English
Karnataka Board PUCPUC Science 2nd PUC Class 12

The rate constant for the decomposition of hydrocarbons is 2.418 × 10^−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?

Advertisements
Advertisements

Question

The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?

Numerical
Advertisements

Solution

Given: k = 2.418 × 10−5 s−1

T = 546 K

Ea = 179.9 kJ mol−1 

According to the Arrhenius equation,

log A = `log k + E_a/(2.303 RT)`

= `log 2.418 × 10^-5 + 179.9/(2.303 xx 8.314 xx 10^-3 xx 546)`

= (−5 + 0.3834) + 17.2081

= −4.6166 + 17.2081

log A = 12.5915

⇒ A = Antilog (12.5915)

= 3.902× 1012 s−1

shaalaa.com
  Is there an error in this question or solution?
Chapter 3: Chemical Kinetics - 'NCERT TEXT-BOOK' Exercises [Page 281]

APPEARS IN

Nootan Chemistry [English] Class 12 ISC
Chapter 3 Chemical Kinetics
'NCERT TEXT-BOOK' Exercises | Q 4.23 | Page 281
NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 3 Chemical Kinetics
Exercises | Q 3.23 | Page 87

RELATED QUESTIONS

The activation energy for the reaction \[\ce{2 HI_{(g)} -> H2_{(g)} + I2_{(g)}}\] is 209.5 kJ mol−1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?


The decomposition of hydrocarbon follows the equation

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

Calculate Ea.


The decomposition of A into product has value of k as 4.5 × 103 s−1 at 10°C and energy of activation 60 kJ mol−1. At what temperature would k be 1.5 × 104 s−1?


The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.


The rate constant of a first order reaction are 0.58 S-1 at 313 K and 0.045 S-1 at 293 K. What is the energy of activation for the reaction?


A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given : log 2 = 0·3010, log 4 = 0·6021, R = 8·314 JK–1 mol–1)


Explain the following terms :

Half life period of a reaction (t1/2)

 

 

Activation energy of a chemical reaction can be determined by ______.


Consider figure and mark the correct option.


During decomposition of an activated complex:

(i) energy is always released

(ii) energy is always absorbed

(iii) energy does not change

(iv) reactants may be formed


Which of the following statements are in accordance with the Arrhenius equation?

(i) Rate of a reaction increases with increase in temperature.

(ii) Rate of a reaction increases with decrease in activation energy.

(iii) Rate constant decreases exponentially with increase in temperature.

(iv) Rate of reaction decreases with decrease in activation energy.


Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.


What happens to most probable kinetic energy and the energy of activation with increase in temperature?


For an endothermic reaction energy of activation is Ea and enthalpy of reaction ΔH (both of there in KJ moI–1) minimum value of Ea will be ______.


The rate constant for a reaction is 1.5 × 10–7 sec–1 at 50°C. What is the value of activation energy?


Arrhenius equation can be represented graphically as follows:

The (i) intercept and (ii) slope of the graph are:


An exothermic reaction X → Y has an activation energy 30 kJ mol-1. If energy change ΔE during the reaction is - 20 kJ, then the activation energy for the reverse reaction in kJ is ______.


It is generally observed that the rate of a chemical reaction becomes double with every 10°C rise in temperature. If the generalisation holds true for a reaction in the temperature range of 298 K to 308 K, what would be the value of activation energy (Ea) for the reaction?


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)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×