# When 5.1 g of solid NH4 HS is introduced into

Question:

When $5.1 \mathrm{~g}$ of solid $\mathrm{NH}_{4} \mathrm{HS}$ is introduced into a two litre evacuated flask at $27^{\circ} \mathrm{C}, 20 \%$ of the solid decomposes into gaseous ammonia and hydrogen sulphide. The $\mathrm{K}_{\mathrm{p}}$ for the reaction at $27^{\circ} \mathrm{C}$ is $\mathrm{x} \times 10^{-2}$.

The value of $x$ is - (Integer answer) $\left[\right.$ Given $\left.\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$

Solution:

moles of $\mathrm{NH}_{4} \mathrm{HS}$ initially taken $=\frac{5.1 \mathrm{~g}}{51 \mathrm{~g} / \mathrm{mol}}$

$=0.1 \mathrm{~mol}$

volume of vessel $=2 \ell$

$\mathrm{NH}_{4} \mathrm{HS}_{(\mathrm{s})} \rightleftharpoons \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{~S}_{(\mathrm{g})}$

$\mathrm{t}=0 \quad 0.1 \mathrm{~mol}$

$\mathrm{t}=\infty \quad 0.1(1-0.2) \quad 0.1 \times 0.2 \quad 0.1 \times 0.2$

$\Rightarrow$ partial pressure of each component

$\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}=\frac{0.1 \times 0.2 \times 0.082 \times 300}{2}$

$=0.246 \mathrm{~atm}$

$\Rightarrow \mathrm{k}_{\mathrm{P}}=\mathrm{P}_{\mathrm{NH}_{3}} \times \mathrm{P}_{\mathrm{H}_{2} \mathrm{~S}}=(0.246)^{2}=0.060516$

$=6.05 \times 10^{-2}$

$\Rightarrow 6$