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For the reaction A + B → products, it is observed that:
(1) On doubling the initial concentration of A only, the rate of reaction is also doubled and
(2) On doubling the initial concentration of both A and B, there is a change by a factor of 8 in the rate of the reaction.
The rate of this reaction is given by
(1) On doubling the initial concentration of A only, the rate of reaction is also doubled and
(2) On doubling the initial concentration of both A and B, there is a change by a factor of 8 in the rate of the reaction.
The rate of this reaction is given by
A
rate = k[A][B]
B
rate = $k[A]^2[B]$
C
rate = $k[A][B]^2$
D
rate = $k[A]^2[B]^2$
Detailed Solution
Let rate = $k[A]^x[B]^y$.
Observation (1): doubling [A] alone doubles the rate: $2^x = 2$, so x = 1.
Observation (2): doubling both [A] and [B] increases the rate 8 times: $2^x\times2^y = 8$
With x = 1: $2\times2^y = 8$, so $2^y = 4$ and y = 2.
Rate law: rate = $k[A][B]^2$
Observation (1): doubling [A] alone doubles the rate: $2^x = 2$, so x = 1.
Observation (2): doubling both [A] and [B] increases the rate 8 times: $2^x\times2^y = 8$
With x = 1: $2\times2^y = 8$, so $2^y = 4$ and y = 2.
Rate law: rate = $k[A][B]^2$