Question

The $\Delta H$ and $\Delta S$ for a reaction at one atmospheric pressure are +30.558 kJ and 0.066 k J $k^{-1}$ respectively. The temperature at which the free energy change will be zero and below this temperature the nature of reaction would be:

• A. 483 K, spontaneous
• B. 443 K, non-spontaneous
• C. 443 K, spontaneous
• D. 463 K, non-spontaneous

Answer 1

The correct option is D

463 K, non-spontaneous

Now, to get T, let's use
$△G$ = $△H-T△S$
Here, $△H=+30.558KJ$
$△S=0.066kJ{k}^{-1}$
$△G=0$
$\Rightarrow 0=30.558-T\times 0.066$
$\Rightarrow T=\frac{30.558}{0.066}=463K$
Now, at 463 $K\Rightarrow △G$ = 0
If we go below 463 K, $T△S$ value will decrease and $△G$ value will become positive.
Therefore, non-spontaneous