If enthalpies of formation of C₂H₄ (g) , CO₂(g) and H₂O_(l) at 25°C and 1atm pressure are 52, – 394 and – 286 kJ/mol respectively, the change in ethalpy is equal to

Enthalpy of formation of C₂H₄, CO₂ and H₂O are 52, – 394 and – 286 kJ/ mol respectively. (Given) 

The reaction is

${\mathrm{C}}_2{\mathrm{H}}_4+3{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$

change in enthalpy,

$\begin{array}{l}\left(\mathrm{\Delta H}\right)={\mathrm{\Delta H}}_{\text{products }}-{\mathrm{\Delta H}}_{\text{reactants }}\\ =2\times (-394)+2\times (-286)-(52+0)\\ =-1412\mathrm{kJ}/\mathrm{mol}\end{array}$