Question

The heat of neutralisation of $HCl$ with $NaOH$ is the same as that of$HN{{O}_{3}}$ with $KOH$. If true, enter $1$ else $0$.

Answer 1

Hint: Heat of neutralisation is calculated by only taking the amount of water formed in the reaction. Reaction that forms water is common to all acids and bases but their individual degrees vary.

Complete step-by-step answer:
Let’s start at the foundation of this question and gradually make our way upwards. The concept being discussed here is the enthalpy of a reaction. Now, what is enthalpy?
Enthalpy is simply the amount of heat that is possessed by a given system at a particular period of time. It is therefore a fundamental thermodynamic quantity. Mathematically, enthalpy is equal to the sum of the internal energy and the product of pressure and volume of a given system. The equation is as follows:
\[H=U+PV\]
The change in enthalpy is defined over the initial and final states in a specified period of time. For example, in a chemical reaction, a change in enthalpy is calculated by subtracting the enthalpy of formation of reactants from the enthalpy of formation of products. This is termed as the “change in enthalpy of reaction” and is abbreviated as$\Delta H{{{}^\circ }_{reaction}}$. Mathematically, it can be equated as-
\[\Delta H{{{}^\circ }_{reaction}}=\sum \Delta H{{{}^\circ }_{\operatorname{products}}}-\sum \Delta H{{{}^\circ }_{reactants}}\]
Now, we can understand the concept of “heat of neutralisation”. It is actually the same as the change in heat or enthalpy of reaction but only in case of neutralisation reactions, that is a reaction between acids and bases. A typical acid-base reaction is as follows:
\[HA+BOH\to AB+{{H}_{2}}O\]
So, neutralisation reactions are always accompanied by the formation of water molecules. The heat or enthalpy of neutralisation is therefore defined as the amount of heat released when one mole of water is formed from the reaction between an acid and base.
As you know already, strong acids and bases dissociate completely inside water, whereas the weak ones are only partially dissociated. Given below are two reactions between the acids and bases of the former category-
\[\begin{align}
  & HCl+NaOH\to NaCl+{{H}_{2}}O \\
 & HN{{O}_{3}}+KOH\to KN{{O}_{3}}+{{H}_{2}}O \\
\end{align}\]
The reaction which is common to both of them is-
\[{{H}^{+}}+O{{H}^{-}}\to {{H}_{2}}O\]
The above reaction favours the formation of water and therefore will continue to take place until all of the protons and hydroxide ions are converted to water. As this happens in all neutralisation reactions between strong acids and bases and the heat of neutralisation is calculated according to the amount of water formed, we can conclude that- The heat of neutralisation of$HCl$with$NaOH$is same as that of$HN{{O}_{3}}$with$KOH$.

The answer is therefore “$1$”.


Note: Only in case of weak acids and bases, the heat of neutralisation differs because they tend to dissociate partially and this tendency is different from one acid or base to another. For this reason, they do not have the same heat of neutralisation.
In case of the strong ones, the presence of impurities can significantly change the heat of neutralisation.