(a) Illustrate graphically the effect of catalyst on activation energy. (b) Catalysts have no effect on the equilibrium constant. Why? (c ) The decomposition of A into product has value of k as 4.5xx10^(3)s^(-1) at 10^(@)C and activation energyis 60 kJ mol^(-1). Calculate the temperature at which the value of kW(J) be 1.5xx10^(4)s^(-1).

(a) Illustrate graphically the effect of catalyst on activation energy

1.	(a) Illustrate graphically the effect of catalyst on activation energy. (b) Catalysts have no effect on the equilibrium constant. Why? (c ) The decomposition of A into product has value of k as 4.5xx10^(3)s^(-1) at 10^(@)C and activation energyis 60 kJ mol^(-1). Calculate the temperature at which the value of kW(J) be 1.5xx10^(4)s^(-1).
Answer» Solution :(a) The catalyst provides an alternate pathways or reaction mechanism by reducing the activation energy between the reactants and products and hence lowering the potential energy barrier as shown in the figure. (b) A catalyst does not change the equilibrium constant of a reaction, rather it helps in attaining the equilibrium faster. That is, it catalysesthe forward as well as backward REACTIONS to the same extent so that the equilibrium state remains same but is REACHED earlier. (C ) Use the relation Given : `"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `k_(1)=4.5xx 10^(3)s^(-1), k_(2)=1.5xx10^(4)s^(-1), T_(1)=10^(@)C=283K, T_(2)=?, E_(a)=60 kJ=60000J` `R=8.314JK^(-1)"MOL"^(-1)` SUBSTITUTING the values in the equation above, we have `"log"(1.5xx10^(4))/(4.5xx10^(3))=(60000)/(2.303xx8.314)[(T_(2)-283)/(283T_(2))] or 0.5228=3133.6269[(T_(2)-283)/(283T_(2))]` or `(T_(2)-283)/(283T_(2))=(0.5228)/(3133.6269) or (T_(2)-283)/(283T_(2))=0.00016684` or `T_(2)-283=0.0472T_(2) or T_(2)(1-0.0472)=283` or `0.9528T_(2)=283` or `T_(2)=283//0.9528=297K or 24^(@)C`