The standard enthalpy of combustion $6_6 \mathrm{H}_6(\mathrm{~g})$ is $-1564.25 \mathrm{~kJ} \mathr

Question

The standard enthalpy of combustion $6_6 \mathrm{H}_6(\mathrm{~g})$ is $-1564.25 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Standard heat of formation of $\mathrm{CO}_2(\mathrm{~g})$ and $\mathrm{H}_2 \mathrm{O}(\ell)$ are $-395 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The absolute value of heat liberated in $\mathrm{kJ}^{-1}$ for the formation of 1 mole of $\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})$ from its constituent elements in standard state is (nearest integer).....

Professor PC Thomas Classes · Answer

$\begin{aligned} & \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\ell) \ & \Delta \mathrm{H}_{\mathrm{t}}^{\circ}=\sum \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(	ext { Products })-\sum \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(	ext { Reactants }) \ & -1564.25=[2 	imes(-395)+3 	imes(-286)]-\left[\Delta \mathrm{H}_{\mathrm{f}}^{\circ} \mathrm{C}_2 \mathrm{H}_6ight] \ & \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \mathrm{C}_2 \mathrm{H}_6=-83.75 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$