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:
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:
C (graphite) → C (diamond)
ΔH? = Δ? H° (graphite) - Δ? H° (Diamond) = -832.8 - [-834.8] = 2kJ/mole
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Chemistry Ncert Solutions Class 11th 2023
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