For the following reaction {CH}₄(g) + {Cl}₂(g) {\ {light}\ } {CH}₃{Cl}(g) + {HCl}(g) the correct statement is [JEE Adv 2021 P2]

Correct Answer is : The reaction is exothermic with ΔH∘=−25 kcal mol−1Δ H^{} = -25\ {kcal mol}^{-1}ΔH∘=−25 kcal mol−1.

Initiation step is exothermic with Δ H^{} = -58\ {kcal mol}^{-1}

Propagation step involving {}^{}{CH}₃ formation is exothermic with Δ H^{} = -2\ {kcal mol}^{-1}.

Propagation step involving {CH}₃{Cl} formation is endothermic with Δ H^{} = +27\ {kcal mol}^{-1}.

The reaction is exothermic with Δ H^{} = -25\ {kcal mol}^{-1}.

For the following reaction {CH}₄(g) + {Cl}₂(g) {\ {light}\ } {CH}₃{Cl}(g) + {HCl}(g) the correct statement is [JEE Adv 2021 P2]

Correct Answer is : The reaction is exothermic with ΔH∘=−25 kcal mol−1Δ H^{} = -25\ {kcal mol}^{-1}ΔH∘=−25 kcal mol−1.

Test Index

JEE Advanced 2021 Paper 2

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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 :

\mathrm{Cl}-\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{\bullet}(\mathrm{g})+\mathrm{Cl}^{\bullet}(\mathrm{g})\;\Delta H^{\circ}=58\;\mathrm{kcal}\;\mathrm{mol}^{-1}

\mathrm{H}_{3}\mathrm{C}-\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{H}_{3}\mathrm{C}^{\bullet}(\mathrm{g})+\mathrm{Cl}^{\bullet}(\mathrm{g})\;\Delta H^{\circ}=85\;\mathrm{kcal}\;\mathrm{mol}^{-1}

\mathrm{H}-\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{H}^{\bullet}(\mathrm{g})+\mathrm{Cl}^{*}(\mathrm{g})\;\Delta H^{\circ}=103\;\mathrm{kcal}\;\mathrm{mol}^{-1}

Section: Chemistry

Question : 33 of 57

Marks: +1, -0

For the following reaction

\mathrm{CH}_4(g) + \mathrm{Cl}_2(g) \xrightarrow{\ \text{light}\ } \mathrm{CH}_3\mathrm{Cl}(g) + \mathrm{HCl}(g)

the correct statement is

[JEE Adv 2021 P2]

Initiation step is exothermic with $\Delta H^{\circ} = -58\ \text{kcal mol}^{-1}$
Propagation step involving ${}^{\bullet}\mathrm{CH}_3$ formation is exothermic with $\Delta H^{\circ} = -2\ \text{kcal mol}^{-1}$ .
Propagation step involving $\mathrm{CH}_3\mathrm{Cl}$ formation is endothermic with $\Delta H^{\circ} = +27\ \text{kcal mol}^{-1}$ .
The reaction is exothermic with $\Delta H^{\circ} = -25\ \text{kcal mol}^{-1}$ .

Validate

Solution:

Reaction :

\mathrm{CH}_4(g) + \mathrm{Cl}_2(g) \xrightarrow{\ \text{light}\ } \mathrm{CH}_3-\mathrm{Cl}(g) + \mathrm{HCl}(g)

\mathrm{CH}_4 \longrightarrow \mathrm{CH}_3^{\bullet} + \mathrm{H}^{\bullet} \qquad \Delta H = 105\ \text{kcal/mol}

\mathrm{Cl}_2 \longrightarrow \mathrm{Cl}^{\bullet} + \mathrm{Cl}^{\bullet} \qquad \Delta H = 58\ \text{kcal/mol}

\mathrm{Cl}^{\bullet} + \mathrm{CH}_3^{\bullet} \longrightarrow \mathrm{CH}_3 + \mathrm{Cl} \qquad \Delta H = -85\ \text{kcal/mol}

\mathrm{Cl}^{\bullet} + \mathrm{H}^{\bullet} \longrightarrow \mathrm{HCl} \qquad \Delta H = -103\ \text{kcal/mol}

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\mathrm{CH}_4 + \mathrm{Cl}_2 \longrightarrow \mathrm{CH}_3-\mathrm{Cl} + \mathrm{HCl} \qquad \Delta H = -25\ \text{kcal/mol}

Initiation step is endothermic, hence option (a) is wrong.

Propagation step invding

{}^{\bullet}\mathrm{CH}_3

formation is endothermic, hence option (b) is wrong.

Propagation step involving

\mathrm{CH}_3\mathrm{Cl}

formation is exothermic, hence option (c) is wrong.

So, overall reaction is exothermic with

\Delta H^{\circ} = -25\ \text{kcal/mol}

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