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

Correct Answer is : The reaction is exothermic with ∆Ho=−25kcalmol−1.

Initiation step is exothermic with ∆ {H}^{∘}=-58 {kcal} {mol}^{-1}

Propagation step involving { }^{●} {CH}_{3} formation is exothermic with ∆ {H}^{ {o}}=-2 {kcal} {mol}^{-1}.

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

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

Basics of Organic Chemistry

Paragraph for next 2 questions
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 :

Cl−Cl(g)⟶Cl∙(g)+Cl∙(g)‌∆H∘=58kcalmol−1
H3C−Cl(g)⟶H3C∙(g)+Cl∙(g)‌∆H∘=85kcalmol−1
H−Cl(g)⟶H∙(g)+Cl*(g)‌∆H∘=103kcalmol−1

Section: Chemistry

Question : 3 of 39

Marks: +1, -0

For the following reaction CH4(g)+Cl2(g)

‌ light ‌

────▶

CH3Cl(g)+HCl(g) the correct statement is

[JEE Adv 2021 P2]

Initiation step is exothermic with ∆H∘=−58kcalmol−1
Propagation step involving ‌●CH3 formation is exothermic with ∆Ho=−2kcalmol−1.
Propagation step involving CH3Cl formation is endothermic with ∆H∘=+27kcalmol−1.
The reaction is exothermic with ∆Ho=−25kcalmol−1.

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