The heats of formation of ${\mathrm{CO}}_2$ , ${\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{C}}_2{\mathrm{H}}_4$ are – 94.05, – 68.2 and 12.6 K.Cals/mole respectively. The heat of combustion of ethylene is

 The reactions of formation of ${\mathrm{CO}}_2$ , ${\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{C}}_2{\mathrm{H}}_4$ are given below:

$2\mathrm{C}+2{\mathrm{H}}_2\to {\mathrm{C}}_2{\mathrm{H}}_4$………..(1) 

The enthalpy of the reaction is 12.6 K.Cal/mole

$\mathrm{C}+{\mathrm{O}}_2\to {\mathrm{CO}}_2$ ……….(2)

The enthalpy of the reaction is -94.05 K.Cal/mole

${\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}$ ………..(3)

The enthalpy of the reaction is -68.2 K.Cal/mole

The reaction we want to get is:

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

To get this equation, we have to add the twice of equation (2) and (3), then subtract this equation from (1).

So, the enthalpy can be calculated as:

$$\begin{gathered} \mathrm{\Delta H}=2[{\mathrm{\Delta H}}_2+{\mathrm{\Delta H}}_3]-{\mathrm{\Delta H}}_1 \\ \mathrm{\Delta H}=2[-94.05+(68.2\left)\right]-12.6 \\ \mathrm{\Delta H}=2[-162.25]-12.6 \\ \mathrm{\Delta H}=-337\text{ K.Cal/mole} \end{gathered}$$

Therefore, the correct option is D.