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The rate of the reaction A rarr products...

The rate of the reaction A `rarr` products, at the initial concentration of `3.24 xx 10^(-2)` M is nine times its rate at another initial concentration of `1.2 xx 10^(-3)` M. The order of the reaction is

A

a) `(1)/(2)`

B

b) `(3)/(4)`

C

c) `(3)/(2)`

D

d) `(2)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
For the reaction A `rarr` products,
let the order of reaction is n. Then
`r = k[A]^(n)`
`= k xx (1.2 xx 10^(-3))^(n)" "...(i)`
At the initial concentration of `3.24 xx 10^(-2)M`, the rate is 9 times. Hence,
`9r = k xx (3.24 xx 10^(-2))^(n)" "...(ii)`
On dividing, `(9r)/(r) = (k xx (3.24 xx 10^(-2))^(n))/(k xx (1.2 xx 10^(-3))^(n))`
`9 = (27)^(n)`
`9 = (3^(3))^(2//3)`
`therefore" "n = (2)/(3)`
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