At 00 C- ice and water are in equilibrium and - H 0-6-00 kJ - mol for the process H 2 O s - H 2 O l - The value of - S 0 for the conversion ofice to liquid water is-A- 10-15 JK -1 mol -1B- 17-25 JK -1 mol -1C- 21-98 JK -1 mol -1D- 30-50 JK -1 mol -1

The correct option is C 21.98 JK^ 1mol 1Given, Δ H^0 for the equlibrium nbsp; H_2O(s)⇋ H_2O(l) is 6.00 kJ/mol. nbsp;Δ S^0=Δ H^0/T ⇒Δ S^0 = 6.00×10^3 Jmol^ ...

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Byju's Answer

Standard XII

Physics

Specific Heat Capacity

At 00 C, ice ...

Question

At $0^{0} C$ , ice and water are in equilibrium and $Δ H^{0} = 6.00 k J / m o l$ for the process $H_{2} O (s) ⇋ H_{2} O (l)$ . The value of $Δ S^{0}$ for the conversion of ice to liquid water is:

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

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

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

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

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Solution

The correct option is C $21.98 J K^{- 1} m o l - 1$
Given, $Δ H^{0}$ for the equlibrium $H_{2} O (s) ⇋ H_{2} O (l)$ is $6.00 k J / m o l$ .
$Δ S^{0} = \frac{Δ H^{0}}{T}$
$\Rightarrow Δ S^{0} = \frac{6.00 \times 10^{3} J m o l^{- 1}}{273 K}$
$\Rightarrow Δ S^{0} = 21.978 J K^{- 1} m o l^{- 1}$

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At 00C, ice and water are in equilibrium and ΔH0=6.00 kJ/mol for the process H2O(s)⇋H2O(l). The value of ΔS0 for the conversion of ice to liquid water is:

The correct option is C 21.98 JK−1mol−1Given, ΔH0 for the equlibrium H2O(s)⇋H2O(l) is 6.00 kJ/mol. ΔS0=ΔH0T ⇒ΔS0=6.00×103 Jmol−1273 K ⇒ΔS0=21.978 JK−1mol−1

At $0^{0} C$ , ice and water are in equilibrium and $Δ H^{0} = 6.00 k J / m o l$ for the process $H_{2} O (s) ⇋ H_{2} O (l)$ . The value of $Δ S^{0}$ for the conversion of ice to liquid water is: