JEE Advanced 2021 Paper 2 Q14 Chemistry Thermodynamics & Thermochemistry Bond Enthalpy Medium

JEE Advanced 2021 Paper 2 · Q14 · Bond Enthalpy

For the following reaction CH$_4$(g) + Cl$_2$(g) $\xrightarrow{\text{light}}$ CH$_3$Cl(g) + HCl(g) the correct statement is

  1. A. Initiation step is exothermic with $\Delta H^\circ$ = $-58$ kcal mol$^{-1}$.
  2. B. Propagation step involving $^\bullet$CH$_3$ formation is exothermic with $\Delta H^\circ$ = $-2$ kcal mol$^{-1}$.
  3. C. Propagation step involving CH$_3$Cl formation is endothermic with $\Delta H^\circ$ = $+27$ kcal mol$^{-1}$.
  4. D. The reaction is exothermic with $\Delta H^\circ$ = $-25$ kcal mol$^{-1}$.
Reveal answer + step-by-step solution

Correct answer:D

Solution

$\Delta H_{rxn}$ = (bonds broken) $-$ (bonds formed) = [BDE(CH$_3$-H) + BDE(Cl-Cl)] $-$ [BDE(CH$_3$-Cl) + BDE(H-Cl)] = [105 + 58] $-$ [85 + 103] = 163 $-$ 188 = $-25$ kcal mol$^{-1}$. So (D) is correct.

(A) Initiation: Cl$_2$ $\xrightarrow{h\nu}$ 2 Cl$^\bullet$. $\Delta H = +58$ kcal mol$^{-1}$ (endothermic, energy supplied by light). WRONG. (B) Propagation $^\bullet$CH$_3$ formation: CH$_4$ + Cl$^\bullet$ $\rightarrow$ $^\bullet$CH$_3$ + HCl. $\Delta H = 105 - 103 = +2$ kcal mol$^{-1}$ (endothermic, not exothermic). WRONG. (C) Propagation CH$_3$Cl formation: $^\bullet$CH$_3$ + Cl$_2$ $\rightarrow$ CH$_3$Cl + Cl$^\bullet$. $\Delta H = 58 - 85 = -27$ kcal mol$^{-1}$ (exothermic, not endothermic). WRONG.

Hence the correct option is (D).

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