The heat of neutralization of which acid and base is close to $-13.6 \ kcal/mol$?

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:

$C + O_2 \to CO_2; \Delta H = X$
$CO + \frac{1}{2} O_2 \to CO_2; \Delta H = Y$
Then the heat of formation of $CO$ is

If bond energy for $H_{2(g)}$,$Br_{2(g)}$ and $HBr_{(g)}$ is $433$,$192$ and $364 \ kJ \ mol^{-1}$ respectively,then $\Delta H^o$ for the reaction,$H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$ is......$kJ$.

The enthalpy of combustion of ${C_6H_6}_{(l)}$ is $-3250 \, kJ/mol$. When $0.39 \, g$ of benzene is burnt in excess of oxygen in an open vessel,the amount of heat evolved is:

What is the enthalpy of formation of $NH_3$ if the bond enthalpies are $(N \equiv N) = 941 \ kJ/mol$,$(H-H) = 436 \ kJ/mol$,and $(N-H) = 389 \ kJ/mol$?