Question

A reaction becomes spontaneous only at 500 K. If $\Delta H$ at 500 K is $3.0kJ$, the change in entropy at 500 K is:

Answer 1

Since reaction becomes spontaneous at $500K$, $\Delta G$ must be changing sign at $500K$.

So, $\Delta G=0$ at $500K$ for the given reaction.

Hence, $\Delta S=\frac{\Delta H}{T}=\frac{3000}{500}=6Jmo{l}^{-1}{K}^{-1}$

Hence, the correct answer is $6$.