If bond enthalpies of $N \equiv N$,$H - H$ and $N - H$ bonds are $x_1, x_2$ and $x_3$ respectively,$\Delta H_f^o$ for $NH_3$ will be

For a reaction $A + B + 30 \ kcal \longrightarrow \text{Product}$,the energy of activation of its backward reaction is $50 \ kcal$. The energy of activation of its forward reaction is :-

If $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H = -X$,and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H = -Y$,calculate $\Delta_f H$ for $CO_{(g)}$ formation.

From the following data,the heat of transition for the conversion of rhombic sulfur $(S_R)$ to monoclinic sulfur $(S_M)$ in $kJ$ is:
$S_R + O_{2(g)} \to SO_{2(g)}; \Delta H = -296.90 \ kJ$
$S_M + O_{2(g)} \to SO_{2(g)}; \Delta H = -299.40 \ kJ$

Identify $I, II, III$ in the given Born-Haber cycle diagram for the dissolution of an ionic solid $AB(s)$:
$AB(s) \xrightarrow{I} A^+(aq) + B^-(aq)$
$AB(s) \xrightarrow{II} A^+(g) + B^-(g)$
$A^+(g) + B^-(g) \xrightarrow{III} A^+(aq) + B^-(aq)$

Given: $S + O_2 \rightarrow SO_2 : \Delta H_1 = -298.2 \ kJ$,$SO_2 + 1/2 \ O_2 \rightarrow SO_3 : \Delta H_2 = -98.7 \ kJ$,$SO_3 + H_2O \rightarrow H_2SO_4 : \Delta H_3 = -130.2 \ kJ$,and $H_2 + 1/2 \ O_2 \rightarrow H_2O : \Delta H_4 = -287.3 \ kJ$. Calculate the heat of formation of $H_2SO_4$ in $kJ$.