Match the following
`{:(,"Column-I",,"Column-II"),(,("Enthalpy change in kcal"),,("Neutralisation")),((A),lt "13.7 kcal",(p),underset(("1 mol"))(HCI)+underset(("1 mol"))(NaOH)),((B),= "13.7 kcal",(q),underset(("1 mol"))(HF)+underset(("1 mol"))(NaOH)),((C),"gt 13.7 kcal",(r),underset(("1 mol"))(NH_(4)OH)+underset(("1 mol"))(HCI)),((D),"= 27.4 kcal",(s),underset(("2 mol"))(NaOH)+underset(("1 mol"))(H_(2)SO_(4))),(,,(t),underset(("1 mol"))(NaOH)+underset(("1 mol"))(CH_(3)COOH)):}`
Assume heat of neutralisation of strong acid with strong base is 13.7 kcal.

Answer

To solve the matching question regarding enthalpy changes during neutralization reactions, we will analyze each case step by step.

### Step-by-Step Solution:

1. **Understanding the Enthalpy of Neutralization**:
- The heat of neutralization for a strong acid reacting with a strong base is approximately **13.7 kcal** per mole of water formed.
- If either the acid or base is weak, the enthalpy change will be **less than 13.7 kcal**.
- If both the acid and base are strong and in a stoichiometric ratio, the enthalpy change will be **13.7 kcal**.

2. **Analyzing Each Pair**:
- **(A) Less than 13.7 kcal**:
- This will match with a reaction involving a weak acid or weak base.
- **R: NH₄OH + HCl** - Here, NH₄OH (ammonium hydroxide) is a weak base. Thus, this reaction will yield less than 13.7 kcal.
- **Match: A - R**.

- **(B) Equal to 13.7 kcal**:
- This will match with a strong acid and a strong base.
- **P: HCl + NaOH** - Both HCl and NaOH are strong, so the enthalpy change is 13.7 kcal.
- **Match: B - P**.

- **(C) Greater than 13.7 kcal**:
- This will match with a reaction involving a weak acid and a strong base.
- **Q: HF + NaOH** - HF is a weak acid, and the enthalpy of neutralization will be greater than 13.7 kcal due to additional energy from solvation.
- **Match: C - Q**.

- **(D) Equal to 27.4 kcal**:
- This will match with a reaction involving two moles of a strong base reacting with a strong acid.
- **S: 2 NaOH + H₂SO₄** - Here, 2 moles of NaOH will neutralize 1 mole of H₂SO₄, resulting in 2 times the enthalpy of neutralization (2 × 13.7 kcal = 27.4 kcal).
- **Match: D - S**.

- **(T) Less than 13.7 kcal**:
- **T: NaOH + CH₃COOH** - Acetic acid (CH₃COOH) is a weak acid, so the enthalpy change will also be less than 13.7 kcal.
- **Match: A - T**.

### Final Matches:
- A - R
- B - P
- C - Q
- D - S

### Summary of Matches:
- A (less than 13.7 kcal) matches with R (NH₄OH + HCl)
- B (equal to 13.7 kcal) matches with P (HCl + NaOH)
- C (greater than 13.7 kcal) matches with Q (HF + NaOH)
- D (equal to 27.4 kcal) matches with S (2 NaOH + H₂SO₄)