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JEE Main 2024 · Chemistry · Chemical Equilibrium · medium

In a closed 1.0 L vessel at a fixed temperature, 1.0 mol H2, 1.0 mol I2 and 2.0 mol HI are mixed. For H2(g)+I2(g)rightleftharpoons 2HI(g), Kc=50. The…

Single correct option

In a closed 1.0 L1.0\ \mathrm{L} vessel at a fixed temperature, 1.01.0 mol H2\mathrm{H_2}, 1.01.0 mol I2\mathrm{I_2} and 2.02.0 mol HI are mixed. For H2(g)+I2(g)⇌2HI(g)\mathrm{H_2(g)+I_2(g)\rightleftharpoons 2HI(g)}, Kc=50K_c=50. The amount of HI at equilibrium is nearly
  1. A. 3.12 mol3.12\ \mathrm{mol}
  2. B. 2.78 mol2.78\ \mathrm{mol}
  3. C. 1.56 mol1.56\ \mathrm{mol}
  4. D. 3.56 mol3.56\ \mathrm{mol}
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Answer: (A) 3.12 mol3.12\ \mathrm{mol}

Qc=221×1=4<KcQ_c=\dfrac{2^2}{1\times1}=4<K_c, so the reaction proceeds forward. Let xx mol of H2\mathrm{H_2} react: [HI]=2+2x[\mathrm{HI}]=2+2x, [H2]=[I2]=1−x[\mathrm{H_2}]=[\mathrm{I_2}]=1-x (no change in moles, so volume cancels). (2+2x)2(1−x)2=50⇒2+2x1−x=50=7.071\dfrac{(2+2x)^2}{(1-x)^2}=50\Rightarrow\dfrac{2+2x}{1-x}=\sqrt{50}=7.071, so x=5.0719.071=0.559x=\dfrac{5.071}{9.071}=0.559 and nHI=2+1.118=3.12n_\mathrm{HI}=2+1.118=3.12 mol. Forgetting the initial HI (solving 4x2/(1−x)2=504x^2/(1-x)^2=50) gives the trap value 1.561.56 mol.
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