Question

For a given reaction:
$xA+yB\to mC+nD$
$△H=-30kJmo{l}^{-1},△S=-100J{K}^{-1}mo{l}^{-1}$
At what temperature the reaction is at equilibrium?

• A. ${50}^{o}C$
• B. ${250}^{o}C$
• C. ${100}^{o}C$
• D. ${27}^{o}C$

Answer 1

The correct option is D ${27}^{o}C$

Gibb's free energy of a reaction is given by:
$△G=△H-T△S$
we know at equilibrium, $△G=0$
$\therefore T=\frac{△H}{△S}=\frac{-30\times 1000}{-100}=300K,or,{27}^{o}C$
hence, the reaction is at equilibrium at ${27}^{o}C$