Chemical and Ionic Equilibrium 1 Question 8
8. $5.1 \mathrm{~g} \mathrm{NH}{4} \mathrm{SH}$ is introduced in $3.0 \mathrm{~L}$ evacuated flask at $327^{\circ} \mathrm{C} .30 %$ of the solid $\mathrm{NH}{4} \mathrm{SH}$ decomposed to $\mathrm{NH}{3}$ and $\mathrm{H}{2} \mathrm{~S}$ as gases. The $K_{p}$ of the reaction at $327^{\circ} \mathrm{C}$ is $\left(R=0.082 \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right.$, molar mass of $\mathrm{S}=32 \mathrm{~g} \mathrm{~mol}^{-1}$, molar mass of $\mathrm{N}=14 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(2019 Main, 10 Jan II)
(a) $0.242 \times 10^{-4} \mathrm{~atm}^{2}$
(b) $0.242 \mathrm{~atm}^{2}$
(c) $4.9 \times 10^{-3} \mathrm{~atm}^{2}$
(d) $1 \times 10^{-4} \mathrm{~atm}^{2}$
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Solution:
- Molar mass of $\mathrm{NH}_{4} \mathrm{SH}=18+33=51 \mathrm{~g} \mathrm{~mol}^{-1}$
Number of moles of $\mathrm{NH}_{4} \mathrm{SH}$ introduced in the vessel
$$ \begin{aligned} & =\frac{\text { Weight }}{\text { Molar mass }}=\frac{5 \cdot 1}{51}=0.1 \mathrm{~mol} \ & \mathrm{NH}{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}{3}(g)+\mathrm{H}{2} \mathrm{~S}(g) \ & \text { Number of } \quad 0.1 \quad 0 \quad 0 \ & \text { moles at } t=0 \ & \text { At } t=t{\text {eq }} \quad 0.1(1-0.03) \quad 30 % \text { of } \quad 30 % \text { of } 0.1 \ & \begin{array}{lll} \begin{array}{l} \text { Active mass } \ \left(\mathrm{mol} \mathrm{L}^{-1}\right) \end{array} & \frac{0.03}{3}=0.01 & \frac{0.03}{3}=0.01 \end{array} \ & K_{C}=\frac{\left[\mathrm{NH}{3}\right]\left[\mathrm{H}{2} \mathrm{~S}\right]}{\left[\mathrm{NH}{4} \mathrm{HS}(s)\right]}=\frac{0.01 \times 0.01}{1}=10^{-4}\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^{2} \ & \Rightarrow \quad K{p}=K_{C}(R T)^{\Delta n_{g}} \ & \text { [where, } \left.\Delta n_{g}=\Sigma n_{\text {product }}-\Sigma n_{\text {reactant }}\right]=2-0=2 \ & \therefore \quad K_{p}=K_{C}(R T)^{2} \ & =10^{-4} \times[0.082 \times(273+327)]^{2} \mathrm{~atm}^{2} \ & =0.242 \mathrm{~atm}^{2} \end{aligned} $$
