Question

What is the enthalpy of neutralization?
A) -57.3 kJ/mol
B) +57.3 kJ/mol
C) 0 kJ/mol
D) None of these

Answer 1

Hint: Neutralization is the process in which an acid and base reacts to form salt and water. The heat evolved/absorbed during this reaction is known as enthalpy of neutralization.

Complete answer:
The enthalpy of neutralization is the change in enthalpy that occurs when one equivalent of acid reacts with one equivalent of base, to undergo a neutralization reaction, to form salt and water. It is also defined as the energy released for formation of 1 mole of water. It is denoted as $\Delta {H_n}$ .
When the reaction is carried out under standard conditions of 298K temperature and 1 atm pressure, which yields one mole of water, it is known as Standard Enthalpy of Neutralization. It is denoted as $\Delta H_n^0$ .
The heat released during a reaction is given by the equation: $Q = m{C_p}dT$
Where m is the mass ${C_p}$ is the specific heat capacity and T is the temperature. The standard enthalpy is given as: $\Delta H = - \dfrac{Q}{n}$
During neutralisation Hydrogen ions (from acid) react with Hydroxide ions from the base to form water. The reaction that occurs is:
 $H_{(aq)}^ + + OH_{(aq)}^ - \to {H_2}{O_{(aq)}}$
Hence, the heat required for the formation of one mole of water is known as enthalpy of neutralisation. This enthalpy ( $\Delta {H_n}$ ) is always negative as the reaction is exothermic.
It is seen that the enthalpy of neutralisation for a reaction between strong acids and strong bases is constant at $ - 57.3kJ/mol$ . But for partially dissociating solutions like ammonia, which is always partially dissociated in solution, some of the enthalpy of neutralisation is used up to dissociate the alkali completely in water. Thus, the heat is slightly less than $ - 57.3kJ/mol$
The heat of neutralisation is the least for weak acids and weak bases reactions. This is because more enthalpy is required to dissociate both acid and alkali in water to produce hydrogen and hydroxide ions. Hence the enthalpy value are always between $ - 57 and -58 kJ/mol$
The correct answer is Option (A).

Note:
The enthalpy of neutralisation also depends on the strength of the acids and the bases. The stronger the acid/base the more it is close to the ideal value of $ - 57.3kJ/mol$ . Remember that for diprotic acids/bases the enthalpy of neutralisation will be doubled, as two moles of water would be formed in that case.