If the enthalpy of combustion of diamond and graphite are -395.4kJ mol^(-1) and -393.6 kJ mol^(-1) what is the enthalpy change for the C (graphite) rarr C (diamond) conversion ?

If the enthalpy of combustion of diamond and graphite are -395.4kJ mol

1.	If the enthalpy of combustion of diamond and graphite are -395.4kJ mol^(-1) and -393.6 kJ mol^(-1) what is the enthalpy change for the C (graphite) rarr C (diamond) conversion ?
Answer» Solution :C(diamond) `+O_(2)(g) rarr CO_(2)(g) Delta= -395.4 kJ MOL^(-1)` C (graphite) `+ O_(2)(g) rarr CO_(2)(g)Delta H = -393.6 kJ mol^(-1)` …(2) C (graphite) `rarr "C(diamond)"_(1)` substracting (1) from (2), we GET C (graphit) `rarr "C(diamond)" DeltaH = -393.6 kJ - (-395.4 kJ)`, `Delta H = +1.8 kJ mol^(-1)`.