The enthalpy of $1 \ mol$ of a compound is equal to its ...... (when it is formed from its constituent elements).

The enthalpy change $(\Delta H)$ for the neutralisation of $1 \ M \ HCl$ by caustic potash in dilute solution at $298 \ K$ is ..... $kJ$.

Calculate the heat of formation of $Ca(OH)_{2(s)}$ at $1.8\,^{\circ}C$ from the following data:
$CaO_{(s)} + H_2O_{(l)} \to Ca(OH)_{2(s)}$; $\Delta H_{1.8\,^{\circ}C} = -15.26\,K\,cal$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H_{1.8\,^{\circ}C} = -68.37\,K\,cal$
$Ca_{(s)} + \frac{1}{2}O_{2(g)} \to CaO_{(s)}$; $\Delta H_{1.8\,^{\circ}C} = -151.80\,K\,cal$

Given two processes:
$\frac{1}{2} P_{4(s)} + 3 Cl_{2(g)} \to 2 PCl_{3(l)} \;; \Delta H = -635 \ kJ$
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)} \;; \Delta H = -137 \ kJ$
The value of the heat of formation of $PCl_{5(s)}$ is ...... $kJ \ mol^{-1}$.

The $\Delta H_f^o$ for $CO_{2(g)}$,$CO_{(g)}$ and $H_2O_{(g)}$ are $-393.5$,$-110.5$ and $-241.8 \ kJ \ mol^{-1}$ respectively. The standard enthalpy change (in $kJ$) for the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$ is

The bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $240 \ kJ \ mol^{-1}$ respectively. If the $\Delta H_f$ (enthalpy of formation) of $HCl$ is $-90 \ kJ \ mol^{-1}$,what is the $H-Cl$ bond energy in $kJ \ mol^{-1}$?