A) \[20\,cal\,{{K}^{-1}}\]
B) \[86.6\,cal\,{{K}^{-1}}\]
C) \[166\,cal\,{{K}^{-1}}\]
D) \[100\,cal\,{{K}^{-1}}\]
Correct Answer: A
Solution :
\[\Rightarrow \] \[\Delta G=\Delta H-T.\Delta S\] or \[T.\Delta S=\Delta H-\Delta G\] \[\therefore \] \[\Delta S=\frac{\Delta H-\Delta G}{T}=\frac{-10-(-16)}{300}\] \[=\frac{6}{300}kcal\,{{K}^{-1}}\] \[=\frac{6\times 1000}{300}\] \[=20\,cal\,{{K}^{-1}}\]
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