At 373 K- steam and water are in equilibrium and Delta H - 40-98 kJ mol-1-What will be Delta S fro conversion of water into steam-H-2O-l- rightarrow H-2O-g-109-8 J K-1-mol-1-31 J K-1-mol-1-21-98 J K-1-mol-1-326 J K-1-mol-1-

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Toppr Admin · Accepted Answer

The correct option is A 109-8 J K-x2212-1mol-x2212-1The given reaction is -H2O-l-x27F6-H2O-g-So- we have-xA0- -xA0- -xA0- -xA0-x25B3-Svap-x25B3-HvapTb-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-40-98-xD7-1000373-109-8JK-x2212-1mol-x2212-1

At 373 K, steam and water are in equilibrium and $Δ H$ = 40.98 kJ $m o l^{- 1}$ .What will be $Δ S$ fro 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 $K^{- 1} m o l^{- 1}$

The correct option is A 109.8 J $K^{- 1} m o l^{- 1}$
The given reaction is :-
$H_{2} O (l) ⟶ H_{2} O (g)$
So, we have
$△ S v a p = \frac{△ H v a p}{T b}$
$= \frac{40.98 \times 1000}{373} = 109.8 {J K}^{- 1} {m o l}^{- 1}$

At 373 K, steam and water are in equilibrium and ΔH = 40.98 kJ mol−1.What will be ΔS fro conversion of water into steam?H2O(l)→H2O(g)109.8 J K−1mol−131 J K−1mol−121.98 J K−1mol−1326 J K−1mol−1

The correct option is A 109.8 J K−1mol−1The given reaction is :-H2O(l)⟶H2O(g)So, we have △Svap=△HvapTb =40.98×1000373=109.8JK−1mol−1

At 373 K, steam and water are in equilibrium and $Δ H$ = 40.98 kJ $m o l^{- 1}$ .What will be $Δ S$ fro 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 $K^{- 1} m o l^{- 1}$

The correct option is A 109.8 J $K^{- 1} m o l^{- 1}$
The given reaction is :-
$H_{2} O (l) ⟶ H_{2} O (g)$
So, we have
$△ S v a p = \frac{△ H v a p}{T b}$
$= \frac{40.98 \times 1000}{373} = 109.8 {J K}^{- 1} {m o l}^{- 1}$