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JEE Main levelNumerical answerDifficulty 2 of 3td079

Using average bond enthalpies in $kJ mo l^{- 1}$ ( $C = C = 614$ , $C - C = 348$ , $Br - Br = 193$ , $C - Br = 276$ ), estimate the enthalpy change in $kJ mo l^{- 1}$ for $CH X_{2} = CH X_{2} (g) + Br X_{2} (g) BrCH X_{2} - CH X_{2} Br (g)$ (nearest integer).

Numerical answer type, answer in kJ/mol, tolerance 0.5.

Answer: -93 kJ/mol

Bonds broken: one

C = C

and one

Br - Br

,

614 + 193 = 807

. Bonds formed: one

C - C

and two

C - Br

,

348 + 2 \times 276 = 900

.

Δ H = 807 - 900 = - 93 kJ mo l^{- 1}

.

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