A certain organic solvent has Δ H\_vap = 28.50\ kJ mol^ -1 and Δ S\_vap = 88.2\ J K^ -1 \ mol^ -1 at 1 bar. Calculate its normal boiling point (where Δ G = 0 ).

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 323.1\ K

At the normal boiling point, the liquid and vapour phases are in equilibrium, which gives

\Delta G = 0

.

Using the Gibbs free energy equation:

\Delta G = \Delta H_{vap} - T_{b} \Delta S_{vap} = 0

T_{b} = \dfrac{\Delta H_{vap}}{\Delta S_{vap}}

Substituting the given values:

T_{b} = \dfrac{28.50 \times 1000}{88.2}

T_{b} = \dfrac{28500}{88.2} = 323.1\ \text{K}

Answer:

323.1\ \text{K}

A certain organic solvent has Δ H\_vap = 28.50\ kJ mol^ -1 and Δ...

Options

Correct answer

Step-by-step solution

More from Thermodynamics (C)