}$+$P_{H_{2}S}$
$P_{Total}$ = $0.1653atm$+$0.6653atm$
$P_{Total}$ =$0.8306atm$