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Thermodynamics — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics2 Marks
Question
The equilibrium constant for a reaction is 10. What will be the value of $\Delta\text{G}^\ominus$? $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~T}=300 \mathrm{~K}$.
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Answer

$\Delta\text{G}^\ominus=-\text{RT ln k}=-2.303\text{RT}\log\text{K}.$$\text{R}=8.314\text{Jk}^{-1} \ \text{mol}^{-1};\text{T}=300\text{K};\text{K}=10$
$\Delta\text{G}^\ominus=-2.303\times8.314\text{Jk}^{-1} \ \text{mol}^{-1}\times(300\text{k})\times\log10 $
$=-5527\text{J} \ \text{mol}^{-1}=-5.527\text{kJ} \ \text{mol}^{-1}.$

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