Question

The free energy change for a reaction having $△H=31400cal/mol,△S=32cal{K}^{-1}mo{l}^{-1}at727{}^{o}C$ is:

• A. - 600 cal
• B. -420 cal
• C. -850 cal
• D. 600 cal

Answer 1

The correct option is A - 600 cal

Given:
$△H=31400cal/mol△S=32cal{K}^{-1}mo{l}^{-1}andT=727{}^{o}C=727+273=1000K$
The expression to calculate Gibbs free energy is given as
$△G=△H-T△S=31400-1000\times 32=-600cal$
hence, the free energy for a reaction is $-600cal$