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 phase transition is at equilibrium, so the change in Gibbs free energy is zero ($\Delta G = 0$).
Key Formula or Approach:
Since $\Delta G = \Delta H - T\Delta S = 0$, the entropy change is $\Delta S = \frac{\Delta H_{vap}}{T_b}$.

Step 2: Detailed Explanation:
Given:
$\Delta H_{vap} = 26.0 \text{ kJ mol}^{-1} = 26,000 \text{ J mol}^{-1}$.
$T_b = 35^\circ C = 35 + 273.15 = 308.15 \text{ K} \approx 308 \text{ K}$.
Calculate entropy change:
\[ \Delta S = \frac{26000}{308} \approx 84.41 \text{ J K}^{-1} \text{ mol}^{-1} \]
Since vaporization involves turning a liquid into a gas, the disorder increases, and $\Delta S$ must be positive.

Step 3: Final Answer:
The entropy change $\Delta S^0$ is $+84.4\text{ JK}^{-1}\text{mol}^{-1}$.