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Q.

$S_{{H_{2(g)}}}^o =130.6 J K^{-1} mol^{-1}$
;
$\large S_{{H_2}{O_{(l)}}}^o =69.9JK^{-1} mol^{-1}$
;
$\large S_{{O_{2(g)}}}^o=205 JK^{-1}mol^{-1}$
, Then the absolute entropy change of
$\large {H_{2(g)}}\; + \;\frac{1}{2}{O_{2(g)}}\; \to \;{H_2}{O_{(l)}}$
is

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a

-303 J mol^-1k^-1

b

-163.2 J mol^-1k^-1

c

+303J mol^-1k^-1

d

+163.2 J mol^-1k^-1

answer is A.

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Detailed Solution

${H_{2(g)}}\; + \;\frac{1}{2}{O_{2(g)}}\; \to \;{H_2}{O_{(l)}}\;\Delta S=?$

$\begin{array}{l} \Delta S = \sum {S_{products}} - \sum {S_{reac\tan ts}}\\\\ \,\,\,\,\,\,\,\,\, = {S_{{H_2}{O_{(l)}}}} - [{S_{{H_2}_{(g)}}} + \frac{1}{2}{S_{{O_2}_{(g)}}}]\\\\ \,\,\,\,\,\,\,\, = 69.9 - [130.6 + \frac{{205}}{2}]\\\\ \,\,\,\,\,\,\, = \, - 163.2\,J/mole/k \end{array}$

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; ; , Then the absolute entropy change of is