Question

For $\left(NH\right)_{4}HS\left(\right.s\left.\right)\rightleftharpoons\left(NH\right)_{3}\left(\right.g\left.\right)+H_{2}S\left(\right.g\left.\right)$ , the observed pressure for reaction mixture in equilibrium is $1.12atm$ at $106^\circ C$ . The value of $K_{p}$ for the reaction is:

• A. $3.136atm^{2}$
• B. $0.3136atm^{2}$
• C. $31.36atm^{2}$
• D. $6.98atm^{2}$

Answer 1

Answer: (B)

$\underset{Pressure at equilibrium}{\left(NH\right)_{4} HS \left(\right. s \left.\right)}\rightleftharpoons\underset{P}{\left(NH\right)_{3} \left(\right. g \left.\right)}+\underset{P}{H_{2} S \left(\right. g \left.\right)}$
$\therefore $ Total pressure at equilibrium $\Rightarrow 2P=1.12atm$
$P=1.12/2atm$
$\because K_{p}=P'_{NH_{3}}\times P'_{H_{2} S}$
$\therefore Kp=\frac{1 . 12}{2}\times \frac{1 . 12}{2}=0.3136atm^{2}$