Question

The enthalpy of vaporization of a liquid is \(30\,\text{kJ mol}^{-1}\) and entropy of vaporization is \(75\,\text{J K}^{-1}\text{ mol}^{-1}\). The boiling point of the liquid at \(1\,\text{atm}\) is

• A. \(250\,\text{K}\)
• B. \(400\,\text{K}\)
• C. \(450\,\text{K}\)
• D. \(600\,\text{K}\)

Hint:

At boiling point: \[ \Delta G = 0 \Rightarrow T_b = \frac{\Delta H_{\text{vap}}}{\Delta S_{\text{vap}}} \] Always convert all quantities into consistent units before substitution.

Answer 1

The Correct Option is B

Step 1: At boiling point, the phase equilibrium condition is: \[ \Delta G = 0 \]
Step 2: Using the relation: \[ \Delta G = \Delta H - T\Delta S \] At equilibrium, \[ \Delta H = T\Delta S \]
Step 3: Convert enthalpy to joules: \[ \Delta H = 30\,\text{kJ mol}^{-1} = 30000\,\text{J mol}^{-1} \]
Step 4: Substitute the given values: \[ T = \frac{\Delta H}{\Delta S} = \frac{30000}{75} = 400\,\text{K} \]
Step 5: Hence, the boiling point of the liquid is: \[ \boxed{400\,\text{K}} \]