Question

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

• A. $ 0.56\, atm^2 $
• B. $ 0.3136\, atm^2 $
• C. $ 1.25 \,atm^2 $
• D. $ 1.12 \,atm^2 $

Answer 1

Answer: (B)

$p_{\text{total}}=1.12 atm$
$K_{p}=p_{NH_3}\times p_{H_2 s} $
$p_{NH_3}=p_{\text{total}} \times x_{NH_3}$
$=1.12\times\frac{x}{2x}=0.56$
$p_{H_2}=p_{\text{total}}\times x_{H_2 s}$
$=1.12\times\frac{x}{2x}=0.56$
$\therefore K_{p}=0.56\times0.56=0.3136\,atm^{2}$