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Chemistry
A certain reaction has a ΔH of 12 kJ and a ΔS of 40 J K-1. The temperature above which the reaction becomes spontaneous is
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Q. A certain reaction has a $ΔH$ of $12\, kJ$ and a $ΔS$ of $40\, J\, K^{-1}$. The temperature above which the reaction becomes spontaneous is
KEAM
KEAM 2013
Thermodynamics
A
$27^°C$
62%
B
$27\,K$
16%
C
$300^°C$
16%
D
$30^°C$
6%
E
$30\, K$
6%
Solution:
A reaction will be spontaneous , if
$\Delta\, H-T \Delta \,S=- ve$ or $\Delta \,H-T \,\Delta \,S<\,0$
or $ T<\,\frac{\Delta \,H}{\Delta\, S}$
Now, $ \frac{\Delta\, H}{\Delta\, S}=\frac{12 \times 10^{3}}{40}=300\, K$
$=300-273=27^{\circ} C$
$\therefore $ Above $27^{\circ} C$, the reactiọn becomes spontaneous.