First slide

2nd Law of Thermodynaimcs

Question

CaCO₃ → CaO+CO₂, ΔH= + 180KJ. If entropies of CaCO₃, CaO & CO₂ respectively are 93, 39 and 213 J/mol/K, then ΔS_(total) [in J/K] at 300 K is :

Moderate

Solution

$\begin{array}{l} \Delta {S_{surroundings}} = - \Delta {S_{system}} = \frac{{ - \Delta {H_{sys.}}}}{T} = \frac{{ - ( + 180 \times 1000)}}{{300}} = - 600J/mole/K\\\\ \Delta {S_{sys}} = \Delta {S_{reaction}}\\\\ \Delta {S_r} = {S_{CaO}} + {S_{C{O_2}}} - {S_{CaC{O_3}}}\\\\ \,\,\,\,\,\,\,\,\, = 39 + 213 - 93 = + 159J \end{array}$
$\begin{array}{l} \Delta {S_{total}} = \Delta {S_{sys}} + \Delta {S_{surr.}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 159 - 600 = - 441J/mole/K \end{array}$

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