If the enthalpy of neutralization of $HCN$ and $NaOH$ is $-12.13 \, kJ/mol$,then the enthalpy of ionization of $HCN$ will be ...... $kJ/mol$.

The total enthalpy change in a chemical reaction is equal to the algebraic sum of the enthalpy changes of the individual steps of the reaction. This statement is associated with which scientist?

The bond energies of $H-H$ and $Cl-Cl$ are $430 \, kJ \, mol^{-1}$ and $240 \, kJ \, mol^{-1}$ respectively. If the value of ${\Delta _f}H$ for $HCl$ is $-90 \, kJ \, mol^{-1}$,then the $H-Cl$ bond energy is ..... $kJ \, mol^{-1}$.

Enthalpy of formation of $CO_{(g)}$ and $CO_{2(g)}$ are $-110 \ kJ \ mol^{-1}$ and $-393 \ kJ \ mol^{-1}$ respectively. The enthalpy of combustion of $CO$ (in $kJ \ mol^{-1}$) is:

Standard molar enthalpy of formation,$\Delta _{f}H^{o}$ is just a special case of enthalpy of reaction,$\Delta _{r}H^{o}$. Is the $\Delta _{r}H^{o}$ for the following reaction same as $\Delta _{f}H^{o}$? Give reason for your answer. $CaO_{(s)} + CO_{2_{(g)}} \to CaCO_{3_{(s)}}$; $\Delta _{r}H^{o} = -178.3 \ kJ \ mol^{-1}$

Given the thermochemical reactions: $C(\text{graphite}) + \frac{1}{2}O_2 \rightarrow CO; \Delta H = -110.5 \, kJ$ and $CO + \frac{1}{2}O_2 \rightarrow CO_2; \Delta H = -283.2 \, kJ$. Calculate the heat of reaction for $C(\text{graphite}) + O_2 \rightarrow CO_2$ in $kJ$.