The heat of neutralization of $HCl$ and $NaOH$ is:

What is bond enthalpy? Explain bond enthalpy in polyatomic molecules like $H_2O$.

The values of bond enthalpy of $H_2$,$Cl_2$ and $HCl$ are $104$,$58$ and $103 \ kcal/mol$ respectively. Find the enthalpy of formation of $HCl_{(g)}$. Reaction: $\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{2(g)} \to HCl_{(g)}$

$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
If graphite is converted into diamond,then the $\Delta H$ for the process is . . . . . . $\text{kJ}$.

Calculate the standard enthalpy change for the reaction,$C_2H_5OH_{(\ell)} + 3O_{2_{(g)}} \rightarrow 2CO_{2_{(g)}} + 3H_2O_{(\ell)}$. Given: $\Delta_{f}H^{\circ}(C_2H_5OH) = -280 \ kJ \ mol^{-1}$,$\Delta_{f}H^{\circ}(CO_2) = -390 \ kJ \ mol^{-1}$,and $\Delta_{f}H^{\circ}(H_2O) = -285 \ kJ \ mol^{-1}$.

The value of the enthalpy of neutralization of an acid and a base is significant only when $......$