Enthalpy of combustion of diamond is 395kJ/mole and Enthalpy of combustion of graphite is 393kJ/mole. Enthalpy change when one mole of diamond converts to one mole of graphite is

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$2 kJ$

$- 2 kJ$

$4 kJ$

$- 4 kJ$

answer is B.

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Detailed Solution

$Combustion is always exothermic, ∆ H = - ve; C (diamond) + O_{2} (g) \to {CO}_{2} (g); ∆ H = - 395 kJ - I; C (graphite) + O_{2} (g) \to {CO}_{2} (g); ∆ H = - 393 kJ - II; Eq I - Eq II C (diamond) \to C (graphite); ∆ H = - 2 kJ / mole$

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