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Chemical Equilibrium question

2020 · 6 Sep · Shift 1 · Q17
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Chemical Equilibrium question

2020 · 6 Sep · Shift 1 · Q17

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
For the reaction Fe2NFe_2NFe2​N(s) + 32{3 \over 2}23​ H2H_2H2​(g) ⇌ 2Fe(s) + NH3NH_3NH3​(g)
  1. A
    KCKCKC = KpKpKp(RTRTRT)1/2
  2. B
    KC = Kp(RT)-1/2
  3. C
    KC = Kp(RT)
  4. D
    KCKCKC = KpKpKp(RTRTRT)3/2
View written solutionFree

Correct answer: A

  1. Write the relation between KpK_pKp​ and KcK_cKc​

For a gaseous equilibrium, Kp=Kc(RT)ΔnK_p = K_c (RT)^{\Delta n}Kp​=Kc​(RT)Δn where Δn=(moles of gaseous products)−(moles of gaseous reactants)\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})Δn=(moles of gaseous products)−(moles of gaseous reactants)

Only gaseous species are counted. Solids do not appear in Δn\Delta nΔn.

  1. Given reaction

Fe2N(s)+32H2(g)⇌2Fe(s)+NH3(g)Fe_2N(s) + \frac{3}{2}H_2(g) \rightleftharpoons 2Fe(s) + NH_3(g)Fe2​N(s)+23​H2​(g)⇌2Fe(s)+NH3​(g)

Gaseous reactants: 32\frac{3}{2}23​ mole of H2H_2H2​

Gaseous products: 111 mole of NH3NH_3NH3​

Therefore, Δn=1−32=−12\Delta n = 1 - \frac{3}{2} = -\frac{1}{2}Δn=1−23​=−21​

  1. Substitute into the formula

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

Rearranging for KcK_cKc​, Kc=Kp(RT)1/2K_c = K_p (RT)^{1/2}Kc​=Kp​(RT)1/2

  1. Match with the options

This corresponds to:

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

So, Option A is correct.

  1. Verification with stored answer

Stored correct answer: A

This matches our derived answer.

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