Question

The heat of neutralisation of ${\text{HCl}}$ and ${\text{KOH}}$ is $13.7{\text{ kCal}}$ . When 49 g of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ neutralizes 37 g of ${\text{Ca}}{\left( {{\text{OH}}} \right)_2}$ in dilute solution, the amount of heat evolved will be:
A.${\text{27}}{\text{.4 kCal}}$
B.$13.7{\text{ kCal}}$
C.${\text{6}}{\text{.85 kCal}}$
D.${\text{20}}{\text{.55 kCal}}$

Answer 1

The heat of neutralisation will remain the same for the same number of equivalents for both strong and weak electrolyte. Calculate the number of moles of sulphuric acid and calcium hydroxide and then calculate the amount of heat evolved.

Complete step by step answer:
${\text{HCl}}$ , that is, hydrochloric acid and ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ , that is, sulphuric acid are strong acids.
${\text{KOH}}$ , that is, potassium hydroxide and ${\text{Ca}}{\left( {{\text{OH}}} \right)_2}$ , that is, calcium hydroxide are strong bases.
The reaction between an acid and a base to form salt and water is known as neutralisation.
Heat of neutralization or enthalpy of neutralization is always constant for strong acids and strong bases. A strong acid and strong base completely dissociates in water. All the strong acids or bases will completely dissociate in water and an equal amount of these acids and base will give equal numbers of moles of hydrogen and hydroxide ions. Hence the enthalpy of neutralization will remain the same for all strong acids and bases.
The molar mass of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is 98.
The number of moles will be $\dfrac{{49}}{{98}} = \dfrac{1}{2}$
But 1 mol of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ gives 2 mol of hydrogen ion, so the number of reacting moles or equivalent will be $\dfrac{1}{2} \times 2 = 1$
The molar mass of ${\text{Ca}}{\left( {{\text{OH}}} \right)_2}$ is 37.
The number of moles will be $\dfrac{{37}}{{74}} = \dfrac{1}{2}$
But 1 mol of ${\text{Ca}}{\left( {{\text{OH}}} \right)_2}$ gives 2 mol of hydroxide ion, so the number of reacting moles or equivalent will be $\dfrac{1}{2} \times 2 = 1$
Heat of neutralisation for 1 mole of ${\text{HCl}}$ and ${\text{KOH}}$ is $13.7{\text{ kCal}}$ so the heat of neutralisation for ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ and ${\text{Ca}}{\left( {{\text{OH}}} \right)_2}$ will also be $13.7{\text{ kCal}}$

Hence, the correct option is B.

Note:
Heat of neutralisation is the amount of energy released when 1 mol of acid and base reacts to form 1 mole of salt and water. The enthalpy of neutralisation is always negative for all acids and bases.