Rate law for the reaction, $A + 2 B \to C$ is found to be
$R a t e = k [A] [B]$
Concentration of reactant 'B' is doubled, keeping the concentration of 'A' constant, the value of rate constant will be ................. .

the same

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doubled

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quadrupled

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halved

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Solution

The correct option is A the same
From the rate law expression, it is clear that the rate w.r.t to B is of the first order.
Rate $\to R_{1} = k [A] [B] . . . (i)$
Rate $\to R_{2} = k [A] [2 B] . . . (i i)$
Dividing eq.(i) by eq. (ii) $= R_{2} = 2 R_{1}$ $. . . (i i i)$

From (iii)
When the concentration of A is kept constant and the concentration of B gets double than the rate will be doubled.

Hence, Option "B" is the correct answer.

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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 ................. .

Rate law for the reaction, $A + 2 B \to C$ is found to be
$R a t e = k [A] [B]$
Concentration of reactant 'B' is doubled, keeping the concentration of 'A' constant, the value of rate constant will be ................. .