The heat of neutralization of four acids $A$,$B$,$C$,and $D$ are $-13.0$,$-12.6$,$-9.2$,and $-11.7 \ KCal/eq$ respectively when neutralized by $NaOH$. The order of acidic strength of the four acids will be:

Given the following data:

Reaction	Energy Change (in $kJ$)
$Li_{(s)} \to Li_{(g)}$	$161$
$Li_{(g)} \to Li^{+}_{(g)}$	$520$
$\frac{1}{2} F_{2(g)} \to F_{(g)}$	$77$
$F_{(g)} + e^- \to F^{-}_{(g)}$	(Electron gain enthalpy)
$Li^{+}_{(g)} + F^{-}_{(g)} \to LiF_{(s)}$	$-1047$
$Li_{(s)} + \frac{1}{2} F_{2(g)} \to LiF_{(s)}$	$-617$

Based on the data provided,the value of electron gain enthalpy of fluorine would be $kJ\ mol^{-1}$.

The heats of formation of $CO_{2(g)}$,$H_2O_{(l)}$,and $CH_{4(g)}$ are $-94.0$,$-68.4$,and $-17.9 \ kcal$ respectively. The heat of combustion of methane is.....$kcal$.

Considering the reaction $3O_2 \rightarrow 2O_3$; $\Delta H = +ve$,what can we conclude about the reaction?

The bond dissociation energy of molecules is.......

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