If the enthalpy of formation of $N_2O$ is $82 \, kJ \, mol^{-1}$,calculate the resonance energy of $N_2O$ in $kJ \, mol^{-1}$.
$N \equiv N \, (946 \, kJ \, mol^{-1}); \, N = N \, (418 \, kJ \, mol^{-1})$
$O = O \, (498 \, kJ \, mol^{-1}); \, N = O \, (607 \, kJ \, mol^{-1})$

Consider the reaction $2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g)$. The magnitude of enthalpy change for the reaction in $\text{kJ mol}^{-1}$ is . . . . . . . (Nearest integer). Given: $\Delta_f H^\circ(H_2S) = -20.1 \text{ kJ mol}^{-1}$, $\Delta_f H^\circ(H_2O) = -286.0 \text{ kJ mol}^{-1}$, $\Delta_f H^\circ(SO_2) = -297.0 \text{ kJ mol}^{-1}$

When $4 \ g$ of iron is burnt to ferric oxide at constant pressure,$29.28 \ kJ$ of heat is evolved. What is the enthalpy of formation of ferric oxide in $kJ \ mol^{-1}$? (At. Wt. of $Fe = 56$)

If the ratio of bond dissociation energies of $XY$,$X_2$,and $Y_2$ is $1 : 1 : 0.5$ and the enthalpy of formation of $XY$ is $\Delta_f H = -200 \ kJ \ mol^{-1}$,then the bond dissociation energy of $X_2$ in $kJ \ mol^{-1}$ is:

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

For the reaction,$3 C_2 H_{2(g)} \longrightarrow C_6 H_{6(g)}$,calculate the standard enthalpy change. The values of $\Delta H_f$ for $C_2 H_2$ and $C_6 H_6$ respectively are $250 \ kJ \ mol^{-1}$ and $90 \ kJ \ mol^{-1}$.