HSC Science (General) 12th Board ExamMaharashtra State Board
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# Calculate C-Cl bond enthalpy from following reaction: CH3Cl(g) + Cl2(g) → Ch2Cl2(g) + HCl(g) ΔH° = -104KJ - HSC Science (General) 12th Board Exam - Chemistry

ConceptChemical Thermodynamics and Energetic Enthalpy of Bond Dissociation

#### Question

Calculate C-Cl bond enthalpy from following reaction:

CH3Cl(g) + Cl2(g) → Ch2Cl2(g) + HCl(g) ΔH° = -104KJ

If C-H, Cl-Cl and H-Cl bond enthalpies are 414, 243 and 431 KJ-Mol-1 respectively.

#### Solution

CH3Cl(g)+Cl2(g) → CH2Cl2(g)HCl(g)
C—H =414 kJ /mol
Cl—Cl =243 kJ /mol
H—Cl =431 kJ /mol
ΔH° =-104 kJ
ΔH°=∑ΔH° (reactant bond) -∑ΔH° (product bond)
=[3*ΔH°(C-H)+ΔH° (C-Cl)+ΔH° (Cl-Cl)] - [2 H (C H) 2 H (C Cl) H (H Cl)]
=[3*414+ΔH°(C-Cl)+243] - [2*414+2*ΔH° (C-Cl)+431]
-104 = 1242+ΔH° (C-Cl)+243-828-2*ΔH° (C-Cl) 431
ΔH° (C-Cl) = 330 kJ

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#### APPEARS IN

2015-2016 (March) (with solutions)
Question 2.3 | 3.00 marks

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Solution Calculate C-Cl bond enthalpy from following reaction: CH3Cl(g) + Cl2(g) → Ch2Cl2(g) + HCl(g) ΔH° = -104KJ Concept: Chemical Thermodynamics and Energetic - Enthalpy of Bond Dissociation.
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