How do I answer this enthalpy question?

Question

The standard enthalpy of combustion for xylene, C8H10(l), is –3908 kJ mol-1. Using this data and the standard enthalpies of formation, : H2O(l) = –285.9 kJ mol-1; CO2(g) = –393.5 kJ mol-1, calculate the standard enthalpy of formation of C8H10(l), in kJ mol-1.

Answer 1

We first write the combustion equation
Enthalpy of Combustion is –3908 kJ mol-1
Enthalpy of Formation of H20 is –285.9 kJ mol-1
Enthalpy of Formation of CO2 is –393.5 kJ mol-1

Let get Start

C8 + 5H2 -> C8H10 Hc = -3908
Enthalpy of Formation of C8H10 = - (Enthalpy of Combustion ) + Enthalpy of Formation of H20 + Enthalpy of Formation of CO2

:. Enthalpy of Formation of C8H10 = -(-3908) + 8(–393.5 ) + 5(–285.9)
:. Enthalpy of Formation of C8H10 = 3908 -3148 -1429.5
= -669.5 kJ mol-1.

Answer 2

C8H10 + 21/2 O2 -> 8CO2 + 5H2O

Heat of formation of C8H10 = -(-3908 + 8*393.5 + 5* 285.9)

= -669.5 kJ/mol