The bond dissociation energies of gaseous $H_2$,$Cl_2$ and $HCl$ are $104$,$58$ and $103 \ kcal$ respectively. The enthalpy of formation of $HCl$ gas would be $... \ kcal$.

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

For strong acid and strong base neutralisation,the net chemical change is $H^{+} + OH^{-} \longrightarrow H_2O_{(l)}$; $\Delta_r H^{\circ} = -55.84 \ kJ \ mol^{-1}$. If the enthalpy of neutralisation of $CH_3COOH$ by $NaOH$ is $-49.86 \ kJ \ mol^{-1}$,then the enthalpy of ionisation of $CH_3COOH$ is:

Bond enthalpies of $A_{2}$,$B_{2}$ and $AB$ are in the ratio $2:1:2$. If the enthalpy of formation of $AB$ is $-100 \ kJ \ mol^{-1}$,the bond enthalpy of $B_{2}$ is:

The heats of combustion of carbon and carbon monoxide are $-394 \, kJ \, mol^{-1}$ and $-285 \, kJ \, mol^{-1}$ respectively. Find the heat of formation of $CO$ in $kJ \, mol^{-1}$.

For the allotropic change represented by the equation $C(\text{diamond}) \to C(\text{graphite})$,the enthalpy change is $\Delta H = -1.89 \ kJ$. If $6 \ g$ of diamond and $6 \ g$ of graphite are separately burnt to yield carbon dioxide,the heat liberated in the first case is: