QJEEAdv21Chbb12: JEE Advanced 2021 Chemistry - Class 11 JEEAdv21Chbb12 Solution: For the following reaction CH 4 (g) + Cl 2 (g) ⟶ l i g h t CH 3 …

Question

Verified Solution • 2025–26 Session moderate

For the following reaction

CH₄(g) + Cl₂(g)

\overset{l i g h t}{⟶}

CH₃Cl(g) + HCl (g)

the correct statement is

A Initiation step is exothermic with $\Delta$ H $^\circ$ = $-$ 58 kcal mol $-$ 1.
B Propagation step involving ${}^ \bullet C{H_3}$ formation is exothermic with $\Delta$ H $^\circ$ = $-$ 2 kcal mol $-$ 1.
C Propagation step involving CH3Cl formation is endothermic with $\Delta$ H $^\circ$ = +27 kcal mol $-$ 1.
D The reaction is exothermic with $\Delta$ H $^\circ$ = $-$ 25 kcal mol $-$ 1.

Accepted Answer

The reaction is exothermic with $\Delta$ H $^\circ$ = $-$ 25 kcal mol^{$-$ 1}

CH₄(g) + Cl₂(g) $\overset{h v}{⟶}$ CH₃Cl(g) + HCl(g)

(a) Initiation step is breaking of Cl $-$ Cl bond which requires energy and hence, this step is endothermic.

JEE Advanced 2021 Paper 2 Online Chemistry - Basics of Organic Chemistry Question 25 English Explanation 1

(b) Propagation step involving formation of CH₃ is also endothermic as bond between C $-$ H is breaking.

JEE Advanced 2021 Paper 2 Online Chemistry - Basics of Organic Chemistry Question 25 English Explanation 2

(c) Propagation step involving formation of CH₃Cl is exothermic as bond is formed between C $-$ Cl

JEE Advanced 2021 Paper 2 Online Chemistry - Basics of Organic Chemistry Question 25 English Explanation 3

(d) $\mathop H\limits^ \bullet$ + $\mathop {Cl}\limits^ \bullet$ $\to$ HCl, $\Delta$ H₄ $^\circ$ = $-$ 103 kcal/mol

Enthalpy of reaction, $\Delta$ H $^\circ$ = $\Delta$ H $_1^o$ + $\Delta$ H $_2^o$ + $\Delta$ H $_3^o$ + $\Delta$ H $_4^o$

= 58 + 105 $-$ 85 $-$ 103

= $-$ 25 kcal mol^{$-$ 1}

Hence, the reaction is exothermic with $\Delta$ H $^o$ = $-$ 25 kcal/mol

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