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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

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

  1. A 1
  2. B 2
  3. C 3
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(C) 3

Step-by-step Solution

Detailed explanation

\[ A: k=A e^{-\frac{F a}{R T}} \] 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. \(D : \ln k=\operatorname{In} A-\frac{E a}{R T}\)
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