Question

Which of the following expressions gives the correct relation between the rate constant and activation energy?

• A. $k=A{e}^{-E_a/RT}$
• B. $lnk=lnA-\frac{E_a}{RT}$
• C. $logk=logA-\frac{E_a}{2.303RT}$
• D. $ln\frac{k}{A}=-\frac{E_a}{RT}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $k=A{e}^{-E_a/RT}$
B $lnk=lnA-\frac{E_a}{RT}$
C $logk=logA-\frac{E_a}{2.303RT}$
D $ln\frac{k}{A}=-\frac{E_a}{RT}$

All of the following expressions give correct relation between the rate constant and activation energy.
$k=A{e}^{-E_a/RT}$
$lnk=lnA-\frac{E_a}{RT}$
$logk=logA-\frac{E_a}{2.303RT}$
$ln\frac{k}{A}=-\frac{E_a}{RT}$
Here, k is the rate constant, A is the pre-exponential factor $E_a$ is the energy of activation, R is the ideal gas constant and T is the absolute temperature.