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One mole of ice is converted into water at 273 K. The entropies of $H_{2} O (s)$ and $H_{2} O (l)$ are $38.20 and 60.01 J m o l^{- 1} K^{- 1}$ respectively. The enthalpy change for the conversion is :

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$59.54 J m o l^{- 1}$

$5954 J m o l^{- 1}$

$595.4 J m o l^{- 1}$

$320.6 J m o l^{- 1}$

answer is B.

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Detailed Solution

$Δ G = Δ H - T Δ S$ ; at equilibrium
$Δ G = 0, ∴ Δ H = T Δ S$
Or $Δ H = 273 \times (60.01 - 38.20) = 5954.13 J m o l^{- 1}$

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