(A) The initial rate is given by $R = k[A]^n[B]^m$. When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is halved $(

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(A) The initial rate is given by $R = k[A]^n[B]^m$. When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is halved $([B]' = \frac{B}{2})$,the new rate $R'$ is: $R' = k[2A]^n[\frac{B}{2}]^m$ $R' = k \cdot 2^n [A]^n \cdot 2^{-m} [B]^m$ $R' = k[A]^n[B]^m \cdot 2^{n - m}$ $R' = R \cdot 2^{n - m}$ Therefore,the ratio of the new rate to the earlier rate is: $\frac{R'}{R} = 2^{n - m}$.

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

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The rate law for a reaction between the substances $A$ and $B$ is given by,$rate = k[A]^n[B]^m$. On doubling the concentration of $A$ and halving the concentration of $B$,the ratio of the new rate to the earlier rate of the reaction will be as

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A
$2^{(n - m)}$
B
$2^{(m - n)}$
C
$2^{(n + m)}$
D
$2^{(m + n)}$

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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The rate law for a reaction between the substances $A$ and $B$ is given by,$rate = k[A]^n[B]^m$. On doubling the concentration of $A$ and halving the concentration of $B$,the ratio of the new rate to the earlier rate of the reaction will be as

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What will be the unit of the rate constant for the following reaction?
$C_2H_{4(g)} + H_{2(g)} \rightarrow C_2H_{6(g)}$