Question

The number of correct statement/s from the following is
A. Larger the activation energy, smaller is the value of the rate constant
B. The higher is the activation energy, higher is the value of the temperature coefficient 
C. At lower temperatures, increase in temperature causes more change in the value of $k$ than at higher temperature 
D. A plot of $\ln k$ vs $\frac{1}{T}$ is a straight line with slope equal to $-\frac{E_a}{R}$

Answer 1

Correct Answer: 3

The correct answer is 3

A:k=AA−RTEa​ 
As Ea increases k decreases 
B : Temperature coefficient =kT​kT+10​​ 
Option (C) is wrong. Δk may be greater or lesser depending on temperature. 
D:lnk=lnA−RTEa​

Answer 2

Now according to Arrhenius equation: k = \(A e^{-\frac{Ea}{RT}}\)
 When E_a increases, the value of  \( e^{-\frac{Ea}{RT}}\) decreases due to the negative sign. This suggests that more activation implies a slower reaction rate. Thus, option A can be eliminated. 
When T→∞ then \(e^{-\frac{Ea}{RT}}\)→0 which means that k→A, then we can see that when the temperature rises, the rate constant rises and approaches the Arrhenius constant.
As we know, the rate of reaction reduces with increasing activation energy and increases with increasing temperature. As a result, for larger activation energies, i.e. if the activation energy is constant, the rate of reaction is only affected by temperature. Regardless of temperature, the pre-exponential factor is always constant for a certain reaction. It reflects the collision frequency at a standard concentration.