The heat of neutralization of a strong acid with a strong base is constant and is equal to:

$H_2 + \frac{1}{2} O_2 \to H_2O; \Delta H = -68.39 \ kcal$
$K + H_2O + \text{water} \to KOH_{(aq)} + \frac{1}{2} H_2; \Delta H = -48 \ kcal$
$KOH + \text{water} \to KOH_{(aq)}; \Delta H = -14 \ kcal$
The heat of formation of $KOH$ is (in $kcal$):

$Fe_2O_{3(s)} + \frac{3}{2} C_{(s)} \to \frac{3}{2} CO_{2(g)} + 2Fe_{(s)}$
$\Delta H^o = +234.1 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
$\Delta H^o = -393.5 \ kJ$
Use these equations and $\Delta H^o$ values to calculate $\Delta H^o$ for this reaction:
$4Fe_{(s)} + 3O_{2(g)} \to 2Fe_2O_{3(s)}$
..... $kJ$

From the given diagram, the enthalpy change $\Delta_r H$ for the reaction $C \rightarrow A$ is:

Which is the best definition of “heat of neutralization”?

The heat of neutralization of a strong acid by a strong base is nearly equal to