ThermochemistryMedium
Question
The bond energies (in kJ/mol) at 25°C are C–C = 346, C–H = 413, H–H = 437, C = C; 611. From these data, the value of ΔH at 25°C for the above reaction is
Options
A.−289 kJ mol–1
B.−124 kJ mol–1
C.124 kJ mol–1
D.289 kJ mol–1
Solution
$\Delta H = \left( B.E._{C - C\pi\text{ bond}} + B.E._{H - H} \right) - \left( 2 \times B.E._{C - H} \right)$
$= \left\lbrack (611 - 346) + 437 \right\rbrack - (2 \times 413) = - 124\text{ kJ}$
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