Correct Answer - Option 2 : 0.242 atm

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**Concept:**

Molar mass of NH_{4}SH = 18 + 33 = 51 g mol^{-1}

Number of moles of NH_{4} SH introduced in the vessel \(=\frac{Weight}{Molar~mass}=\frac{5.1}{5}=0.1~mol\)

NH_{4}SH (s) ⇌ NH_{3} (g) + H_{2}S (g)

Moles at t = 0 0.1 0 0

At \(t={{t}_{eq}}\) 0.1(1-0.3) 30%of 30% of 0.1

0.1 = 0.03 = 0.03

Active mass (mol L^{-1}) \(\frac{0.03}{3}=0.01\text{ }\!\!~\!\!\text{ }\frac{0.03}{3}=0.01\)

**Calculation:**

\({{K}_{C}}=\frac{\left[ \text{N}{{\text{H}}_{3}} \right]\left[ {{\text{H}}_{2}}\text{S} \right]}{\left[ \text{N}{{\text{H}}_{4}}\text{HS}\left( s \right) \right]}=\frac{0.01\times 0.01}{1}\)

= 10^{-4} (mol L^{-1})^{2}

\(\Rightarrow {{K}_{p}}={{K}_{C}}{{(RT)}^{\Delta {{n}_{g}}}}\)

[where, Δn_{g} = Σn_{product }– Σn_{reactant}] = 2 – 0 = 2

∴ Kp = K_{c}(RT)^{2}

= 10^{-4 }× [0.082 × (273 + 327)]^{2} atm^{2}

= 0.242 atm

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