The heat of neutralisation will be highest in

Calculate the standard enthalpy change for the reaction,$C_2H_5OH_{(\ell)} + 3O_{2_{(g)}} \rightarrow 2CO_{2_{(g)}} + 3H_2O_{(\ell)}$. Given: $\Delta_{f}H^{\circ}(C_2H_5OH) = -280 \ kJ \ mol^{-1}$,$\Delta_{f}H^{\circ}(CO_2) = -390 \ kJ \ mol^{-1}$,and $\Delta_{f}H^{\circ}(H_2O) = -285 \ kJ \ mol^{-1}$.

Given $C + O_2 \rightarrow CO_2 + 94.2 \ kcal$,$\Delta H = -94.2 \ kcal$; $H_2 + 1/2 O_2 \rightarrow H_2O + 68.3 \ kcal$,$\Delta H = -68.3 \ kcal$ and $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + 210.8 \ kcal$,$\Delta H = -210.8 \ kcal$. Calculate the heat of formation of methane $(CH_4)$ in $kcal$.

Calculate the enthalpy of formation of nitric oxide $(NO)$ in $kJ \, mol^{-1}$ from the following data:
$NO_{(g)} + CO_{(g)} \rightarrow \frac{1}{2} N_{2(g)} + CO_{2(g)}; \Delta H^o = -372.2 \, kJ \, mol^{-1}$
$\Delta H_f^o (CO) = -110.5 \, kJ \, mol^{-1}$
$\Delta H_f^o (CO_2) = -393.5 \, kJ \, mol^{-1}$

The enthalpy of neutralization of a strong acid by a strong base is $-57.32 \ kJ/mol$. The enthalpy of formation of water is $-285.84 \ kJ/mol$. The enthalpy of formation of hydroxyl ion is......$kJ/mol$. (Assume $\Delta H_{f}^{\circ}(H^{+}_{(aq)}) = 0 \ kJ/mol$)

The enthalpy change at $298 \ K$ for the decomposition of water is given in the following two steps:
Step $I$: $H_2O_{(g)} \to H_{(g)} + OH_{(g)}$; $\Delta H = 498 \ kJ \ mol^{-1}$
Step $II$: $OH_{(g)} \to H_{(g)} + O_{(g)}$; $\Delta H = 428 \ kJ \ mol^{-1}$
The average bond enthalpy of the $O-H$ bond is $.... \ kJ \ mol^{-1}$