The temperature coefficient of the saponification reaction of an ester by $NaOH$ is $1.75$. The activation energy of the reaction is .......... $kcal \ mol^{-1}$. (in $.21$)

  • A
    $10$
  • B
    $11$
  • C
    $9$
  • D
    $12$

Explore More

Similar Questions

$A \rightarrow B$
The rate constants of the above reaction at $200 \, K$ and $300 \, K$ are $0.03 \, min^{-1}$ and $0.05 \, min^{-1}$ respectively. The activation energy for the reaction is $.... \, J$ (Nearest integer).
(Given: $\ln 10 = 2.3$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\log 5 = 0.70$,$\log 3 = 0.48$,$\log 2 = 0.30$)

In a reaction,the rate of reaction doubles with a $10^\circ C$ rise in temperature. If the temperature is increased from $10^\circ C$ to $100^\circ C$,by how many times will the rate of reaction increase?

The activation energy of a reaction is zero. Its rate constant at $280 \ K$ is $1.6 \times 10^{-6} \ s^{-1}$,the rate constant at $300 \ K$ is

$A$ reaction takes place in three steps with individual rate constants and activation energies. The overall rate constant is given by $k = (\frac{k_1 k_2}{k_3})^{2/3}$. The overall activation energy of the reaction in $kJ/mol$ is:
$Step$ $Rate\ Constant\ /\ Activation\ energy$
$Step\ 1$ $k_1, E_{a_1} = 180\ kJ/mol$
$Step\ 2$ $k_2, E_{a_2} = 80\ kJ/mol$
$Step\ 3$ $k_3, E_{a_3} = 50\ kJ/mol$

Difficult
View Solution

With an increase in temperature,the rate constant $k$:

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