Question

At 1 atm pressure, for a reaction ΔH=+30.558 kJ mol−1 and ΔS=0.066 kJ mol−1. The temperature at which free energy is equal to zero and the nature of the reaction below this temperature is:

• 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

$\Delta G=\Delta H-T\Delta S=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{+30.558kJmo{l}^{-1}}{0.066kJ{K}^{-1}mo{l}^{-1}}=463K$

$\Delta G=\Delta H-T\Delta S=(30.558-T\times 0.066)kJmol-1$

At $T<463K$,

$0.066T<463\times 0.066$

or, $0.066T<30.558$

or, $0<30.558-0.066T$

or, $0<\Delta G$

Thus, at $T<463K,\Delta G>0$ i.e. , process is non-spontaneous.

Hence, the correct option is $\text{D}$