\(A+B \longrightarrow \text { Products }\)

On doubling the initial concentration of \(A\) only, the rate of the reaction is also doubled, therefore

\(\text {Rate } \propto[A]^{1}\) ..... \((i)\)

Let initial rate law is

\(\text {Rate}=k[A][B]^{y}\) ..... \((ii)\)

If concentration of \(A\) and \(B\) both are doubled, the rate gets changed by a factor of \(8\)

\(8 \times \text {rate}=k[2 A][2 B]^{y}\) ..... \((iii)\)

\(\left[\therefore \text { Rate } \propto[A]^{1}\right]\)

Dividing Eq. \((iii)\) by Eq. \((ii)\), we get

\({8=2 \times 2^{y}} \)

\({4=2^{y}} \)

\({(2)^{2}=(2)^{y}} \)

\({y=2}\)

Hence, rate law is, rate \(=k[A][B]^{2}\)