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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?
- −250 kJ
- +125 kJ
- −125 kJ
- +250 kJ
Correct answer: +125 kJ
Solution
The enthalpy change for the reaction can be calculated by subtracting the total bond enthalpies of the bonds broken from those formed. In this case, breaking the C=C and H—H bonds requires energy, while forming two C—H bonds releases energy, resulting in a net positive enthalpy change of +125 kJ.
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