English

The Rate Constant for the First-order Decomposition of H2O2 is Given by the Following Equation

Advertisements
Advertisements

Question

The rate constant for the first-order decomposition of H2O2 is given by the following equation:

`logk=14.2-(1.0xx10^4)/TK`

Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes.

(Given: R = 8.314 JK–1 mol–1)

Advertisements

Solution

Given:

Order of the reaction = First order

t1/2 = 200 minutes = 200 × 60 = 12,000 seconds

The relation between t1/2  and k is given by

t1/2 = 0.693/k

k = 0.693/12000 = 5.7 × 10−5

The rate constant for the first-order decomposition of H2O2 is given by

`logk=14.2-(1.0xx10^4)/TK...................(i)`

By Arrhenius equation

`log k=logA-E_a/(2.303RT)...............(ii)`

Comparing (i) and (ii), we get

Ea = 1.91 × 105

shaalaa.com
  Is there an error in this question or solution?
2015-2016 (March) Delhi Set 3

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

 

Consider the reaction

`3I_((aq))^-) +S_2O_8^(2-)->I_(3(aq))^-) + 2S_2O_4^(2-)`

At particular time t, `(d[SO_4^(2-)])/dt=2.2xx10^(-2)"M/s"`

What are the values of the following at the same time?

a. `-(d[I^-])/dt`

b. `-(d[S_2O_8^(2-)])/dt`

c. `-(d[I_3^-])/dt`

 

 

The decomposition of hydrocarbon follows the equation

k = `(4.5 xx 10^11 s^-1) e^(-28000 K//T)`

Calculate Ea.


What is the effect of adding a catalyst on Activation energy (Ea)


 Write a condition under which a bimolecular reaction is kinetically first order. Give an example of  such a reaction. (Given : log2 = 0.3010,log 3 = 0.4771, log5 = 0.6990).


The chemical reaction in which reactants require high amount of activation energy are generally ____________.


Why does the rate of a reaction increase with rise in temperature?


Total number of vibrational degrees of freedom present in CO2 molecule is


The equation k = `(6.5 xx 10^12 "s"^(-1))"e"^(- 26000 " K"//"T")` is followed for the decomposition of compound A. The activation energy for the reaction is ______ kJ mol-1. (Nearest integer) (Given: R = 8.314 JK-1 mol-1)


The rate of a reaction quadruples when temperature changes from 27°C to 57°C calculate the energy of activation. 

(Given: R = 8. 314 J K−1 mol−1, log 4 = 0.6021)


Assertion (A): A reaction can have zero activation energy.

Reason (R): The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to the threshold value is called activation energy.

In the light of the above statements, choose the correct answer from the options given below:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×