Question

${5.1g \; NH_4SH }$ is introduced in $3.0\, L$ evacuated flask at ${327^{\circ}C}$. 30% of the solid ${NH4SH}$ decomposed to ${NH_3}$ and ${H2S}$ as gases. The Kp of the reaction at ${327^{\circ}C}$ is {(R = 0.082 \; L \; atm \; mol^{-1} K^{-1}}}, Molar mass of ${S = 32 \; g \; mol/^{01}}$, molar mass of ${N = 14g \, mol^{-1}}$)

• A. $1 \times 10^{-4} \; atm^2$
• B. $4.9 \times 10^{-3} \; atm^2$
• C. $0.242 \; atm^2$
• D. $0.242 \times 10^{-4} \; atm^2$

Answer 1

Answer: (C)

$0.242 \; atm^2$

Answer 2

${NH SH(s) <=> NH (g) H_2S(g)}$
$\begin{matrix}n = \frac{5.1}{51} =.1 \text{mole}&0&0\\\ .1\left(-1-\alpha\right)&.1\alpha&.1\alpha\end{matrix}$
$\alpha=30\% =.3$
so number of moles at equilibrium
$\begin{matrix}&.1\left(1-3\right)&.1\times.3&.1\times.3\\\ =&.07&=.03&=.03\end{matrix}$
Now use PV = nRT at equilibrium
$P_{\text{total}}\times3 \, \text{lit} =\left(.03 +.03\right) \times.082 \times600$
$P_{\text{total}} = .984 \text{atm}$
At equilibrium
$P_{NH_3} = P_{H_2S} = \frac{P_{\text{total}}}{2} = .492$
$k_{P} =P_{NH_3} . P_{H_2S} = \left(.492\right)\left(.492\right)$
$k_{p} = .242 \text{atm}^{2}$