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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

KEAMKEAM 2013Thermodynamics

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.