If the enthalpy change for the transition of liquid water of steam is 30 kJ mol-1 at 27°C, the entropy change for the process would be :

Question

If the enthalpy change for the transition of liquid water of steam is 30 kJ mol-1 at 27°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

Tardigrade Admin · Accepted Answer

Δ S vap =(Δ H vap / T )=(30 KJ mol -1/300 K ) =100 J mol -1 K -1

Q. If the enthalpy change for the transition of liquid water of steam is $30 k J m o l^{- 1}$ at $2 7^{\circ} C$ , the entropy change for the process would be :

$0.1 J m o l^{- 1} K^{- 1}$

$100 J m o l^{- 1} K^{- 1}$

$10 J m o l^{- 1} K^{- 1}$

$1.0 J m o l^{- 1} K^{- 1}$

$Δ S_{v a p} = \frac{Δ H _{v a p}}{T} = \frac{30 K J m o l ^{- 1}}{300 K}$

$= 100 J m o l^{- 1} K^{- 1}$

Q. If the enthalpy change for the transition of liquid water of steam is 30kJmol−1 at 27∘C, the entropy change for the process would be :

0.1Jmol−1K−1

100Jmol−1K−1

10Jmol−1K−1

1.0Jmol−1K−1