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At 0°C, ice and water are in equilibrium,
$H_{2} O_{(s)} ⇌ H_{2} O_{(l)}, ΔH = 6 kJ {mol}^{- 1}$
The values of $∆$ S and $∆$ G for conversion of ice into liquid water at 0°C are

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$0 .219 {JK}^{- 1} {mol}^{- 1} and 0$

$21 .9 {JK}^{- 1} {mol}^{- 1} and 0$

$0 .0219 {JK}^{- 1} {mol}^{- 1} and 0$

$- 21 .9 {JK}^{- 1} {mol}^{- 1} and 0$

answer is C.

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Detailed Solution

$\begin{array}{l} ΔG = 0, \\ ∴ ΔH - TΔS = 0 \\ ΔS = \frac{ΔH}{T} = \frac{6000}{273} = 21 .9 {JK}^{- 1} {mol}^{- 1} \end{array}$

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