In a reaction, A + B `rarr` product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentration of both the reactants (A and B) are doubled, rate law for the reaction can be written as

Options

A. rate = `k[A][B]^2`
B. rate = `k[A]^2[B]^2`
C. rate = k[A][B]
D. rate = `k[A]^2`[B]

Answer

D

Explanation

`{:("[A]","[B]","Rate",""),(x,y,R,...(i)),(x,2y,R,...(ii)),(2x,2y,8R,...(iii)):}`
Let the rate law, rate = `k[A]^a[B]^b`
From data given, `(x)^a(y)^b = R` …(iv)
`(x)^a(2y)^b=2R` …(v)
Dividing eqn. (v) by (iv), `((2y)^b)/((y)^b)=(2R)/(R)` or `((2)^b)` = 2
Thus b=1
From data of (iii) experiment,
`(2x)^a(2y)^b=8R` .........(vi)
Dividing eqn. (vi) by (v),`((2x)^a)/((x)^a)=(8R)/(2R)" or "((2)^a)=4`
Thus a=2
By replacing the values of a and b in rate law , rate=`k[A]^2[B]`