Correct Answer - Option 2 : 0.242 atm
2
Concept:
Molar mass of NH4SH = 18 + 33 = 51 g mol-1
Number of moles of NH4 SH introduced in the vessel \(=\frac{Weight}{Molar~mass}=\frac{5.1}{5}=0.1~mol\)
NH4SH (s) ⇌ NH3 (g) + H2S (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, Δng = Σnproduct – Σnreactant] = 2 – 0 = 2
∴ Kp = Kc(RT)2
= 10-4 × [0.082 × (273 + 327)]2 atm2
= 0.242 atm
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