$AB$,$A_2$ and $B_2$ are diatomic molecules. If the bond enthalpies of $A_2$,$AB$ and $B_2$ are in the ratio $1:1:0.5$ and enthalpy of formation of $AB$ from $A_2$ and $B_2$ is $-100 \, kJ \, mol^{-1}$,what is the bond energy of $A_2$ in $kJ \, mol^{-1}$?

Find the enthalpy of formation of the $OH^-$ ion in $KJ$ at $25^\circ C$ from the following data:
$H_2O_{(l)} \to H^+_{(aq)} + OH^-_{(aq)} ; \Delta H = 57.32 \ KJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.20 \ KJ$

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

Calculate the standard enthalpy of formation of $ICl_{(g)}$ based on the following reactions. The standard states of iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$ respectively.
$(i)$ $Cl_{2(g)} = 2Cl_{(g)}$,$\Delta H = 242.3 \text{ kJ mol}^{-1}$
$(ii)$ $I_{2(g)} = 2I_{(g)}$,$\Delta H = 151.0 \text{ kJ mol}^{-1}$
$(iii)$ $ICl_{(g)} = I_{(g)} + Cl_{(g)}$,$\Delta H = 211.3 \text{ kJ mol}^{-1}$
$(iv)$ $I_{2(s)} = I_{2(g)}$,$\Delta H = 62.76 \text{ kJ mol}^{-1}$
Result in $\text{kJ mol}^{-1}$:

$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
If graphite is converted into diamond,then the $\Delta H$ for the process is . . . . . . $\text{kJ}$.

On the basis of the following reactions,which one is correct?
$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$
$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$
$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$