Answer
$\Delta H^{\circ}_{f} =+ 1.9 kJ/mol $
Work Step by Step
The equation for the enthalpy of formation of diamond is
$C_{graphite} → C_{diamond}$.
Equation 1: $C_{(graphite)} + O_{2(g)} →CO_{2(g)}$
$ H^{\circ}rxn = - 393.5 kJ/mol$
Equation 2: $C_{(diamond)} + O_{2(g)} →CO_{2(g)}$
$ H^{\circ}rxn = - 395.4 kJ/mol$
By adding equation 1 with the reversed equation of 2, we get the enthalpy of formation equation $C_{graphite} → C_{diamond}$.
$\Delta H^{\circ}_{f} = - 393.5 – (- 395.4) kJ/mol = + 1.9 kJ/mol $
