At 373 K, steam and water are in equilibrium and Δ H = 40.98 kJ mol-1. What will be Δ S for conversion of water into steam? H2O(l) → H2O(g)

Question

At 373 K, steam and water are in equilibrium and Δ H = 40.98 kJ mol-1. What will be Δ S for conversion of water into steam? H2O(l) → H2O(g) (A) 109.8 J K-1 mol-1 (B) 31 J K-1 mol-1 (C) 21.98 J K-1 mol-1 (D) 326 J K-1 mol-1

Tardigrade Admin · Accepted Answer

Δ Svap = (Δ Hvap/Tb) = (40.98 × 1000/373) = 109.8 J K-1 mol-1

Q. At $373 K$ , steam and water are in equilibrium and $Δ H = 40.98 k J m o l^{- 1}$ . What will be $Δ S$ for conversion of water into steam?
$H_{2} O_{(l)} \to H_{2} O_{(g)}$

$109.8 J K^{- 1} m o l^{- 1}$

$31 J K^{- 1} m o l^{- 1}$

$21.98 J K^{- 1} m o l^{- 1}$

$326 J K^{- 1} m o l^{- 1}$

$Δ S_{v a p} = \frac{Δ H _{v a p}}{T _{b}} = \frac{40.98 \times 1000}{373}$
$= 109.8 J K^{- 1} m o l^{- 1}$

Q. At 373K, steam and water are in equilibrium and ΔH=40.98kJmol−1. What will be ΔS for conversion of water into steam? H2​O(l)​→H2​O(g)​

109.8JK−1mol−1

31JK−1mol−1

21.98JK−1mol−1

326JK−1mol−1