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Chemistry Question on Enthalpy change

If the enthalpy of vaporization of water is 186.5 kJ mol−1186.5 \,kJ \, mol^{-1}, the entropy if its vaporization will be :

A

0.5 k JK−1 mol−10.5\,k \, JK^{-1} \, mol^{-1}

B

1.0 k JK−1 mol−11.0 \,k \, JK^{-1} \, mol^{-1}

C

1.5k JK−1 mol−11.5 k \, JK^{-1} \, mol^{-1}

D

2.0 k JK−1 mol−12.0\,k \, JK^{-1} \, mol^{-1}

Answer

0.5 k JK−1 mol−10.5\,k \, JK^{-1} \, mol^{-1}

Explanation

Solution

Given enthalpy of vaporization,
ΔH=186⋅5 kJ mol−1\Delta H =186 \cdot 5\, kJ\, mol ^{-1}
Boiling point of water
=100∘C=100+273=373 K=100^{\circ} C =100+273=373\, K
Entropy change,
ΔS=ΔHT=186⋅5 kJ mol−1373 k\Delta S =\frac{\Delta H }{ T }=\frac{186 \cdot 5\, kJ\, mol ^{-1}}{373\, k }
=0⋅5 kJ mol−1K−1=0 \cdot 5\, kJ\, mol ^{-1} K ^{-1}