### Chemical Kinetics Question 1

#### Question 1 - 24 January - Shift 1

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 $\ln kvs \frac{1}{T}$ is a straight line with slope equal to $-\frac{E a}{R}$

## Show Answer

#### Answer: (3)

#### Solution:

$\mathbf{A}: k=Ae^{-\overline{RT}}$

As Ea increases k decreases

B : Temperature coefficient $=\frac{k _{T+10}}{k_T}$

Option (C ) is wrong. $\Delta k$ may be greater or lesser depending on temperature.

$D: \ln k=\ln A-\frac{Ea}{RT}$