d
Let the order of reaction with respect to \(\mathrm{A}\) is \(\mathrm{x}\)

and with respect to \(\mathrm{B}\) is \(y\).

Thus,

rate \(=\mathrm{k}[A]^{x}[B]^{y}\)

(\(x\) and \(y\) are stoichiometric coefficient)

For the given cases,

\(I\). rate \(=\mathrm{k}(0.1)^{x}(0.1)^{y}=6.0 \times 10^{-3}\)

\(II\). rate \(=\mathrm{k}(0.3)^{x}(0.2)^{y}=7.2 \times 10^{-2}\)

\(III\). rate \(=\mathrm{k}(0.3)^{x}(0.40)^{y}=2.88 \times 10^{-1}\)

\(IV\). rate \(=\mathrm{k}(0.34)^{x}(0.1)^{y}=2.40 \times 10^{-2}\)

Dividing Eq. \((I)\) by Eq. \((IV)\), we get

\(\left(\frac{0.1}{0.4}\right)^{x}\left(\frac{0.1}{0.1}\right)^{y}\)

\(\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}\)

\(\left(\frac{1}{4}\right)^{x} =\left(\frac{1}{4}\right)^{1}\)

\(\therefore x=1\)

On dividing Eq. \((II)\) by Eq. \((III)\), we get

\(\left(\frac{0.3}{0.3}\right)^{x}\left(\frac{0.2}{0.4}\right)^{y}=\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}\)

\(\left(\frac{1}{2}\right)^{y}=\frac{1}{4}\)

\(\left(\frac{1}{2}\right)^{y}=\left(\frac{1}{2}\right)^{2}\)

\(\therefore y=2\)

Thus, rate law is,

\(\text {rate} =k[A]^{1}[B]^{2}\)

\(=k[A][B]^{2}\)