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One mole of ice is converted in to water at 273K. The entropies of H₂O_(s) and H₂O_(l) are 38.20 and 60.01 J mole^-1K^-1 respectively. The enthalpy change for this conversion is

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595.4 J mol^-1

59.54 J mol^-1

320.6 J mol^-1

5954 J mol^-1

answer is D.

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Detailed Solution

$begin{array}{l} {H_2}{O_{(s)}} to {H_2}{O_{(l)}}\\ Delta {S_r} = sum {S_{products}} - sum {S_{reactan ts}}\ ,,,,,,,,, = {S_{{H_2}{O_{(l)}}}} - {S_{{H_2}{O_{(s)}}}}\ ,,,,,,,, = ,60.01 - 38.20 = 21.81J/mole/K end{array}$

$Delta H=TDelta S$ = 273 X 21.81 = 5954J

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