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JEE Advanced 2021Paper 2
Exam
JEE Advanced
Year
2021
Paper
Paper 2
Subject
Chemistry
Chapter
Haloalkanes and Haloarenes

JEE Advanced 2021 · Paper 2 · Chemistry · Haloalkanes and Haloarenes

Paragraph: The amount of energy required to break a bond is same as...

Paragraph: 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 ss-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below: Question: CH4(g)+Cl2(g)⟶ light CH3Cl(g)+HCl(g)\mathrm{CH}_{4}(\mathrm{ g})+\mathrm{Cl}_{2}(\mathrm{ g}) \stackrel{\text{ light }}{\longrightarrow} \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) the correct statement is

Options

  1. AInitiation step is exothermic with ΔH°=−58kcalmol−1\Delta H ° = - 58kcal{mol}^{- 1}.
  2. BPropagation step involving CH3∙CH₃ ∙ formation is exothermic with ΔH°=−2kcalmol−1\Delta H ° = - 2kcal{mol}^{- 1}
  3. CPropagation step involving CH3Cl{CH}_{3}Cl formation is endothermic with ΔH°=+27kcalmol−1\Delta H ° = + 27kcal{mol}^{- 1}
  4. DThe reaction is exothermic with ΔH°=−25kcalmol−1\Delta H ° = - 25kcal{mol}^{- 1}.

Correct answer

D. The reaction is exothermic with ΔH°=−25kcalmol−1\Delta H ° = - 25kcal{mol}^{- 1}.

Step-by-step solution

CH4(g)+Cl2(g)→lightCH3Cl(g)+HCl(g){CH}_{4}\left(g\right) + {Cl}_{2}\left(g\right)\overset{light}{\to}{CH}_{3}Cl\left(g\right) + HCl\left(g\right)

ΔHr∘=∑(BDE)reactans−∑(BDE)products{\Delta H}_{r}^{∘} = ∑ {\left(BDE\right)}_{reactans} - ∑ {\left(BDE\right)}_{products}

=EC−H+ECl−Cl−EC−Cl−EH−Cl= {E}_{C - H} + {E}_{Cl - Cl} - {E}_{C - Cl} - {E}_{H - Cl}

=+105+58−85−103= + 105 + 58 - 85 - 103

=−25kcalmol−1= - 25kcal{mol}^{- 1}

reaction is exothermic.

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