Following data is given for the reaction,$\mathrm{X}+\mathrm{Y}\to \mathrm{P}$

Experiment	$\left[\mathrm{X}\right] \mathrm{mol} {\mathrm{L}}^{-1}$	$\left[\mathrm{Y}\right] \mathrm{mol} {\mathrm{L}}^{-1}$	Initial rate $\left(\mathrm{mol} {\mathrm{L}}^{-1}{\mathrm{s}}^{-1}\right)$
I	$2.5\times {10}^{-4}$	$6\times {10}^{-5}$	$5\times {10}^{-4}$
II	$5\times {10}^{-4}$	$12\times {10}^{-5}$	$4\times {10}^{-3}$
III	$1\times {10}^{-3}$	$12\times {10}^{-5}$	$1.6\times {10}^{-2}$

The order of reaction with respect to X and Y respectively are

detailed solution

Correct option is B

Let the order of reaction with respect ta X and Y be a and b, the rate law expression r=k[X]a[Y]bSimilarly,  Exp. I. 5×10−4=k2.5×10−4a6×10−5b            ………(i)                 Exp. II. 4×10−3=k5×10−4a12×10−5b            ………(ii)                 Exp. III. 1.6×10−2=k1×10−3a12×10−5b       ………(iii)From Eqs. (i) and (ii),            4×10−35×10−4=5×10−42.5×10−4a12×10−56×10−5b                                           8=(2)a(2)b                                         ……..(iv)From Eq. (ii) and (iii),            1.6×10−24×10−3=1×10−35×10−4a12×10−512×10−5b                             4=(2)a(2)2=(2)a                          ∴ a=2       On substituting value of a in Eq. (iv), we get                                    8=(2)2(2)b                                 (2)=(2)b or b=1Thus, order of reaction with respect ta X is 2 and with respect to Y is 1.