Difficult
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$3.2 \times 10^{-6} \ s^{-1}$
- BZero
- C$1.6 \times 10^{-6} \ s^{-1}$
- D$1.6 \times 10^{-5} \ s^{-1}$
Explore More
Similar Questions
According to the Arrhenius equation,a plot of the logarithm of the rate constant $(\log \, k)$ versus which of the following yields a straight line?
Easy
View SolutionThe following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$
For the gaseous reactions:
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.34 T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.34 T}$
Calculate the approximate temperature at which $k_1 = k_2$ in $K$.
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.34 T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.34 T}$
Calculate the approximate temperature at which $k_1 = k_2$ in $K$.
Difficult
View SolutionThe rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \times e^{-2000/T}$ and $10^{15} \times e^{-1000/T}$ respectively. The temperature at which $k_1 = k_2$ is
MediumKCET 2023
View SolutionActivation energy for the acid catalysed hydration of sucrose is $6.22 \ kJ \ mol^{-1}$,while the activation energy is only $2.15 \ kJ \ mol^{-1}$ when hydrolysis is catalysed by the enzyme sucrase. Explain.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo