The effect of increase of the enthalpy of vaporisation of a liquid (in joules `mol^(-1)`) to its boiling point (in degrees K) is equal to ………….. . This rule is known as …………………..

Answer

To solve the question, we need to understand the relationship between the enthalpy of vaporization of a liquid and its boiling point, as well as the specific rule that describes this relationship.

### Step-by-Step Solution:

1. **Understanding Enthalpy of Vaporization**:
- The enthalpy of vaporization (ΔH_vap) is the amount of energy required to convert one mole of a liquid into vapor at its boiling point. It is expressed in joules per mole (J/mol).

2. **Boiling Point and Temperature**:
- The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. It is measured in degrees Kelvin (K).

3. **Relationship Between Enthalpy and Boiling Point**:
- According to the Trouton's Rule, there is a specific relationship between the enthalpy of vaporization and the boiling point of a liquid. This rule states that the increase in enthalpy of vaporization (in J/mol) to its boiling point (in K) is approximately equal to a constant value.

4. **Applying Trouton's Rule**:
- Trouton's Rule states that the value of the enthalpy of vaporization (ΔH_vap) divided by the boiling point (T_b) in Kelvin is approximately 85 to 88 J/mol/K. Thus, we can express this relationship mathematically as:
\[ \frac{\Delta H_{vap}}{T_b} \approx 85 \text{ to } 88 \text{ J/mol/K} \]

5. **Final Answer**:
- Therefore, the effect of the increase of the enthalpy of vaporization of a liquid to its boiling point is equal to approximately **88 J/mol/K**. This rule is known as **Trouton's Rule**.

### Summary of the Solution:
- The effect of increase of the enthalpy of vaporization of a liquid (in joules `mol^(-1)`) to its boiling point (in degrees K) is equal to **88 J/mol/K**. This rule is known as **Trouton's Rule**.