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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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