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Karnataka Board PUCPUC Science 2nd PUC Class 12

Rate law for the reaction A+2B⟶C is found to be Rate = k [A][B]. Concentration of reactant ‘B’ is doubled, keeping the concentration of ‘A’ constant, the value of rate constant will be ______.

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Question

Rate law for the reaction \[\ce{A + 2B -> C}\] is found to be Rate = k [A][B]. Concentration of reactant ‘B’ is doubled, keeping the concentration of ‘A’ constant, the value of rate constant will be ______.

Options

  • the same

  • doubled

  • quadrupled

  • halved

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Solution

Rate law for the reaction \[\ce{A + 2B -> C}\] is found to be Rate = k [A][B]. Concentration of reactant ‘B’ is doubled, keeping the concentration of ‘A’ constant, the value of rate constant will be the same.

Explanation:

The rate of concentration of a reaction does not depend upon the concentrations of the reactants. Hence, it will remain the same. Even if the equation shows the double concentration level, even the rate concentration doubles so it's the same throughout.

Following with the equation \[\ce{A + 2B -> C}\] 

If rate considered as (1) Rate1 = k[A][B]

If rate considered as 2

Then, Rate1 = k[A][2B]

Rate2 = 2Rate1 

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Chapter 4: Chemical Kinetics - Exercises [Page 51]

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NCERT Exemplar Chemistry Exemplar [English] Class 12
Chapter 4 Chemical Kinetics
Exercises | Q I. 14. | Page 51

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