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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

Calculate the entropy change for the reaction \[\] $CO(g) + \large\frac{1}{2}$ $O_2(g) \rightarrow CO_2(g);\: \: \: \: \: \Delta S^o_{CO2} = 213.6\: J/Kmol, \Delta S^o_{CO }= 197.6\: J/Kmol, \Delta S^o_{O2} = 205.03 \: J/Kmol$

$\begin {array} {1 1} (a)\;-189.03\: J/Kmol & \quad (b)\;86.5\: J/Kmol \\ (c)\; -86.5\: J/Kmol & \quad (d)\;189.03\: J/Kmol \end {array}$

 

1 Answer

$ \Delta S^o = \sum S^o$(products) $- \sum S^o$(reactants)
$= S^o_{CO2} – \bigg(S^o_{CO} + \large\frac{1}{2}$$ S^o_{O2} \bigg)$
$= 213.6 –\bigg (197.6 +\large\frac{1}{2}$$ \times 205.03 \bigg) = 213.6 – 300.1 = -86.5\: J/Kmol$
Ans : (c)
answered Mar 13, 2014 by thanvigandhi_1
 

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