Question

What is the enthalpy of neutralization for CH₃COOH + NaOH?

• A. Less than 57.1 kJ/mol
• B. 57.1 kJ/mol
• C. More than 57.1 kJ/mol
• D. Zero

Hint:

The enthalpy of neutralization for a strong acid and strong base is approximately -57.1 kJ/mol (exothermic). For reactions involving a weak acid or a weak base, the magnitude of the enthalpy of neutralization will be less than 57.1 kJ/mol because some energy is absorbed for the ionization of the weak electrolyte.

Answer 1

The Correct Option is A

Step 1: Define Enthalpy of Neutralization.
The enthalpy of neutralization is the heat change when one mole of water forms from an acid-base reaction. For a strong acid and strong base (e.g., HCl + NaOH), the reaction is H⁺(aq) + OH^-(aq) → H₂O(l). The enthalpy change for this is about -57.1 kJ/mol (a constant). The negative sign indicates an exothermic reaction.
Step 2: Analyze the reaction: CH₃COOH + NaOH.
Here, CH₃COOH (acetic acid) is a weak acid, and NaOH (sodium hydroxide) is a strong base.
The neutralization of a weak acid and strong base involves two steps:
1. Ionization of the weak acid: This is endothermic; energy is needed to break the bond and release H⁺.
CH₃COOH(aq) ↔ CH₃COO^-(aq) + H⁺(aq)
2. Neutralization of H⁺ with OH^-: H⁺ from the weak acid reacts with OH^- from the strong base to form water. This is highly exothermic, about -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 change is the sum of the enthalpy change for the ionization of the weak acid and the formation of water.
Since the ionization of a weak acid is endothermic (requires energy, positive ΔH), some heat from water formation is used for ionization.
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 (positive value).
ΔH_{neutralization} = ΔH_ionization + ΔH_{water formation}
ΔH_{neutralization} = ΔH_ionization + (-57.1 kJ/mol)
Since ΔH_ionization > 0, then ΔH_{neutralization} > -57.1 kJ/mol (in magnitude, less heat is released). 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: Consistent with the analysis; energy is absorbed for weak acid ionization.
• Option (2): 57.1 kJ/mol: This value is for a strong acid with a strong base.
• Option (3): More than 57.1 kJ/mol: Incorrect.
• Option (4): Zero: Incorrect; 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 acetic acid.
\[ \boxed{\text{Less than 57.1 kJ/mol}} \]