Question

A certain reaction has a $\Delta H$ of $12kJ$ and a $\Delta S$ of $40J{K}^{-1}$. The temperature above which the reaction becomes spontaneous is:

• A. ${27}^{\circ }C$
• B. $27K$
• C. ${300}^{\circ }C$
• D. ${30}^{\circ }C$
• E. $30K$

Answer 1

Answer: (A)

${27}^{\circ }C$

A reaction will be spontaneous, if

$\Delta H-T\Delta S=-ve$ or $\Delta H-T\Delta S<0$

or $T<\frac{\Delta H}{T_a\Delta S}$

Now, $\frac{\Delta H}{\Delta S}=\frac{12\times {10}^3}{40}=300K$

$=300-273={27}^{\circ }C$

$\therefore$ Above ${27}^{\circ }C$, the reaction becomes spontaneous.