The enthalpy change of a reaction does not depend upon:

${\mathrm{H}}_2{\mathrm{SO}}_4\left(\ell \right)+$ aqueous $\to {\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)$ then the heat of reaction is called:

Enthalpy of neutralisation of ${\mathrm{NH}}_4\mathrm{OH}$ and $\mathrm{HCl},$ is numerically:

Calculate the resonance energy of ${\mathrm{N}}_2\mathrm{O}$ form the following data$: \Delta {\mathrm{H}}_{\mathrm{f}}°$ of ${\mathrm{N}}_2\mathrm{O}=82 \mathrm{kJ} {\mathrm{mol}}^{-1}$ and calculated heat of formation is $170 \mathrm{kJ} {\mathrm{mol}}^{-1}:$

Enthalpy of atomisation of ${\mathrm{NH}}_3$ and ${\mathrm{N}}_2{\mathrm{H}}_4$ are $\mathrm{x} \mathrm{KCal} {\mathrm{mol}}^{-1}$ and $\mathrm{y} \mathrm{KCal} {\mathrm{mol}}^{-1}$ respectively. Calculate average bond energy of $\mathrm{N}-\mathrm{N}$ bond:

Given the following standard enthalpies of reaction: [Thermodynamics & Thermochemistry]
(i) Enthalpy of formation of water $=-68.3 \mathrm{kCal}/\mathrm{mol}$
(ii) Enthalpy of combustion of acetylene $=-310.6 \mathrm{kCal}/\mathrm{mol}$
(iii) Enthalpy of combustion of ethylene $=-337.2 \mathrm{kCal}/\mathrm{mol}$

Calculate the heat of reaction for the hydrogenation of acetylene into ethylene (in $\mathrm{kCal}/\mathrm{mol}$) at constant pressure. Report your answer in terms of magnitude, rounding it off to the nearest whole number.