Rate of reaction for A + B $\to$ Product

Rate = k[A]^x[B]^y …(1)

where, x and y are order w.r.t. A and B respectively. When the concentration of only B is doubled, the rate is doubled, so

R’ = k [A]^x [2B]^y = 2R …(2)

If concentration of both the reactants A and B are doubled then the rate increases by a factor of 8 so

R’’ = k[2A]^x[2B]^y = 8R ...(3)

= k2^x 2^y [A]^x [B]^y = 8R …(4)

From equation (1) and (2), we get

${{2R} \over R} = {{{{\left[ A \right]}^x}{{\left[ {2B} \right]}^y}} \over {{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}$

$\Rightarrow$ 2 = 2^y

$\Rightarrow$ y = 1

From equation (1) and (4), we get

${{8R} \over R} = {{{2^x}{2^y}{{\left[ A \right]}^x}{{\left[ B \right]}^y}} \over {{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}$

$\Rightarrow$ 8 = ${{2^x}{2^y}}$

Substituting the value of y gives

${{2^x}{2^1}}$ = 8

$\Rightarrow$ ${{2^x}}$ = 4

$\Rightarrow$ x = 2

By replacing the values of x and y in

rate law; rate = k[A]²[B]