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JEE Advance - Chemistry (2021 - Paper 2 Online - No. 14)

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) $$\to$$ Cl$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 58 kcal mol$$-$$1

H3C $$-$$ Cl (g) $$\to$$ H3C$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 85 kcal mol$$-$$1

H $$-$$ Cl (g) $$\to$$ H$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 103 kcal mol$$-$$1
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) $$\to$$ Cl$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 58 kcal mol$$-$$1

H3C $$-$$ Cl (g) $$\to$$ H3C$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 85 kcal mol$$-$$1

H $$-$$ Cl (g) $$\to$$ H$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 103 kcal mol$$-$$1
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) $$\to$$ Cl$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 58 kcal mol$$-$$1

H3C $$-$$ Cl (g) $$\to$$ H3C$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 85 kcal mol$$-$$1

H $$-$$ Cl (g) $$\to$$ H$$_{(g)}^ \bullet $$ + Cl$$_{(g)}^ \bullet $$ $$\Delta$$H$$^\circ$$ = 103 kcal mol$$-$$1
For the following reaction

CH4(g) + Cl2(g) $$\buildrel {light} \over \longrightarrow $$ CH3Cl(g) + HCl (g)

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

Explanation

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

CH4(g) + Cl2(g) $$\buildrel {hv} \over \longrightarrow $$ 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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