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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 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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प्रश्न

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 ______.

पर्याय

  • the same

  • doubled

  • quadrupled

  • halved

MCQ
रिकाम्या जागा भरा
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उत्तर

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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पाठ 4: Chemical Kinetics - Exercises [पृष्ठ ५१]

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एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 12
पाठ 4 Chemical Kinetics
Exercises | Q I. 14. | पृष्ठ ५१

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