Question

A certain organic solvent has \(\Delta H_{\text{vap}}=28.50\ \text{kJ mol}^{-1}\) and \(\Delta S_{\text{vap}}=88.2\ \text{J K}^{-1}\ \text{mol}^{-1}\) at 1 bar. Calculate its normal boiling point (where \(\Delta G=0\)).

• A. \(298.5\ \text{K}\)
• B. \(323.1\ \text{K}\)
• C. \(355.0\ \text{K}\)
• D. \(372.8\ \text{K}\)

Hint:

At the boiling point, delta G is zero, so T equals delta H divided by delta S, with units made consistent.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
At the normal boiling point liquid and vapour are in equilibrium, so \(\Delta G = 0\). Then \(\Delta H - T\Delta S = 0\), which gives \(T_{b} = \dfrac{\Delta H_{vap}}{\Delta S_{vap}}\).

Step 2: Convert units
\(\Delta H_{vap} = 28.50\ \text{kJ mol}^{-1} = 28500\ \text{J mol}^{-1}\).

Step 3: Calculation
\(T_{b} = \dfrac{28500}{88.2} = 323.1\ \text{K}\).

Final Answer:
The normal boiling point is 323.1 K. \[ \boxed{\text{(B) }323.1\ \text{K}} \]