The heat of neutralisation of $\mathrm{HCl}$ by $\mathrm{NaOH}$ is $-55.9 \mathrm{kJ} {\mathrm{mole}}^{-1}$. If the heat of neutralisation of $\mathrm{HCN}$ by $\mathrm{NaOH}$ is $-12.1 \mathrm{kJ} {\mathrm{mole}}^{-1}$, the energy of dissociation of $\mathrm{HCN}$ is:

The enthalpy of formation of ammonia is $-46.0 \mathrm{kJ} {\mathrm{mol}}^{-1}.$ The enthalpy change for the reaction $4{\mathrm{NH}}_3\left(\mathrm{g}\right)\to 2{\mathrm{N}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\left(\mathrm{g}\right)$ is

Assuming the water vapour is an ideal gas, the internal energy change $\left(∆\mathrm{U}\right)$ when $1 \mathrm{mole}$ of water is vaporised at $1 \mathrm{bar}$ pressure and $100{}^{\circ }\mathrm{C}$ will be

$\left(\text{Given: molar enthalpy of vaporisation of} {\mathrm{H}}_2\mathrm{O} \text{at} 1 \mathrm{bar} \text{and} 373 \mathrm{K}=41 \mathrm{kJ} {\mathrm{mol}}^{-1}\right)$