Question

The absolute enthalpy of neutralization of the reaction will be
$Mg{O_{(s)}} + 2HC{l_{(aq)}} \to MgC{l_{2(aq)}} + {H_2}{O_{(l)}}$
A.${\text{ + 57}}{\text{.33kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
B.$- 57.33$ ${\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
C.Greater than ${\text{ - 57}}{\text{.33kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
D.Less than ${\text{ - 57}}{\text{.33kJmo}}{{\text{l}}^{{\text{ - 1}}}}$

Answer 1

We first need to know that the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to generate water and a salt is known as the enthalpy of neutralization. It is a specific case of reaction enthalpy. It is defined as the amount of energy released when one mole of water is formed. Under standard conditions, it is called the standard enthalpy of neutralization.

Complete answer:
We have to remember that when a strong acid reacts with a strong base, they undergo a neutralization reaction to form salt into one mole of water. The enthalpy of neutralization is ${\text{ - 57}}{\text{.33kJmo}}{{\text{l}}^{{\text{ - 1}}}}$. And, we know that $Mg{O_{(s)}}$ is a weak base and $HC{l_{(aq)}}$is a strong acid. But, $Mg{O_{(s)}}$ being a weak base, requires some heat in neutralisation and thus consumes a part of the heat. This results in a decrease in the net amount of heat that is released. In other words, the absolute enthalpy of neutralisation of this reaction, that is, reaction between a weak base and a strong acid will be lower than that of the enthalpy of neutralisation of a strong acid and a strong base.
This leads us to the conclusion that the absolute enthalpy of neutralisation of the given reaction will be less than - 57.33$$$${\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}.

Thus, the correct option is D.

Note:
We must be noted that when a strong acid, such as$^{}{\text{HA}}$, reacts with a strong base, such as${\text{BOH}}$, the acid and base are entirely dissociated, and neither the cation ${{\text{B}}^ + }$ nor the anion ${{\text{A}}^ - }$ are involved in the neutralisation process. At ${25^ \circ }C$, the enthalpy change of this reaction is approximately ${\text{ - 57}}{\text{.33kJmo}}{{\text{l}}^{{\text{ - 1}}}}$. The enthalpy of neutralisation for weak acids or bases is $pH$dependent.