For the following reaction,
$C (diamond) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -97.6 \ kcal$
$C (graphite) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -94.3 \ kcal$
The heat change for the conversion of $1 \ g$ of $C (diamond) \rightarrow C (graphite)$ is (in $kcal$)

The heat of formation of $CO_{(g)}$ and $CO_{2(g)}$ are $-26.4 \ kcal$ and $-94.0 \ kcal$ respectively. The heat of combustion of carbon monoxide will be $... \ kcal$.

The heat of neutralization for the reaction $NaOH + HCl \to NaCl + H_2O$ is $57.1 \ kJ \ mol^{-1}$. What will be the heat released when $0.25 \ mol$ of $NaOH$ is titrated against $0.25 \ mol$ of $HCl$ (in $kJ$)?

Which substance has a standard molar enthalpy of formation equal to zero at $298 \ K$?

The formation enthalpies,$\Delta H_{f}^{\ominus}$ for $H_{(g)}$ and $O_{(g)}$ are $220.0$ and $250.0 \ kJ \ mol^{-1}$,respectively,at $298.15 \ K$,and $\Delta H_{f}^{\ominus}$ for $H_2O_{(g)}$ is $-242.0 \ kJ \ mol^{-1}$ at the same temperature. The average bond enthalpy of the $O-H$ bond in water at $298.15 \ K$ is $.......... \ kJ \ mol^{-1}$ (nearest integer).

The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus the amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction: $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$? Given that the bond energy of $H_2$,$Br_2$,and $HBr$ is $435 \ kJ \ mol^{-1}$,$192 \ kJ \ mol^{-1}$,and $368 \ kJ \ mol^{-1}$ respectively.