$A_2 + B_2 \rightarrow 2AB$; $\Delta H_{r}^0 = -400\,kJ\,mol^{-1}$. $AB$,$A_2$ 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).

$AB$,$A_2$,and $B_2$ are diatomic molecules. Enthalpies of dissociation of $AB$,$A_2$,and $B_2$ are in the ratio of $1:1:0.5$. Enthalpy of formation of $AB$,$\Delta_f H = -100 \ kJ \ mol^{-1}$. Find the dissociation enthalpy of $A_2$?
Reaction : $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$

Given the following thermochemical equations:
$1) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -787 \ kJ$
$2) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)}, \Delta H = -286 \ kJ$
$3) \ C_2H_{2(g)} + \frac{5}{2} O_{2(g)} \rightarrow 2CO_{2(g)} + H_2O_{(l)}, \Delta H = -1310 \ kJ$
Calculate the enthalpy of formation of acetylene $(C_2H_{2(g)})$ in $kJ \ mol^{-1}$.

Using the given reaction enthalpies,find the enthalpy of formation of $H_2O_2(l)$ in $kJ/mol$.
$(i) N_2H_4(l) + 2H_2O_2(l) \rightarrow N_2(g) + 4H_2O(l); \Delta_r H_1^\circ = -818 \, kJ/mol$
$(ii) N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(l); \Delta_r H_2^\circ = -622 \, kJ/mol$
$(iii) H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l); \Delta_r H_3^\circ = -285 \, kJ/mol$

Enthalpy change for a reaction does not depend upon

The heat of combustion of solid benzoic acid at constant volume is $-321.30 \ kJ$ at $27^{\circ} C$. Its heat of combustion at constant pressure is :