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 a ln(k) versus 1/T (K) plot for the uncatalyzed reaction? Assume that both rate laws are first-order. Explain

Short answer

For the catalyzed reaction has a less negative slope than the uncatalyzed reaction. In catalysis, the reduction in the activation energy potential expresses that the reaction is catalyzed by catalysts.

Step-by-step solution

Explaining the Arrhenius equation and collision theory

The collision theory states that the particles of the substrate are assumed to be spheres colliding with other particles of the substrate with a specific orientation and minimum energy, which results in the formation of products.

The Arrhenius equation is the relationship between the temperature and rate constant of reaction (k) based on the collision theory. The rate constant depends upon the collision factor, activation energy, and temperature.

The Arrhenius equation is as follows –

$k=A{e}^{-\frac{E}{RT}}$

Where k is the reaction rate constant,

A is Arrhenius or collision factor,

E is the activation energy in kJ/mol,

R is gas constant,

T is the temperature in K.

Explain the role of the catalyst in the catalyzed and uncatalyzed reactions using the Arrhenius equation.

The catalyst is necessary for the reaction to proceed by lowering the activation energy in acatalyzed reaction. During the uncatalyzed reaction, the activation energy is not reduced.

The reaction is favorable if the reactant molecules overtake the activation energy barrier for catalyzed reactions. If the energy barrier is minimized, the rate is increased.

Therefore, in the Arrhenius equation, $\left(\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.\right)$term should be minimum for a larger rate constant. As increases in the term, the exponential value will be more in the denominator which leads to the rate constant$\left(k\right)$ decreases and vice-versa.