Enthalpy of combustion of diamond is 395 kJ/mole and Enthalpy of combustion of graphite is 393 kJ/mole. Enthalpy change when 24 grams of graphite converts to 24 grams of diamond is

A
$2 kJ$
B
$- 2 kJ$
C
$4 kJ$
D
$- 4 kJ$

Solution:

$Combustion is always exothermic, ∆ H = - ve; C (diamond) + O_{2} (g) \to {CO}_{2} (g); ∆ H = - 395 kJ - I; C (graphite) + O_{2} (g) \to {CO}_{2} (g); ∆ H = - 393 kJ - II; Eq II - Eq I C (graphite) \to C (diamond); ∆ H = + 2 kJ / mole 12 grams 2 kJ 24 grams 4 kJ$

Solution:

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