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5.1 g NH4SH is introduced in 3.0L evacuated flask at 327°C. 30% of the solid NH4 SH decomposed to NH3 and H2 S as gases. The Kp of the reaction at 327°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)
1. 0.242 × 10-4) atm2
2. 0.242 atm2
3. 4.9 × 10-3 atm2
4. 1 × 10-4 atm2

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Correct Answer - Option 2 : 0.242 atm2

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 atm2

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