Question

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 K_p 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

Answer 1

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, Δn_g = Σn_product – Σn_reactant] = 2 – 0 = 2

∴ Kp = K_c(RT)²

= 10^-4 × [0.082 × (273 + 327)]² atm²

= 0.242 atm²