Question

Suppose a reaction has $\Delta H=-40kJ$ and $\Delta S=-50J/K$. At what temperature range will reaction become spontaneous from non-spontaneous?

• A. 801 K to 799 K
• B. 799 K to 800 K
• C. 800 K to 801 K
• D. 799 k to 801 K

Answer 1

The correct option is A 801 K to 799 K

Solution:-

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

$\Delta H=-40kJ=-40000J$

$\Delta S=-50J/K$

if $\Delta G<0$, process is spontaneous

if $\Delta G>0$, process is non spontaneous

if $\Delta G=0$, process is in equilibrium

For T$=801$K

$\Delta G=-40000-(801\times -50)=+50$

The process is non-spontaneous

For T$=799$K

$\Delta G=-40000-(799\times -50)=50$

The process is in spontaneous.

Hence from $801K$ to $799K$, the reaction will become spontaneous from non-spontaneous.