In a reaction, A+B arrow Product, rate is doubled when the...

Let the order of reaction with respect to A and B$is$x and y$respectively. So, the rate law can be given as$R=k[A]^x[B]^y$When the concentration of only$B$is doubled, the rate is doubled, so$R_1=k[A]^x[2 B]^y=2 R$If concentrations of both the reactants A and B$ are doubled, the rate increases by a factor of 8 , so $$\begin{aligned} R^{\prime \prime}=k[2 A]^x[2 B]^y & =8 R \\ \Rightarrow k 2^x 2^y[A]^x[B]^y & =8 R \end{aligned}$$ From Eq. (i) and (ii), we get $$\begin{array}{rlrl} & \Rightarrow & \frac{2 R}{R} & =\frac{[A]^x[2 B]^y}{[A]^x[B]^y} \\ & \therefore & =2^y \\ & y & =1 \end{array}$$ From Eq. (i) and (iv), we get $\Rightarrow \frac{8 R}{R}=\frac{2^x 2^y[A]^x[B]^y}{[A]^x[B]^y} or 8=2^x 2^y Substitution of the value of$y$gives,$$\begin{aligned} & 8=2^x 2^1 \\ & 4=2^x \end{aligned}$$\begin{aligned} & & (2)^2 & =(2)^x \\ \therefore & & x & =2 \end{aligned}$$$Substitution$ of the value ofx and y$in Eq. (i) gives,$R=k[A]^2[B]$