JEE Challenger
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ChemMedAdv 2021 • Paper 2

Enthalpy Change for Radical Chain Steps in Methane Chlorination

Consider the given photochemical chlorination reaction: CH4(g)+Cl2(g)→lightCH3Cl(g)+HCl(g)\text{CH}_4(\text{g}) + \text{Cl}_2(\text{g}) \xrightarrow{\text{light}} \text{CH}_3\text{Cl}(\text{g}) + \text{HCl}(\text{g}) Select the true statement regarding the energetics of this system.

Options

A

Initiation step is exothermic with ΔH∘=−58 kcal mol−1\Delta H^\circ = -58 \text{ kcal mol}^{-1}.

B

Propagation step involving ∙CH3{^\bullet}\text{CH}_3 formation is exothermic with ΔH∘=−2 kcal mol−1\Delta H^\circ = -2 \text{ kcal mol}^{-1}.

C

Propagation step involving CH3Cl\text{CH}_3\text{Cl} formation is endothermic with ΔH∘=+27 kcal mol−1\Delta H^\circ = +27 \text{ kcal mol}^{-1}.

D

The overall reaction is exothermic with ΔH∘=−25 kcal mol−1\Delta H^\circ = -25 \text{ kcal mol}^{-1}.

Correct

Topics & Concepts

HydrocarbonsAlkanes, Alkenes, AlkynesEnthalpy

Step-by-Step Solution

To determine the correct statement regarding the energetics of the photochemical chlorination of methane, we analyze the bond dissociation energies (BDE\text{BDE}) and enthalpy changes (ΔH∘\Delta H^\circ) for each step of the mechanism:

CH4(g)+Cl2(g)→hνCH3Cl(g)+HCl(g)\text{CH}_4(\text{g}) + \text{Cl}_2(\text{g}) \xrightarrow{h\nu} \text{CH}_3\text{Cl}(\text{g}) + \text{HCl}(\text{g})

1. Chain Initiation Step

Homolytic cleavage of the chlorine-chlorine bond occurs upon exposure to light: Cl2(g)→hν2Cl∙(g)\text{Cl}_2(\text{g}) \xrightarrow{h\nu} 2\text{Cl}^\bullet(\text{g})

  • Bond broken: Cl−Cl\text{Cl}-\text{Cl} (BDE=+58 kcal mol−1\text{BDE} = +58\text{ kcal mol}^{-1})
  • Enthalpy change: ΔHinitiation∘=+58 kcal mol−1\Delta H_{\text{initiation}}^\circ = +58\text{ kcal mol}^{-1} Since energy is absorbed to break the bond, this step is endothermic. Thus, (A) is incorrect.

2. Propagation Steps

First Propagation Step (Formation of ∙CH3{^\bullet}\text{CH}_3):

CH4+Cl∙→∙CH3+HCl\text{CH}_4 + \text{Cl}^\bullet \rightarrow {^\bullet}\text{CH}_3 + \text{HCl}

  • Bond broken: C−H\text{C}-\text{H} in CH4\text{CH}_4 (BDE≈+105 kcal mol−1\text{BDE} \approx +105\text{ kcal mol}^{-1})
  • Bond formed: H−Cl\text{H}-\text{Cl} (BDE≈103 kcal mol−1\text{BDE} \approx 103\text{ kcal mol}^{-1})
  • Enthalpy change: ΔHprop 1∘=BDE(C−H)−BDE(H−Cl)=+105−103=+2 kcal mol−1\Delta H_{\text{prop } 1}^\circ = \text{BDE}(\text{C}-\text{H}) - \text{BDE}(\text{H}-\text{Cl}) = +105 - 103 = +2\text{ kcal mol}^{-1} This step is endothermic. Thus, (B) is incorrect.

Second Propagation Step (Formation of CH3Cl\text{CH}_3\text{Cl}):

∙CH3+Cl2→CH3Cl+Cl∙{^\bullet}\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet

  • Bond broken: Cl−Cl\text{Cl}-\text{Cl} (BDE=+58 kcal mol−1\text{BDE} = +58\text{ kcal mol}^{-1})
  • Bond formed: C−Cl\text{C}-\text{Cl} (BDE≈85 kcal mol−1\text{BDE} \approx 85\text{ kcal mol}^{-1})
  • Enthalpy change: ΔHprop 2∘=BDE(Cl−Cl)−BDE(C−Cl)=+58−85=−27 kcal mol−1\Delta H_{\text{prop } 2}^\circ = \text{BDE}(\text{Cl}-\text{Cl}) - \text{BDE}(\text{C}-\text{Cl}) = +58 - 85 = -27\text{ kcal mol}^{-1} This step is exothermic. Thus, (C) is incorrect (since it claims the step is endothermic with +27 kcal mol−1+27\text{ kcal mol}^{-1}).

3. Overall Reaction Energetics

The overall enthalpy change for the reaction is the sum of the enthalpy changes of the two propagation steps:

ΔHoverall∘=ΔHprop 1∘+ΔHprop 2∘\Delta H_{\text{overall}}^\circ = \Delta H_{\text{prop } 1}^\circ + \Delta H_{\text{prop } 2}^\circ ΔHoverall∘=(+2 kcal mol−1)+(−27 kcal mol−1)=−25 kcal mol−1\Delta H_{\text{overall}}^\circ = (+2\text{ kcal mol}^{-1}) + (-27\text{ kcal mol}^{-1}) = -25\text{ kcal mol}^{-1}

Since ΔHoverall∘<0\Delta H_{\text{overall}}^\circ < 0, the overall reaction is exothermic with ΔH∘=−25 kcal mol−1\Delta H^\circ = -25\text{ kcal mol}^{-1}.

Conclusion

The correct statement is (D).

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