1. Tardigrade
  2. Question
  3. Chemistry
  4. The entropy (So) of the following substances are: CH4 (g) 186.2 J K-1 mol-1 O2 (g) 205.0 J K-1 mol-1 CO2 (g) 213.6 J K-1 mol-1 H2O (g) 69.9 J K-1 mol-1 The entropy change (Δ So) for the reaction CH4(g) + 2O2(g) arrow CO2(g) + 2H2O(l) is :

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Q. The entropy (S$^{o}$) of the following substances are :
$CH_{4}\,\left(g\right) 186.2\,J\,K^{-1}\,mol^{-1}$
$O_{2}\,\left(g\right) 205.0\,J\,K^{-1}\,mol^{-1}$
$CO_{2}\,\left(g\right) 213.6\,J\,K^{-1}\,mol^{-1}$
$H_{2}O\,\left(g\right) 69.9\,J\,K^{-1}\,mol^{-1}$
The entropy change $\left(\Delta S^{o}\right)$ for the reaction
$CH_{4}\left(g\right) + 2O_{2}\left(g\right) \rightarrow CO_{2}\left(g\right) + 2H_{2}O\left(l\right)$ is :

JEE MainJEE Main 2014Thermodynamics

Solution:

$\underset{186.2}{CH_{4(g)}} + \underset{205}{2O_{2(g)}} \rightarrow \underset{213.6}{CO_{2(g)}} + \underset{69.9}{2H_2O(\ell)}$

$\Delta S^{\circ} = \sum(s^{\circ}_P - \sum(s^{\circ})_R$

$ =[213.6 + (2\times 69.9)] - [186.2 + 2 \times 205]$

$ = 353.4 - 596.2$

$= -242.8\,J/mol\,K$