JEE Main · 2019 · Shift-IIhardCEQ-051

5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327 °C. 30% of the solid NH4SH decomposes to NH3 and H2S as…

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

5.1 g NH4SH\mathrm{NH_4SH} is introduced in 3.0 L evacuated flask at 327 °C. 30% of the solid NH4SH\mathrm{NH_4SH} decomposes to NH3\mathrm{NH_3} and H2S\mathrm{H_2S} as gases. The KPK_P of the reaction at 327 °C is

(R=0.082 L atm mol−1 K−1(R = 0.082\ \mathrm{L\ atm\ mol^{-1}\ K^{-1}}; Molar mass of S = 32 g mol−1^{-1}, N = 14 g mol−1^{-1}

Options
  1. a

    0.242 atm20.242\ \mathrm{atm^2}

    ✓
  2. b

    0.242×10−4 atm20.242 \times 10^{-4}\ \mathrm{atm^2}

  3. c

    1×10−4 atm21 \times 10^{-4}\ \mathrm{atm^2}

  4. d

    4.9×10−3 atm24.9 \times 10^{-3}\ \mathrm{atm^2}

Correct Answera

0.242 atm20.242\ \mathrm{atm^2}

Detailed Solution

Step 1 — Moles of NH4SH\mathrm{NH_4SH}: Molar mass = 14 + 4 + 1 + 32 + 1 = 51 g/mol n=5.151=0.1 moln = \frac{5.1}{51} = 0.1\ \mathrm{mol}

Step 2 — Moles decomposed (30%): ndecomp=0.03 moln_{decomp} = 0.03\ \mathrm{mol}

Reaction: NH4HS(s)⇌NH3(g)+H2S(g)\mathrm{NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)}

Moles of NH3\mathrm{NH_3} = moles of H2S\mathrm{H_2S} = 0.03 mol

Step 3 — Partial pressures (V = 3 L, T = 600 K): PNH3=PH2S=0.03×0.082×6003=1.4763=0.492 atmP_{\mathrm{NH_3}} = P_{\mathrm{H_2S}} = \frac{0.03 \times 0.082 \times 600}{3} = \frac{1.476}{3} = 0.492\ \mathrm{atm}

Step 4 — KPK_P: KP=PNH3×PH2S=0.492×0.492=0.242 atm2K_P = P_{\mathrm{NH_3}} \times P_{\mathrm{H_2S}} = 0.492 \times 0.492 = 0.242\ \mathrm{atm^2}

Answer: Option (1) — 0.242 atm20.242\ \mathrm{atm^2}

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