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At 298 K, the bond enthalpies of C—H, C—C, C=C and H—H bonds are 414, 347, 615 and 435 kJ mol−1, respectively. What is the enthalpy change for the reaction H2C=CH2(g) + H2(g) → H3C—CH3(g) at 298 K?

1. −250 kJ
2. +125 kJ
3. −125 kJ
4. +250 kJ

Correct answer: −125 kJ

Solution

Broken: C=C + 4 C-H + H-H = 615 + 4(414) + 435 = 2706 kJ. Formed: C-C + 6 C-H = 347 + 6(414) = 2831 kJ. dH = 2706 - 2831 = -125 kJ.

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