Exams › JEE Main › Chemistry
Given that the bond enthalpies at 298 K for C−H, C−C, C=C, and H−H 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?
- −88 kJ
- −66 kJ
- −62 kJ
- −44 kJ
Correct answer: −62 kJ
Solution
The enthalpy change for the reaction can be calculated using the bond enthalpies of the bonds broken and formed. In this case, breaking the C=C and H−H bonds requires energy, while forming C−C and C−H bonds releases energy, resulting in an overall enthalpy change of −62 kJ.
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