At 373 K- steam and water are in equilibrium and - H - 40-98 kJ mol-1-What will be - S fro conversion of water into steam- H2Ol- H2Og

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Liquid-Vapour Equillibrium

At 373 K, ste...

Question

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}

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31 J

K^{- 1} m o l^{- 1}

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21.98 J

K^{- 1} m o l^{- 1}

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326 J

K^{- 1} m o l^{- 1}

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△_{v a p} H = 40.98 k J m o l^{- 1} .

H_{2} O_{(l)} \to H_{2} O_{(g)}

△_{v a p} H = 40.98 k J m o l^{- 1} .

H_{2} O_{(l)} \to H_{2} O_{(g)}

X \to Y + Z

K^{- 1} m o l^{- 1}

K^{- 1} m o l^{- 1}

K^{- 1} m o l^{- 1}

X \to Y + Z

X, Y

Z

120 J K^{- 1} m o l^{- 1}, 213.8 J K^{- 1} m o l^{- 1} a n d 197.9 J K^{- 1} m o l^{- 1}

S_{H_{2 (g)}}^{O} = 130.6

K^{- 1}

m o l 6 - 1

S_{H_{2} O_{(l)}}^{O} = 69.9

K^{- 1} m o l^{- 1}

S_{O_{2 (g)}}^{O} = 205

K^{- 1}

m o l^{- 1}

H_{2_{(g)}} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}

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