The relation between the fraction of molecules $\left( \frac{N_E}{N_T} \right)$ and temperature is explained by the Boltzmann and Maxwell distribution graph.

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(N/A) When the temperature is raised,the maximum of the curve moves to a higher energy value and the curve broadens out,i.e.,it spreads to the right such that there is a greater proportion of molecules with much higher energies.
Mole fraction and Temperature:
The area under the curve must be constant since the total probability must be one at all times. According to the figure,increasing the temperature of the substance increases the fraction of molecules that collide with energies greater than $E_a$. It is clear from the diagram that in the curve at $(t+10)$,the area showing the fraction of molecules having energy equal to or greater than the activation energy increases significantly,leading to an increase in the rate of reaction.

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Similar Questions

In a bimolecular reaction,the steric factor $P$ was experimentally determined to be $4.5$. The correct option$(s)$ among the following is(are)
$[A]$ The activation energy of the reaction is unaffected by the value of the steric factor
$[B]$ Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation
$[C]$ Since $P=4.5$,the reaction will not proceed unless an effective catalyst is used
$[D]$ The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally

The rate of a reaction:

Collision theory is used to explain how chemical species undergo a reaction. Using this theory and the kinetic molecular model,which of the following does $NOT$ influence the rate of a chemical reaction?

The temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$

The rate constant of a reaction at $25\,^oC$ is $1 \times 10^{-3}\,s^{-1}$. If the rate of the reaction doubles when the temperature is increased to $35\,^oC$,the activation energy of the reaction is .......... $kJ\, mol^{-1}$.

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