JEE Main · 2019 · Shift-ImediumCEQ-050

What are the values of KpKc for the following reactions at 300 K respectively? (RT = 24.62\ dm3\ atm\ mol-1) (i) N2(g)…

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

What are the values of KpKc\frac{K_p}{K_c} for the following reactions at 300 K respectively?

(RT=24.62 dm3 atm mol−1)(RT = 24.62\ \mathrm{dm^3\ atm\ mol^{-1}})

(i) N2(g)+O2(g)⇌2NO(g)\mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)} (ii) N2O4(g)⇌2NO2(g)\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)} (iii) N2(g)+3H2(g)⇌2NH3(g)\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}

Options
  1. a

    24.63 dm3 atm mol−124.63\ \mathrm{dm^3\ atm\ mol^{-1}}, 606.0 dm6 atm2 mol−2606.0\ \mathrm{dm^6\ atm^2\ mol^{-2}}, 1.65×10−3 dm−6 atm−2 mol21.65 \times 10^{-3}\ \mathrm{dm^{-6}\ atm^{-2}\ mol^2}

  2. b

    11, 24.62 dm6 atm3 mol−124.62\ \mathrm{dm^6\ atm^3\ mol^{-1}}, 606.0 dm6 atm2 mol−2606.0\ \mathrm{dm^6\ atm^2\ mol^{-2}}

    ✓
  3. c

    11, 24.62 dm3 atm mol−124.62\ \mathrm{dm^3\ atm\ mol^{-1}}, 1.65×10−3 dm−6 atm−2 mol21.65 \times 10^{-3}\ \mathrm{dm^{-6}\ atm^{-2}\ mol^2}

  4. d

    11, 4.1×10−2 dm−3 atm−1 mol4.1 \times 10^{-2}\ \mathrm{dm^{-3}\ atm^{-1}\ mol}, 606 dm6 atm2 mol−2606\ \mathrm{dm^6\ atm^2\ mol^{-2}}

Correct Answerb

11, 24.62 dm6 atm3 mol−124.62\ \mathrm{dm^6\ atm^3\ mol^{-1}}, 606.0 dm6 atm2 mol−2606.0\ \mathrm{dm^6\ atm^2\ mol^{-2}}

Detailed Solution

Formula: KpKc=(RT)Δng\frac{K_p}{K_c} = (RT)^{\Delta n_g}

Step 1 — Calculate Δng\Delta n_g for each:

(i) N2+O2⇌2NO\mathrm{N_2 + O_2 \rightleftharpoons 2NO}: Δng=2−2=0KpKc=(RT)0=1\Delta n_g = 2 - 2 = 0 \frac{K_p}{K_c} = (RT)^0 = 1

(ii) N2O4⇌2NO2\mathrm{N_2O_4 \rightleftharpoons 2NO_2}: Δng=2−1=1KpKc=RT=24.62 dm3 atm mol−1\Delta n_g = 2 - 1 = 1 \frac{K_p}{K_c} = RT = 24.62\ \mathrm{dm^3\ atm\ mol^{-1}}

(iii) N2+3H2⇌2NH3\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}: Δng=2−4=−2KpKc=(RT)−2=1(24.62)2=1606≈1.65×10−3 dm−6 atm−2 mol2\Delta n_g = 2 - 4 = -2 \frac{K_p}{K_c} = (RT)^{-2} = \frac{1}{(24.62)^2} = \frac{1}{606} \approx 1.65 \times 10^{-3}\ \mathrm{dm^{-6}\ atm^{-2}\ mol^2}

Answer: Option (2) — 11, 24.6224.62, 1.65×10−31.65 \times 10^{-3}

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