(B) Let the rate law be $Rate = k[A]^n$. When concentration is doubled,$Rate' = k[2A]^n = 4 \times Rate$. So,$2^n = 4$,which implies $n = 2$. When con

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(B) Let the rate law be $Rate = k[A]^n$. When concentration is doubled,$Rate' = k[2A]^n = 4 \times Rate$. So,$2^n = 4$,which implies $n = 2$. When concentration is tripled,$Rate'' = k[3A]^n = 9 \times Rate$. So,$3^n = 9$,which implies $n = 2$. Thus,the rate is proportional to the square of the concentration of $A$ $([A]^2)$.

Class 12 Chemistry Chemical Kinetics Rate law , Rate constant , Order of Reaction and Molecularity

Medium

If doubling the concentration of a reactant $A$ increases the rate $4$ times and tripling the concentration of $A$ increases the rate $9$ times,the rate is proportional to

AIIMS 1991 All AIIMS

A
Concentration of $A$
B
Square of concentration of $A$
C
Under root of the concentration of $A$
D
Cube of concentration of $A$

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Class 12

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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AIIMS 1991 Paper All AIIMS years →

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DifficultJEE MAIN 2016

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If doubling the concentration of a reactant $A$ increases the rate $4$ times and tripling the concentration of $A$ increases the rate $9$ times,the rate is proportional to

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When the initial concentration of a reactant is doubled in a reaction,its half-life period is not affected. The order of the reaction is:

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The rate law for the reaction below is given by the expression $Rate = k[A][B]$.$A + B \to \text{Product}$If the concentration of $B$ is increased from $0.1 \ M$ to $0.3 \ M$,keeping the concentration of $A$ constant at $0.1 \ M$,the rate constant $(k)$ will be:

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The rate law for the reaction below is given by the expression $Rate = k[A][B]$.
$A + B \to \text{Product}$
If the concentration of $B$ is increased from $0.1 \ M$ to $0.3 \ M$,keeping the concentration of $A$ constant at $0.1 \ M$,the rate constant $(k)$ will be: