In the reaction $H_2 + Cl_2 \rightarrow 2HCl$,heat is released. The bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $242 \ kJ \ mol^{-1}$ respectively. If the enthalpy of reaction is $-182 \ kJ \ mol^{-1}$,the bond energy of $H-Cl$ is . . . . . . $kJ \ mol^{-1}$.

The enthalpy change for the reaction,$H_{2(g)} + C_2H_{4(g)} \to C_2H_{6(g)}$ is $......$ $kcal \ mol^{-1}$. The bond energies are,$[e_{H-H} = 103, e_{C-H} = 99, e_{C-C} = 80]$ and $[e_{C=C} = 145] \ kcal \ mol^{-1}$.

The enthalpy of combustion of ${C_6H_6}_{(l)}$ is $-3250 \, kJ/mol$. When $0.39 \, g$ of benzene is burnt in excess of oxygen in an open vessel,the amount of heat evolved is:

Which of these species has a standard enthalpy of formation equal to zero?

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 $\text{kJ mol}^{-1}$:

If the enthalpy of formation and enthalpy of solution of $HCl(g)$ are $-92.3 \ kJ/mol$ and $-75.14 \ kJ/mol$ respectively,then find the enthalpy of formation of $Cl^{-}(aq)$. [Assume $\Delta H_{f}(H^{+}) = 0 \ kJ/mol$]