For the equilibrium, PCl_(5(g))iffPCl_(3(g))+Cl_(2(g)) at 298K. K_(C)=1.8xx10^(-7) what is DeltaG^(0) for the reaction? (R=8.314JK^(-1)mol^(-1))

For the equilibrium, PCl_(5(g))iffPCl_(3(g))+Cl_(2(g)) at 298K. K_(C)=

1.	For the equilibrium, PCl_(5(g))iffPCl_(3(g))+Cl_(2(g)) at 298K. K_(C)=1.8xx10^(-7) what is DeltaG^(0) for the reaction? (R=8.314JK^(-1)mol^(-1))
Answer» Solution :`K_(P)=K_(C)(RT)^(Deltan)Deltan=n_(p)-n_(r)=2-1=1` `T=298KR=8.314JK^(-1)mol^(-1)` `K_(P)=(1.8xx10^(-7))(8.314xx298)=4.46xx10^(-4)` `Delta_(r)G^(0)=-2.303RTlogK_(P)` `=2-2.303xx8.314xx298xxlog(4.46xx10^(-4))` `=19118.6Jmol^(-1)=19.12KJmol^(-1)`