If the bond energies of $H-H$,$Br-Br$ and $H-Br$ are $433$,$192$ and $364 \ kJ \ mol^{-1}$ respectively,then $\Delta H^{\circ}$ for the reaction: $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$ is:

$H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}, \Delta H = -44 \ kcal$
$2Na_{(s)} + 2HCl_{(g)} \to 2NaCl_{(s)} + H_{2(g)}, \Delta H = -152 \ kcal$
For the reaction $Na_{(s)} + \frac{1}{2}Cl_{2(g)} \to NaCl_{(s)}, \Delta H = \dots \ kcal$

The bond enthalpies of heavy hydrogen $(D-D)$, oxygen $(O=O)$, and heavy water $(D-O)$ are $+400$, $+498$, and $+490 \ kJ \ mol^{-1}$, respectively. The $\Delta_{r} H^{\circ}$ of the reaction to produce $D_2O$ is:

Hess's law of constant heat summation is based on

Find the value $\Delta H_f^o[NH_{3(g)}]$ if $\Delta H_r$ for the reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is $-\,183.6 \ kJ/mol$ of $N_{2(g)}$.

Enthalpy is an extensive property. In general,if enthalpy of an overall reaction $A \to B$ along one route is $\Delta_r H$ and $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \dots$ represent enthalpies of intermediate reactions leading to product $B$. What will be the relation between $\Delta_r H$ for overall reaction and $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \dots$ etc. for intermediate reactions?