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Chemistry Question on Spontaneity

The entropy of vaporization of a liquid is 58 JK−1 mol−158 \, JK^{-1} \,mol^{-1}. If 100 g100\, g of its vapour condenses at its boiling point of 123∘C123^{\circ}C, the value of entropy change for the process is (Molar mass of the liquid = 58 g mol−1)58 \,g \,mol^{-1})

A

−100 JK−1 - 100 \, JK^{-1}

B

100 JK−1 100 \, JK^{-1}

C

−123 JK−1 - 123 \,JK^{-1}

D

123 JK−1 123 \, JK^{-1}

Answer

−100 JK−1 - 100 \, JK^{-1}

Explanation

Solution

Given entropy of vaporisation =58 JK−1 mol−1=58 \,J K ^{-1}\, mol ^{-1} i.e., entropy change when 1 mole of the liquid vapourises =58 JK−1=58 \,JK ^{-1} ∴\therefore Entropy change when 11 mole of the liguid condenes =−58JK−1  (∵=-58 JK ^{-1}\,\, (\because Condensation and vaporisation are opposite process). Moles of the liquid used in the reaction =10058=1.72 mol=\frac{100}{58}=1.72\, mol ∴\therefore Entropy change when 1.721.72 mole (or 100 g100\, g ) of the liquid condenses =−58×1.721=\frac{-58 \times 1.72}{1} =−100 JK−1=-100\, JK ^{-1}