In thermodynamic reaction,

$\Delta H=\Delta E+\Delta nRT$

Where, $\Delta H$ is the enthalpy change, $\Delta E$ is the internal energy change, $\Delta n$ is the number of gaseous moles of product - number of gaseous moles of reactant. And $R$ and $T$ are the gas constant and temperature respectively.

The reaction given is ${\mathrm{C}}_2{\mathrm{H}}_{4\left(\mathrm{ }\mathrm{g}\right)}+3{\mathrm{O}}_{2\left(g\right)}\to 2{\mathrm{CO}}_{2\left(g\right)}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

The number of gaseous moles of product $=2$ and the number of gaseous moles of reactant $=$ $3+1=4$

So, $\Delta n=2-4=-2$

So, the equation for the given reaction is $\Delta H=\Delta E-2RT$.

Therefore, option (D) is correct.