English
Tamil Nadu Board of Secondary EducationHSC Science Class 12

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200 K to 400 K? (R = 8.314 JK−1mol−1) - Chemistry

Advertisements
Advertisements

Question

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200 K to 400 K? (R = 8.314 JK−1mol−1)

Options

  • 234.65 kJ mol1

  • 434.65 kJ mol−1

  • 2.305 kJ mol1

  • 334.65 J mol1

MCQ

Solution

2.305 kJ mol1

shaalaa.com
Factors Affecting the Reaction Rate
  Is there an error in this question or solution?
Chapter 7: Chemical Kinetics - Evaluation [Page 228]

APPEARS IN

Samacheer Kalvi Chemistry - Volume 1 and 2 [English] Class 12 TN Board
Chapter 7 Chemical Kinetics
Evaluation | Q 11. | Page 228

RELATED QUESTIONS

In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively −x kJ mol−1 and y kJ mol−1. Therefore , the energy of activation in the backward direction is


If the initial concentration of the reactant is doubled, the time for half reaction is also doubled. Then the order of the reaction is ____________.


In a homogeneous reaction \[\ce{A -> B + C + D}\], the initial pressure was P0 and after time t it was P. Expression for rate constant in terms of P0, P and t will be


The correct difference between first and second order reactions is that


Explain the effect of catalyst on reaction rate with an example.


The rate law for a reaction of A, B and C has been found to be rate = `"k" ["A"]^2["B"] ["L"]^(3/2)`. How would the rate of reaction change when

  1. Concentration of [L] is quadrupled
  2. Concentration of both [A] and [B] are doubled
  3. Concentration of [A] is halved
  4. Concentration of [A] is reduced to `(1/3)` and concentration of [L] is quadrupled.

The rate of formation of a dimer in a second order reaction is 7.5 × 10−3 mol L−1s−1 at 0.05 mol L−1 monomer concentration. Calculate the rate constant.


For a reaction \[\ce{x + y + z -> products}\] the rate law is given by rate = `"k" ["x"]^(3/2) ["y"]^(1/2)` what is the overall order of the reaction and what is the order of the reaction with respect to z.


How do concentrations of the reactant influence the rate of reaction?


How do nature of the reactant influence rate of reaction?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×