Chemical and Ionic Equilibrium - Result Question 8
8. $5.1$ $ g$ $ NH _4 SH$ is introduced in $3.0 L$ evacuated flask at $327^{\circ} C . $ $30 %$ of the solid $NH _4 SH$ decomposed to $NH _3$ and $H _2 S$ as gases. The $K _p$ of the reaction at $327^{\circ} C$ is $(R=0.082 $ $atm $ $mol^{-1} K^{-1}$, molar mass of $S=32$ $ g $ $mol^{-1}$, molar mass of $N=14$ $ g $ $mol^{-1}$ )
(2019 Main, 10 Jan II)
(a) $0.242 \times 10^{-4}$ $ atm^{2}$
(b) $0.242$ $ atm^{2}$
(c) $4.9 \times 10^{-3}$ $ atm^{2}$
(d) $1 \times 10^{-4}$ $ atm^{2}$
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Answer:
Correct Answer: 8. (b)
Solution:
- Molar mass of $NH _4 SH=18+33=51$ $ g $ $mol^{-1}$
Number of moles of $NH _4 SH$ introduced in the vessel
$=\frac{\text { Weight }}{\text { Molar mass }}=\frac{5 \cdot 1}{51}=0.1 $ $mol $
$ K _C=\frac{\left[NH _3\right]\left[H _2 S\right]}{\left[NH _4 HS(s)\right]}=\frac{0.01 \times 0.01}{1}=10^{-4}\left(mol 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} atm^{2}$
$ =0.242 $ $atm^{2}$
BETA
