$H^{+}_{(aq)} + OH^{-}_{(aq)} \rightarrow H_2O_{(l)} + 56 \text{ kJ/mol}$. The heat of neutralization for the complete neutralization of $1 \text{ mole}$ of $H_2SO_4$ will be ...... $\text{kJ}$.

Find the standard enthalpy of formation of ammonia from the following reaction:
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} ; \Delta_{r}H^0 = -92.0 \ kJ$

Calculate the standard enthalpy of formation of $ICl_{(g)}$ based on the following reactions. The standard states of iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$ respectively.
$(i)$ $Cl_{2(g)} = 2Cl_{(g)}$,$\Delta H = 242.3 \text{ kJ mol}^{-1}$
$(ii)$ $I_{2(g)} = 2I_{(g)}$,$\Delta H = 151.0 \text{ kJ mol}^{-1}$
$(iii)$ $ICl_{(g)} = I_{(g)} + Cl_{(g)}$,$\Delta H = 211.3 \text{ kJ mol}^{-1}$
$(iv)$ $I_{2(s)} = I_{2(g)}$,$\Delta H = 62.76 \text{ kJ mol}^{-1}$
Result in $\text{kJ mol}^{-1}$:

The bond enthalpies of $H_2$,$X_2$ and $HX$ are in the ratio of $2 : 1 : 2$. If the enthalpy for formation of $HX$ is $-50 \ kJ \ mol^{-1}$,the bond enthalpy of $H_2$ is ..... $kJ \ mol^{-1}$

The enthalpy of the reaction,$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}$ is $\Delta H_1$ and that of $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$ is $\Delta H_2$. Then:

For the reaction $2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}$,the enthalpy change is $\Delta H = -571 \ kJ$. If the $H-H$ bond energy is $435 \ kJ \ mol^{-1}$ and the $O=O$ bond energy is $498 \ kJ \ mol^{-1}$,calculate the average bond energy of the $O-H$ bond in $kJ \ mol^{-1}$.