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 $.... \ kJ \ mol^{-1}$

$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \dots \dots(I) \quad \Delta H = -393 \, kJ \, mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O_{(l)} \dots \dots(II) \quad \Delta H = -287.3 \, kJ \, mol^{-1}$
$2CO_{2(g)} + 3H_{2}O_{(l)}$ $\rightarrow C_{2}H_{5}OH_{(l)} + 3O_{2(g)} \dots \dots(III) \quad \Delta H = 1366.8 \, kJ \, mol^{-1}$
Find the standard enthalpy of formation of $C_{2}H_{5}OH_{(l)}$.

The enthalpy of neutralization of any strong acid and strong base is constant because......

$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}$; $\Delta H_{298K} = -68.32 \, kcal$. The enthalpy of vaporization of water at $25 \, ^\circ C$ and $1 \, atm$ pressure is $10.52 \, kcal$. Calculate the standard enthalpy of formation of $1 \, mole$ of water vapor at $25 \, ^\circ C$ (in $kcal$).

The standard heat of formation $\left(\Delta_{f} H_{298}^{0}\right)$ of ethane in $kJ/mol$,if the heat of combustion of ethane,hydrogen and graphite are $-1560$,$-286$,and $-393.5 \; kJ/mol,$ respectively is ........... $kJ/mol$.

The $\Delta H_f^o$ for $CO_{2(g)}$,$CO_{(g)}$ and $H_2O_{(g)}$ are $-393.5$,$-110.5$ and $-241.8 \ kJ \ mol^{-1}$ respectively. The standard enthalpy change (in $kJ$) for the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$ is