Clearly, if $\mathrm{E}_a=0, \mathrm{~K}$ is temperature independent if $\mathrm{E}_a>0, \mathrm{~K}$ increase with increase in temperature if $\mathrm{E}_a<0, \mathrm{~K}$ decrease with increase in temperature

• Rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy.


• Higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant.


• The pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.
  
  a. A high activation energy usually implies a slow reaction.
  
  b. $k=\mathrm{P} \times \mathrm{Z} \times \mathrm{e}^{-\mathrm{E}_a / \mathrm{RT}}$
  
  c. The pre-exponential factor $(\mathrm{A}=\mathrm{P} \times \mathrm{Z})$ is independent of the activation energy and the energy of molecules.