Chemical and Ionic Equilibrium 1 Question 8
8. $5.1 \mathrm{~g} \mathrm{NH}{4} \mathrm{SH} 3.0 \mathrm{~L} 327^{\circ} \mathrm{C} .30 % \mathrm{NH}{4} \mathrm{SH} \mathrm{NH}{3} \mathrm{H}{2} \mathrm{~S} K_{p} 327^{\circ} \mathrm{C} \left(R=0.082 \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right. \mathrm{S}=32 \mathrm{~g} \mathrm{~mol}^{-1} \mathrm{N}=14 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(2019 Main, 10 Jan II)
(a)
(b)
(c)
(d)
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Solution:
- Molar mass of
Number of moles of
$$
\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 \
&