Question

If the enthalpy change for the transition of liquid water of steam is $30 \,kJ\, mol^{-1}$ at $27^{\circ}C$, the entropy change for the process would be :

• A. $0.1 \,J\, mol^{-1} K^{-1}$
• B. $100 \,J\, mol^{-1} K^{-1}$
• C. $10 \,J\, mol^{-1} K^{-1}$
• D. $1.0\, J\, mol^{-1} K^{-1}$

Answer 1

Answer: (B)

$\Delta S _{ vap }=\frac{\Delta H _{ vap }}{ T }=\frac{30\, KJ\,mol ^{-1}}{300\, K } $

$=100 \,J \,mol ^{-1}\, K ^{-1}$