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Given that ∆c H° (combustion enthalpy) for Diamond is -834.8 kJ mole and ∆c H° for graphite is -832.8 kJ/mole, the value of ∆f H°(Diamond) i.e. [C(graphite) → C(diamond)] [in kJ/mole] is:

0 3 Views | Posted 4 weeks ago
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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 weeks ago

    C (graphite) → C (diamond)
    ΔH? = Δ? H° (graphite) - Δ? H° (Diamond) = -832.8 - [-834.8] = 2kJ/mole

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