मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Consider figure and mark the correct option.

Advertisements
Advertisements

प्रश्न

Consider figure and mark the correct option.

पर्याय

  • Activation energy of forward reaction is E1 + E2 and product is less stable than reactant.

  • Activation energy of forward reaction is E1 + E2 and product is more stable than reactant.

  • Activation energy of both forward and backward reaction is E1 + E2 and reactant is more stable than product.

  • Activation energy of backward reaction is E1 and product is more stable than reactant.

MCQ
Advertisements

उत्तर

Activation energy of forward reaction is E1 + E2 and product is less stable than reactant.

Explanation:

Ea (forward) = E1 + E2

Since energy of reactants is less than products and the product is less stable than the reactant.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Chemical Kinetics - Exercises [पृष्ठ ४७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 12
पाठ 4 Chemical Kinetics
Exercises | Q I. 4. | पृष्ठ ४७

संबंधित प्रश्‍न

 

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 rate constant for the decomposition of N2O5 at various temperatures is given below:

T/°C 0 20 40 60 80
105 × k/s−1 0.0787 1.70 25.7 178 2140

Draw a graph between ln k and `1/T` and calculate the values of A and Ea. Predict the rate constant at 30º and 50ºC.


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?


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?


Define activation energy.


Calculate activation energy for a reaction of which rate constant becomes four times when temperature changes from 30 °C to 50 °C. (Given R = 8.314 JK−1 mol−1). 


 Predict the main product of the following reactions:


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


The rate of chemical reaction becomes double for every 10° rise in temperature because of ____________.


Which of the following graphs represents exothermic reaction?

(a)  

(b)  

(c)  


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


Oxygen is available in plenty in air yet fuels do not burn by themselves at room temperature. Explain.


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


The activation energy in a chemical reaction is defined as ______.


The activation energy in a chemical reaction is defined as ______.


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 ______.


What happens to the rate constant k and activation energy Ea as the temperature of a chemical reaction is increased? Justify.


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×