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At $27^\circ C$ the enthalpy change for the transition of liquid water to steam is 30 kJ/mol. What would be the entropy change for the process?

(a) 0.1 $Jmol^{-1}K^{-1}$
(b) 100 $Jmol^{-1}K^{-1}$
(c) 10 $Jmol^{-1}K^{-1}$
(d) 1.0 $Jmol^{-1}K^{-1}$

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Answer: 100 $Jmol^{-1}K^{-1}$
Liquid water $\rightarrow $ steam
$\Delta G^{\circ} = \Delta H^\circ - T \Delta S^\circ = 30 \times 10^3 - T \Delta S$
$\therefore \Delta S = \frac{30\times 10^3}{300}$ = 100 $Jmol^{-1}K^{-1}$
answered Feb 28, 2014 by mosymeow_1

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