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# Heat of neutralisation at$\text{ HCl }$ against $\text{NaOH}$ is 13.7 $\text{Kcal e}{{\text{q}}^{-1}}$. What will be the ionisation energy of $\text{C}{{\text{H}}_{3}}\text{COOH}$ in $\text{Kcal e}{{\text{q}}^{-1}}$ if its heat of neutralisation is $\text{11}\text{.7 Kcal e}{{\text{q}}^{-1}}$ ?A. 4B. 2C. 6D. 8

Last updated date: 11th Aug 2024
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Hint: It is the amount of heat that will release when 1 equivalent of acid reacts with 1 equivalent of a base to yield salt and water. The removal of an electron is dependent on the ionisation energy, so the heat of neutralisation for a weak acid is less than the strong acid so the value of ionisation energy will be more for the weak acid because it cannot completely dissociate. So, if we subtract the heat of neutralisation of the weak acid from the strong acid then we will get the ionisation energy.

Complete step by step answer:
- It is given in the question that the heat of neutralisation of strong acid and strong base is 13.7 $\text{Kcal e}{{\text{q}}^{-1}}$ i.e. of $\text{HCl}$and $\text{NaOH}$ .
- But the heat of neutralisation of weak base and strong base is $\text{11}\text{.7 Kcal e}{{\text{q}}^{-1}}$ i.e. for $\text{C}{{\text{H}}_{3}}\text{COOH}$ and $\text{NaOH}$.
- When a strong base reacts with a weak base then some extra amount of energy is consumed for the ionisation of acetic acid i.e. weak base.
- This is because acetic acid is a weak electrolyte and it cannot dissociate completely in the solution, that’s why an extra amount of energy is required.
So, to calculate the extra energy we will subtract heat of neutralisation of strong base and strong acid with the heat of neutralisation of strong base and weak base i.e.
$\text{13}\text{.7 - 11}\text{.7 = 2 Kcal e}{{\text{q}}^{-1}}$.
Therefore, option B is the correct answer.

Note: Heat of neutralisation can also be carried out in the standard condition also such a temperature and pressure and known as standard heat of neutralisation. The value of heat of neutralisation will be negative if the reaction is exothermic.