The enthalpy of vaporisation of a liquid is `30 kJ mol^(-1)` and entropy of vaporisation is `75 J mol^(-1) K^(-1)`. The boiling point of the liquid at `1atm` is :

Options

A. 250 K
B. 400 K
C. 450 K
D. 600 K

Answer

To find the boiling point of the liquid at 1 atm, we can use the relationship between enthalpy of vaporization (ΔH) and entropy of vaporization (ΔS). The formula we will use is:

\[ \Delta H = T \Delta S \]

Where:
- ΔH is the enthalpy of vaporization,
- T is the temperature (boiling point in Kelvin),
- ΔS is the entropy of vaporization.

### Step-by-Step Solution:

1. **Convert ΔH from kJ/mol to J/mol**:
Given that the enthalpy of vaporization (ΔH) is 30 kJ/mol, we need to convert this to Joules.
\[ \Delta H = 30 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 30000 \, \text{J/mol} \]

**Hint**: Remember to convert kJ to J by multiplying by 1000.

2. **Use the given value of ΔS**:
The entropy of vaporization (ΔS) is given as 75 J/mol·K.

3. **Rearrange the equation to solve for T**:
We can rearrange the equation ΔH = TΔS to find T:
\[ T = \frac{\Delta H}{\Delta S} \]

4. **Substitute the values into the equation**:
Now substitute the values of ΔH and ΔS into the equation:
\[ T = \frac{30000 \, \text{J/mol}}{75 \, \text{J/mol·K}} \]

5. **Calculate T**:
Performing the division:
\[ T = \frac{30000}{75} = 400 \, \text{K} \]

6. **Conclusion**:
The boiling point of the liquid at 1 atm is 400 K.

### Final Answer:
The boiling point of the liquid at 1 atm is **400 K**.

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