Question

The enthalpy of vapourisation of diethyl ether is $26.0\text{ kJ mol}^{-1}$ and its normal boiling point is $35^\circ C$. What is the value of $\Delta S^0$ for the conversion of liquid diethyl ether to vapour at $35^\circ C$?

• A. -84.4 $\text{JK}^{-1}\text{mol}^{-1}$
• B. +742.9 $\text{JK}^{-1}\text{mol}^{-1}$
• C. -8.44 $\text{JK}^{-1}\text{mol}^{-1}$
• D. +84.4 $\text{JK}^{-1}\text{mol}^{-1}$

Hint:

Always convert kJ to J before calculating entropy, as the final unit is usually in J/K. Also, ensure temperature is in Kelvin.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
At the boiling point, the liquid and vapour phases are in equilibrium. This means the Gibbs free energy change (\( \Delta G \)) for the process is zero.
Step 2: Key Formula or Approach:
For phase transition at equilibrium:
\( \Delta G = \Delta H - T\Delta S = 0 \implies \Delta S = \frac{\Delta H}{T} \).
Temperature (T) must be in Kelvin.
Step 3: Detailed Explanation:
Given:
\( \Delta H_{\text{vap}} = 26.0 \text{ kJ mol}^{-1} = 26000 \text{ J mol}^{-1} \).
\( T = 35 + 273 = 308 \text{ K} \).
Calculate entropy change (\( \Delta S \)):
\[ \Delta S = \frac{26000}{308} \] \[ \Delta S \approx 84.41 \text{ J K}^{-1}\text{mol}^{-1} \] Since liquid is converting to gas, the randomness increases, so \( \Delta S \) must be positive.
Step 4: Final Answer:
The value of \( \Delta S^0 \) is +84.4 JK{-1}mol{-1}.