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JEE Advanced Chemistry Haloalkanes and Haloarenes 2021 JEE Advanced 2021 (Paper 2)

JEE Advanced Chemistry Question (2021) — Solution

Question

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 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: Question: CH _ 4 ( ~g )+ Cl _ 2 ( ~g ) light CH _ 3 Cl ( g )+ HCl ( g ) the correct statement is

Options

  1. A. Initiation step is exothermic with ΔH ° = - 58   kcal   mol - 1 .
  2. B. Propagation step involving CH 3 ∙ formation is exothermic with  ΔH ° = - 2   k   cal   mol - 1
  3. C. Propagation step involving CH 3 Cl formation is endothermic with  ΔH ° = + 27   k   cal   mol - 1
  4. D. The reaction is exothermic with ΔH ° = - 25   k   cal   mol - 1 .

Answer

D. The reaction is exothermic with ΔH ° = - 25   k   cal   mol - 1 .

Step-by-step solution

CH 4 g + Cl 2 g → light CH 3 Cl g + HCl g ΔH r ∘ = ∑ BDE reactans − ∑ BDE products = E C − H + E Cl − Cl − E C − Cl − E H − Cl = + 105 + 58 − 85 − 103 = − 25   k   cal   mol − 1 reaction is exothermic.

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Related: Chemistry — Haloalkanes and Haloarenes · All PYQ Banks