Question

Equal volumes of equimolar ${{HCl}}$ and ${{{H}}_2}{{S}}{{{O}}_4}$ are separately neutralised by dilute ${{NaOH}}$ solution, then heat liberated are ${{x}}$ ${{K}}{{.Cal}}$ and ${{y}}$ ${{K}}{{.Cal}}$ respectively. Which of the following is true?
A.${{x = y}}$
B.${{x = }}\dfrac{y}{2}$
C.${{x = 2y}}$
D.${{x = }}\dfrac{y}{3}$

Answer 1

Hint:When a chemical reaction occurs or a chemical process occurs, the heat involved during the reaction or process is called the change in enthalpy of the reaction. Here, the reactions of two different acids, hydrochloric acid and sulphuric acid with a dilute base is the chemical process that is happening.

Complete answer:
When an acid reacts with a base, it is called a neutralisation reaction. The enthalpy of neutralisation or neutralisation energy or heat of neutralisation is the change in enthalpy that occurs when one equivalent of an acid reacts with one equivalent of a base to form salt and water as the neutralisation products. Heat of neutralisation or enthalpy of neutralisation is represented as $\Delta {{H}}$ .
We can consider the given reaction to find out the heat of neutralisation or enthalpy of neutralisation.
Let us consider the first reaction as a reaction of hydrochloric acid with sodium hydroxide base.
Heat of neutralisation $ = $ Heat of formation of water
${{HCl + NaOH }} \to {{ NaCl + }}{{{H}}_2}{{O}}$ $\Delta {{H}}$ $ = $ $ - {{x}}$ ${{K}}{{.Cal}}$
${{{H}}^ + }{{ + O}}{{{H}}^ - }{{ }} \to {{ }}{{{H}}_2}{{O}}$ $\Delta {{H}}$ $ = $ $ - {{x}}$ ${{K}}{{.Cal}}$
The negative sign before ${{x}}$ indicates that so much amount of energy is released when the reaction occurs.

Let us consider the first reaction as a reaction of hydrochloric acid with sodium hydroxide base.
Heat of neutralisation $ = $ Heat of formation of water
${{{H}}_2}{{S}}{{{O}}_4}{{ + 2NaOH }} \to {{ N}}{{{a}}_2}{{S}}{{{O}}_4}{{ + 2}}{{{H}}_2}{{O}}$ $\Delta {{H}}$ $ = $ $ - {{y}}$ ${{K}}{{.Cal}}$
${{2}}{{{H}}^ + }{{ + 2O}}{{{H}}^ - }{{ }} \to {{ 2}}{{{H}}_2}{{O}}$ $\Delta {{H}}$ $ = $ $ - {{y}}$ ${{K}}{{.Cal}}$
This reaction indicates that the reaction of two moles of hydrogen ions with two moles of hydroxyl ions gives two moles of water. The heat of this neutralisation or heat liberated during this reaction is ${{y}}$ ${{K}}{{.Cal}}$ . So, when one mole of hydrogen ions react with one moles of hydroxyl ions gives one mole of water, the enthalpy of neutralisation will be reduced by half.
${{{H}}^ + }{{ + O}}{{{H}}^ - }{{ }} \to {{ }}{{{H}}_2}{{O}}$ $\Delta {{H}}$ $ = $ $ - \dfrac{{{y}}}{2}$ ${{K}}{{.Cal}}$
Since we are considering the reaction of equal volumes of equimolar concentration of ${{HCl}}$ and ${{{H}}_2}{{S}}{{{O}}_4}$ reacting with dilute ${{NaOH}}$ solution, their heat of neutralisation or enthalpy of neutralisation will be the same or they will be equal.
Therefore, we can equate the two heats of neutralisation.
$ - {{x}}$ $ = $ $ - \dfrac{{{y}}}{2}$
We can cancel out the negative signs on both the sides.
${{x}}$ $ = $ $\dfrac{{{y}}}{2}$

Hence , option (B) is the correct answer.

Note:
A reaction which occurs at a standard temperature of $298{{ K}}$ , a pressure of $1{{ atm}}$ leading to the formation of $1{{ mole}}$ of water is called standard enthalpy of neutralisation. It is represented as $\Delta {{{H}}^ \circ }$ . Enthalpy of a reaction will be positive if heat is absorbed and it will be negative if heat is liberated during the reaction.