JEE Main · 2020 · Shift-ImediumCEQ-008

For the reaction Fe2N(s) + 32H2(g) 2Fe(s) + NH3(g)

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

For the reaction Fe2N(s)+32H2(g)2Fe(s)+NH3(g)\mathrm{Fe_2N(s) + \frac{3}{2}H_2(g) \rightleftharpoons 2Fe(s) + NH_3(g)}

Options
  1. a

    Kc=Kp(RT)K_c = K_p(RT)

  2. b

    Kc=Kp(RT)1/2K_c = K_p(RT)^{-1/2}

  3. c

    Kc=Kp(RT)1/2K_c = K_p(RT)^{1/2}

  4. d

    Kc=Kp(RT)3/2K_c = K_p(RT)^{3/2}

Correct Answerc

Kc=Kp(RT)1/2K_c = K_p(RT)^{1/2}

Detailed Solution

Formula: KP=KC(RT)ΔngKC=KP(RT)ΔngK_P = K_C (RT)^{\Delta n_g} \Rightarrow K_C = K_P (RT)^{-\Delta n_g}

Step 1 — Identify gaseous species only (solids excluded from K expression):

  • Gaseous reactants: 32H2\frac{3}{2}\mathrm{H_2} \rightarrow 32\frac{3}{2} moles
  • Gaseous products: NH3\mathrm{NH_3} \rightarrow 1 mole

Step 2 — Calculate Δng\Delta n_g: Δng=132=12\Delta n_g = 1 - \frac{3}{2} = -\frac{1}{2}

Step 3 — Relationship: KC=KP(RT)Δng=KP(RT)+1/2=KP(RT)1/2K_C = K_P \cdot (RT)^{-\Delta n_g} = K_P \cdot (RT)^{+1/2} = K_P(RT)^{1/2}

Answer: Option (3)

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