The enthalpies of combustion ofC(graphite) and C( diamond) are −393.5 kJ mol−1 and −395.3 kJ mol−1, respectively. The enthalpy change of the reaction C(graphite) __, C(diamond) is

The enthalpies of combustion ofC(graphite) and C( diamond) are $- 393.5 kJ {mol}^{- 1}$ and $- 395.3 kJ {mol}^{- 1}$ , respectively. The enthalpy change of the reaction C(graphite) __, C(diamond) is

A
$- 1.80 kJ {mol}^{- 1}$
B
$1.80 kJ {mol}^{- 1}$
C
$3.60 kJ {mol}^{- 1}$
D
$- 3.60 kJ {mol}^{- 1}$

Solution:

$(a) C (graphite) + O_{2} (g) \to {CO}_{2} (g) Δ H_{a} = - 393.5 kJ {mol}^{- 1}$

$(b) C (diamond) + O_{2} (g) \to {CO}_{2} (g) Δ H_{b} = - 395.3 kJ {mol}^{- 1}$

The given transformation C(graphite) $\to$ C(diamond) is obtained by subtracting Eq. (b) from Eq. (a). Hence

$Δ H = Δ H_{b} - Δ H_{a} = [- 393.5 - (- 395.3)] kJ {mol}^{- 1} = 1.80 kJ {mol}^{- 1}$

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