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}

Practice this question with progress tracking

Want timed practice with adaptive difficulty? Solve this question (and hundreds more from Chemical Equilibrium) inside The Crucible, our adaptive practice platform.

Practice in The Crucible →Browse Chemical Equilibrium chapter →

More JEE Main Chemical Equilibrium PYQs

2024easy

The ratio KPKC for the reaction: CO(g) + 12O2(g) CO2(g) is:

2024medium

The equilibrium constant for the reaction SO3(g) SO2(g) + 12O2(g) is Kc = 4.9 10-2. The value of Kc for the reaction 2SO2(g) + O2(g) 2SO3(g) is:

2024easy

For the given reaction, choose the correct expression of KC: Fe3+(aq) + SCN-(aq) (FeSCN)2+(aq)

2024easy

For the given hypothetical reactions, the equilibrium constants are: X Y;\ K1 = 1.0 Y Z;\ K2 = 2.0 Z W;\ K3 = 4.0 The equilibrium constant for the reaction X W is

2024hard

A(g) B(g) + C2(g). The correct relationship between KP, and equilibrium pressure P is

View all 31 Chemical Equilibrium PYQs →