Hess's law deals with

$A_2 + B_2 \rightarrow 2AB$; $\Delta H_{r}^0 = -400\,kJ\,mol^{-1}$. $AB$,$A_2$ and $B_2$ are diatomic molecules. If the bond enthalpies of $A_2$,$B_2$ and $AB$ are in the ratio $1:0.5:1$,then the bond enthalpy of $A_2$ is $......\,kJ\,mol^{-1}$ (Nearest integer).

Diborane is formed from the elements as shown in equation $(i)$:
$2 B_{(s)} + 3 H_{2(g)} \longrightarrow B_2H_{6(g)} \dots (i)$
Given that:
$H_2O_{(l)} \longrightarrow H_2O_{(g)}, \quad \Delta H_1^{\circ} = 44 \, kJ$
$2 B_{(s)} + \frac{3}{2} O_{2(g)} \longrightarrow B_2O_{3(s)}, \quad \Delta H_2^{\circ} = -1273 \, kJ$
$B_2H_{6(g)} + 3 O_{2(g)} \longrightarrow B_2O_{3(s)} + 3 H_2O_{(g)}, \quad \Delta H_3^{\circ} = -2035 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}, \quad \Delta H_4^{\circ} = -286 \, kJ$
The $\Delta H^{\circ}$ for the reaction $(i)$ is $..... \, kJ$.

Calculate the heat change for the reaction $4NH_3(g) + 3O_2(g) \rightarrow 2N_2(g) + 6H_2O(l)$ in $kJ$. The enthalpies of formation of $NH_3(g)$ and $H_2O(l)$ at $298 \ K$ are $-46.0 \ kJ \ mol^{-1}$ and $-286.0 \ kJ \ mol^{-1}$ respectively.

Average bond enthalpy of water is $464.5 \text{ kJ mol}^{-1}$. If the energy required to break the first $O-H$ bond is $502 \text{ kJ mol}^{-1}$,how much energy per mol is required to break the second $O-H$ bond?

Which of the following compounds will absorb the maximum quantity of heat when dissolved in the same amount of water? The heats of solution of these compounds at $25 \, ^\circ C$ in $kJ/mole$ of each solute are given in brackets.