For reaction, `2K_((g))+L_((g))rarr2M_((g)),DeltaU^(@)=-"10.5 KJ and "DeltaS^(@)=-"44.1 J K"^(-1)`. Calculate `DeltaG^(@)` for the reaction and predict whether the reaction will be spontaneous or non-spontaneous?A. `DeltaG=+"0.16 kJ, non-spontaneous"`B. `DeltaG=-"0.16 kJ, spontaneous"`C. `DeltaG=+"26.12 kJ, non-spontaneous"`D. `DeltaG=-"26.12 kJ, spontaneous"`

For reaction, `2K_((g))+L_((g))rarr2M_((g)),DeltaU^(@)=-"10.5 KJ and "

1.	For reaction, `2K_((g))+L_((g))rarr2M_((g)),DeltaU^(@)=-"10.5 KJ and "DeltaS^(@)=-"44.1 J K"^(-1)`. Calculate `DeltaG^(@)` for the reaction and predict whether the reaction will be spontaneous or non-spontaneous?A. `DeltaG=+"0.16 kJ, non-spontaneous"`B. `DeltaG=-"0.16 kJ, spontaneous"`C. `DeltaG=+"26.12 kJ, non-spontaneous"`D. `DeltaG=-"26.12 kJ, spontaneous"`
Answer» Correct Answer - A `2K+L rarr 2M` `Deltan_(g)=2-3=-1` `DeltaH=DeltaU+Deltan_(g)RT=-10.5xx10^(3)+(-1xx8.314xx298)` `=-10500+(-2477.572)=-12977.57J=-12.98kJ` `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)=-12.98-298(-44.1xx10^(-3))` `=-12.98+13.14=0.16kJ` Since `DeltaG^(@)` is `+ve` hence it is non-spontaneous.

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