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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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