First slide

2nd Law of Thermodynaimcs

Question

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

Easy

Solution

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

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App