The bond dissociation energies of $XY$,$X_2$,and $Y_2$ (all diatomic molecules) are in the ratio $1 : 1 : 0.5$. If the enthalpy of formation of $XY$ is $\Delta_fH = -200 \ kJ \ mol^{-1}$,find the bond dissociation energy of $X_2$ in $kJ \ mol^{-1}$.

The heat of neutralisation of $NaOH$ and $HCl$ is $-57.46 \, kJ/eq$. What is the heat of ionisation of water in $kJ/mol$?

Using the given reaction enthalpies,find the enthalpy of formation of $H_2O_2(l)$ in $kJ/mol$.
$(i) N_2H_4(l) + 2H_2O_2(l) \rightarrow N_2(g) + 4H_2O(l); \Delta_r H_1^\circ = -818 \, kJ/mol$
$(ii) N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(l); \Delta_r H_2^\circ = -622 \, kJ/mol$
$(iii) H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l); \Delta_r H_3^\circ = -285 \, kJ/mol$

Calculate the amount of methane formed by the liberation of $149.6 \ kJ$ of heat using the following equation:
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ/mol$ (in $g$)

Explain the change in enthalpy related to a chemical reaction.

If the $\Delta H_{fusion}$ of a substance is $'x'$ and $\Delta H_{vap}$ is $'y'$,then $\Delta H_{sublimation}$ will be: