The reason for almost doubling the rate of reaction on increasing the temperature of the reaction system by $10 \, ^\circ C$ is

  • A
    The value of threshold energy increases
  • B
    Collision frequency increases
  • C
    The fraction of the molecule having energy equal to threshold energy or more increases
  • D
    Activation energy decreases

Explore More

Similar Questions

$A$ reaction having equal energies of activation for forward and reverse reactions has

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: $\ln 10 = 2.303$

For a first-order reaction $A \rightarrow P$,the rate constant $k$ depends on temperature $(T)$ according to the equation $\log \, k = - \frac{2000}{T} + 0.6$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:

The activation energy of a chemical reaction can be determined by:

What is the role of a catalyst in a chemical reaction?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo