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Question: What will be \(\Delta H\)for the reaction. \(CH_{2}Cl_{2} \rightarrow C + 2H + 2Cl\)? (B.E. of C – ...

What will be ΔH\Delta Hfor the reaction.

CH2Cl2→C+2H+2ClCH_{2}Cl_{2} \rightarrow C + 2H + 2Cl? (B.E. of C – H and C – Cl bonds are 416 kJ mol−1mol^{- 1}and 325 kJ mol−1mol^{- 1} respectively)

A

832 kJ

B

1482 kJ

C

650 kJ

D

1855 kJ

Answer

1482 kJ

Explanation

Solution

:C+2H+2ClC + 2H + 2Cl

ΔH=∑B.E.R−∑B.E.P\Delta H = \sum B.E._{R} - \sum B.E._{P}

=2×B.E.(C−H)+2×B.E.(C−Cl)= 2 \times B.E.(C - H) + 2 \times B.E.(C - Cl)

=2×416+2×325=1482kJ= 2 \times 416 + 2 \times 325 = 1482kJ