The enthalpy changes at $298 \ K$ in successive breaking of $O-H$ bonds of $H_2O$ are:
$H_2O_{(g)} \to H_{(g)} + OH_{(g)}, \Delta H = 498 \ kJ \ mol^{-1}$
$OH_{(g)} \to H_{(g)} + O_{(g)}, \Delta H = 428 \ kJ \ mol^{-1}$
The bond enthalpy of the $O-H$ bond is ..... $kJ \ mol^{-1}$.

On the basis of the following equations,the heat of dimerisation of $NO_2$ will be:
$(i) \ N_2 + 2O_2 \to 2NO_2, \Delta H = 67.9 \ kJ$
$(ii) \ N_2 + 2O_2 \to N_2O_4, \Delta H = 9.3 \ kJ$

Which of the following reactions is endothermic?

What will be the enthalpy of formation of benzene in $kJ$ for the reaction: $6C_{(s)} + 3H_{2(g)} \rightarrow C_6H_{6(l)}$? Given that the enthalpy of combustion of benzene is $-3268 \ kJ$, the enthalpy of formation of $CO_{2(g)}$ is $-393.5 \ kJ$, and the enthalpy of formation of $H_2O_{(l)}$ is $-285.8 \ kJ$.

The enthalpy change for the transition of carbon from diamond to graphite is $\Delta H = -453.5 \ \text{cal}$. What does this indicate?

Given $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$ and $2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$,calculate $\Delta H$ for the reaction $Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$ in $Kcal$.