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

Consider the following cases of standard enthalpy of reaction $\Delta H_{r}^{\circ}$ in $kJ \ mol^{-1}$:
$C_{2}H_{6(g)} + \frac{7}{2} O_{2(g)} \rightarrow 2 CO_{2(g)} + 3 H_{2}O(\ell)$,$\Delta H_{1}^{\circ} = -1550$
$C(\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H_{2}^{\circ} = -393.5$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O(\ell)$,$\Delta H_{3}^{\circ} = -286$
The magnitude of $\Delta H_{f, C_{2}H_{6(g)}}^{\circ}$ is $........... kJ \ mol^{-1}$ $(Nearest \ integer)$.

From the following data,the heat of transition for the conversion of rhombic sulfur $(S_R)$ to monoclinic sulfur $(S_M)$ in $kJ$ is:
$S_R + O_{2(g)} \to SO_{2(g)}; \Delta H = -296.90 \ kJ$
$S_M + O_{2(g)} \to SO_{2(g)}; \Delta H = -299.40 \ kJ$

Given that the molar combustion enthalpies of benzene,cyclohexane,and hydrogen are $x, y$,and $z$ respectively,the molar enthalpy of hydrogenation of benzene to cyclohexane is

Calculate the heat of combustion (in $kJ$) of methane from the following data:
$(i)$ $C_{\text{(graphite)}} + 2H_{2(g)} \rightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ$
(ii) $C_{\text{(graphite)}} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
(iii) $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -286.2 \ kJ$

The value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$. Then,the formation of one mole of water in the gaseous state from $H_{(g)}$ and $O_{(g)}$ atoms is accompanied by: