Enthalpy Change

Step 1: Define Enthalpy of Neutralization.
The enthalpy of neutralization is the heat change when one mole of water is formed from the reaction of an acid and a base. For the neutralization of a strong acid with a strong base (e.g., HCl + NaOH), the reaction is essentially H⁺(aq) + OH^-(aq) → H₂O(l). The enthalpy change for this reaction is approximately -57.1 kJ/mol, which is a constant value. The negative sign indicates an exothermic reaction.
Step 2: Analyze the given reaction: CH₃COOH + NaOH.
Here, CH₃COOH (acetic acid) is a weak acid, and NaOH (sodium hydroxide) is a strong base.
When a weak acid reacts with a strong base, the neutralization process involves two steps:
1. Ionization of the weak acid: The weak acid first needs to ionize. This process is endothermic, as energy is required to break the bond and release the H⁺ ion.
CH₃COOH(aq) ↔ CH₃COO^-(aq) + H⁺(aq)
2. Neutralization of H⁺ with OH^-: The H⁺ ions released from the weak acid then react with OH^- ions from the strong base to form water. This step is highly exothermic, approximately -57.1 kJ/mol.
H⁺(aq) + OH^-(aq) → H₂O(l) (ΔH ≈ -57.1 kJ/mol)
Step 3: Determine the overall enthalpy change.
The overall enthalpy of neutralization for a weak acid and a strong base is the sum of the enthalpy change for the ionization of the weak acid and the enthalpy change for the formation of water
Since the ionization of a weak acid is an endothermic process (requires energy, positive ΔH), some of the heat released during the formation of water is consumed for the ionization of the weak acid.
Therefore, the net heat released during the neutralization of a weak acid with a strong base will be less than 57.1 kJ/mol.
Let ΔH_ionization be the enthalpy of ionization of the weak acid (positive value).
ΔH_{neutralization} = ΔH_ionization + ΔH_{water formation}
ΔH_{neutralization} = ΔH_ionization + (-57.1 kJ/mol)
Since ΔH_ionization > 0, it follows that ΔH_{neutralization} > -57.1 kJ/mol (in terms of magnitude, it means less heat is released). When we talk about the value of enthalpy of neutralization, it is usually reported as a positive magnitude. So, the magnitude of heat released will be less than 57.1 kJ/mol.
Step 4: Evaluate the options.

• Option (1): Less than 57.1 kJ/mol: This is consistent with our analysis, as some energy is absorbed for the ionization of the weak acid.
• Option (2): 57.1 kJ/mol: This value is for the neutralization of a strong acid with a strong base.
• Option (3): More than 57.1 kJ/mol: This is incorrect.
• Option (4): Zero: This would mean no heat change, which is incorrect as neutralization reactions are typically exothermic.

Step 5: Conclusion.
The enthalpy of neutralization for CH₃COOH + NaOH will be less than 57.1 kJ/mol due to the endothermic ionization of the weak acetic acid.