Question

would the slope of a ln(k) versus 1/T (K) plot for a catalyzed reaction be more or less negative than the slope of the ln(k) versus 1/T (K) for the uncatalyzed reaction? Explain. Assume both rate laws for the ccatalyze and uncatalyzed reaction are all first- order overall.

Answer 1

Arrhenius equation:
K= -A e^E_a/RT ---> ln K = -E_a/R * (1/T) + A

equation of straight line y= m * x + c

A catalyst increases the rate of a reaction by providing reactants with an alternate pathway (mechanism) to convert to products. This alternate pathway has a lower activation energy, thus increasing the rate of the reaction.

As the value of E_a decreases the quantity -E_a/R (slope ) negative value decreases hence we can say that

he slope of a ln(k) versus 1/T (K) plot for a catalyzed reaction be less negative than the slope of the ln(k) versus 1/T (K) for the uncatalyzed reaction .