Question

If the enthalpy of vaporization of water is \(18.6\,\text{kJ mol}^{-1}\), the entropy of its vaporization will be:

• A. \(0.5\,\text{kJ K}^{-1}\text{mol}^{-1}\)
• B. \(1.0\,\text{kJ K}^{-1}\text{mol}^{-1}\)
• C. \(1.5\,\text{kJ K}^{-1}\text{mol}^{-1}\)
• D. \(2.0\,\text{kJ K}^{-1}\text{mol}^{-1}\)

Hint:

Entropy change = enthalpy change divided by absolute temperature.

Answer 1

The Correct Option is A

Step 1: \[ \Delta S = \frac{\Delta H}{T} \]
Step 2: At boiling point \(T=373\,K\): \[ \Delta S=\frac{18.6}{373}\approx0.05\,\text{kJ K}^{-1} \]