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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Calculate the standard entropy change for a reaction $XY$ if the value of $ \Delta H^{\circ} = 28.4 \: kJ$ and equilibrium constant is $1.8 \times 10^{-7}\: at\: 298\: K$.

$\begin {array} {1 1} (A)\;-33.8 \: J/Kmol & \quad (B)\;38.484 \: J/Kmol \\ (C)\;-38.484 \: J/Kmol & \quad (D)\;33.8 \: J/Kmol \end {array}$

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$ \Delta G^{\circ} = -2.303 \: RT\: log \: K_{eq}$
$= -2.303 \times (8.314 \: J/Kmol)(298\: K)(log\: K_{eq})$
$= -2.303 \times 8.314 \times 298 \times log(1.8 \times 10^{-7} ) = 38.484\: kJ/mol$
$ \Delta G^{\circ} - T \Delta S^{\circ} = \Delta H^{\circ}$
$ \Rightarrow \Delta S^{\circ} = \large\frac{\Delta H^{\circ}- \Delta G^{\circ} }{ T}$$ = \large\frac{(28.4 – 38.498)}{298}$$ = -33.8\: J/Kmol $
Ans : (A)
answered Mar 16, 2014 by thanvigandhi_1
 

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