The rate constant of a reaction of a reaction increases by 5% whenthe temperature of the reaction is increased from 300 to 301 K wheres equilibrium constant increases only by 2%. Calculate the activation energy for the forward as well as backward reaction.

The rate constant of a reaction of a reaction increases by 5% whenthe

1.	The rate constant of a reaction of a reaction increases by 5% whenthe temperature of the reaction is increased from 300 to 301 K wheres equilibrium constant increases only by 2%. Calculate the activation energy for the forward as well as backward reaction.
Answer» Solution :ACCORDING to Arrhenius EQUATION, `log""(k_(2))/(k_(1))=(E_(a,f))/(2.303"R")((T_(2)-T_(1))/(T_(1)T_(2)))` If `k_(1)=k" at "300" K, then at "301" K,"k_(2)=k+(5)/(100)k=1.05k` `:.log""(1.05k)/(k)=(E_(a,f))/(2.303xx8.314" JK"^(-1)MOL^(-1))((301" K"-300" K")/(300" K"xx301" K"))` `:.E_(a,f)=(log1.05)xx2.303xx(8.314" JK"^(-1)mol^(-1))xx300xx301" (K) "=366554" J mol"^(-1)=36.65" kJ mol"^(-1)` According to van't Hoff equation (giving the effect of temperature on equilibrium constant), `log""(K_(2))/(K_(1))=(DeltaH^(@))/(2.303"R")((T_(2)-T_(1))/(T_(1)T_(2)))` If `K=K_(1)" at "300" K, then at "301" K","K"_(2)="K"+(2)/(100)"K"=1.02" K"` `:.log""(1.02"K")/("K")=(DeltaH^(@))/(2.303xx8.314" JK"^(-1)mol^(-1))((301"K"-300"K")/(300"K"XX301"K"))` or `DeltaH^(@)=" Activation energy for forward reaction-Activation energy for backward reaction"=E_(a,f)-E_(a,b)` `:.E_(a,b)=E_(a,f)-DeltaH^(@)=36.65-14.87" kJ mol"^(-1)=21.78" kJ mol"^(-1).`