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JEE Advanced 2021 · Paper 2, Chemistry question 14

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The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below:

  • HX3C−H(g)→HX3CX ∙ (g)+HX ∙ (g)\ce{H3C-H(g) -> H3C^.(g) + H^.(g)}, ΔH∘=105\Delta H^\circ = 105 kcal mol−1^{-1}
  • Cl−Cl(g)→ClX ∙ (g)+ClX ∙ (g)\ce{Cl-Cl(g) -> Cl^.(g) + Cl^.(g)}, ΔH∘=58\Delta H^\circ = 58 kcal mol−1^{-1}
  • HX3C−Cl(g)→HX3CX ∙ (g)+ClX ∙ (g)\ce{H3C-Cl(g) -> H3C^.(g) + Cl^.(g)}, ΔH∘=85\Delta H^\circ = 85 kcal mol−1^{-1}
  • H−Cl(g)→HX ∙ (g)+ClX ∙ (g)\ce{H-Cl(g) -> H^.(g) + Cl^.(g)}, ΔH∘=103\Delta H^\circ = 103 kcal mol−1^{-1}

For the following reaction
CHX4(g)+ClX2(g)→lightCHX3Cl(g)+HCl(g)\ce{CH4(g) + Cl2(g) ->[light] CH3Cl(g) + HCl(g)}
the correct statement is

Choose one option
Answer and worked solution

Official answer: D

Step enthalpies from the BDEs:

  • Initiation, ClX2→2 ClX ∙ \ce{Cl2 -> 2Cl^.}: +58+58 kcal mol−1^{-1} (endothermic).
  • Propagation 1, ClX ∙ +CHX4→X ∙ X22 ∙ CHX3+HCl\ce{Cl^. + CH4 -> ^.CH3 + HCl}: 105−103=+2105-103=+2 kcal mol−1^{-1}.
  • Propagation 2, X ∙ X22 ∙ CHX3+ClX2→CHX3Cl+ClX ∙ \ce{^.CH3 + Cl2 -> CH3Cl + Cl^.}: 58−85=−2758-85=-27 kcal mol−1^{-1}.
  • Overall: 105+58−85−103=−25105+58-85-103=-25 kcal mol−1^{-1}.

Check each option.

  • (A) The initiation step is endothermic, not exothermic. False.
  • (B) Propagation 1 is +2+2, not −2-2. False.
  • (C) Propagation 2 is exothermic (−27-27), not endothermic. False.
  • (D) The reaction is exothermic with ΔH∘=−25\Delta H^\circ=-25 kcal mol−1^{-1}. True.

(D)

Topics:Bond enthalpyReactive intermediates
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