If enthalpies of formation of $${C_2}{H_4}\left( g \right),C{O_2}\left( g \right)$$    and $${H_2}O\left( l \right)$$  at $${25^ \circ }C$$  and $$1\,atm$$  pressure are $$52,-394$$   and $$ - 286\,kJ/mol,$$   the enthalpy of combustion of ethene is equal to

Solution :
$${\text{Combustion of hydrocarbon,}}$$
$${C_2}{H_4}\left( g \right) + 3{O_2}\left( g \right) \to $$     $$2C{O_2}\left( g \right) + 2{H_2}O\left( l \right)$$
$${\Delta _r}H = \sum {{\Delta _f}{H_p} - \sum {{\Delta _f}{H_r}} } $$
         $$ = \left\{ {2 \times {\Delta _f}H\left( {C{O_2}} \right) + 2 \times {\Delta _f}H\left( {{H_2}O} \right)} \right\}$$        $$ - \left\{ {{\Delta _f}H\left( {{C_2}{H_4}} \right) + 3{\Delta _f}H\left( {{O_2}} \right)} \right\}$$
         $$ = \left( {2 \times - 394 + 2 \times - 286} \right) - \left( {52 + 3 \times 0} \right)$$
         $$ = - 788 - 572 - 52$$
         $$ = - 1412\,kJ\,mo{l^{ - 1}}$$