g12 chem enthalpy change question?

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  • 1 month ago
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    20.

    H₂C=CH₂ + H₂(g) → CH₃CH₃(g)    ΔH° = ? kJ/mol

    Each H₂C=CH₂ molecule contains 4 C-H bonds and 1 C=C double bond.

    Each H₂ molecule contains 1 H-H single bond.

    Each CH₃CH₃ molecule contains 6 C-H bonds and 1 C-C single bond.

    Enthalpy absorbed for bond breaking of the reactants

    = 4 E(C-H) + E(C=C) + E(H-H)

    = [4 (410) + (611) + (436)] kJ/mol

    = 2687 kJ/mol

    Enthalpy released for bond formation of the product

    = 6 E(C-H) + E(C-C)

    = [6 (410) + (350)] kJ/mol

    = 2810 kJ/mol

    ΔH°

    = (2687 - 2810) kJ/mol

    = -123 kJ/mol

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