Hess's law deals with

The bond energies of $H-H$ and $Cl-Cl$ are $430 \, kJ \, mol^{-1}$ and $242 \, kJ \, mol^{-1}$ respectively. $\Delta H_f$ for $HCl$ is $-91 \, kJ \, mol^{-1}$. The bond energy of $HCl$ will be ............. $kJ \, mol^{-1}$.

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 $kJ\ mol^{-1}$.

Find the enthalpy of formation of the $OH^-$ ion in $KJ$ at $25^\circ C$ from the following data:
$H_2O_{(l)} \to H^+_{(aq)} + OH^-_{(aq)} ; \Delta H = 57.32 \ KJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.20 \ KJ$

The bond dissociation enthalpies of $H_2$, $Cl_2$, and $HCl$ are $434 \text{ kJ mol}^{-1}$, $242 \text{ kJ mol}^{-1}$, and $431 \text{ kJ mol}^{-1}$ respectively. Calculate the enthalpy of formation of $HCl$.

The atomisation enthalpy of $CH_4$ is $1660 \ kJ \ mol^{-1}$. The $C-H$ bond enthalpy of each successive step in $CH_4$ $\rightarrow CH_3$ $\rightarrow CH_2$ $\rightarrow CH$ are $+15, +30$ and $+45 \ kJ \ mol^{-1}$ higher than the mean bond enthalpy of $C-H$ bonds,respectively. The bond enthalpy of the last $C-H$ unit is