NEET UG 2010 — Chemistry Physical Chemistry

Let the order of reaction with respect to $\mathrm{A}$A is $x$x and with respect to $B$B is $y$y. Thus, $$\text { rate }=\mathrm{k}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$$ rate =k[A]x[B]y For the given cases, (I) rate $=\mathrm{k}(0.1)^{\mathrm{x}}(0.1)^{\mathrm{y}}=6.0 \times 10^{-3}$=k(0.1)x(0.1)y=6.0×10−3 (II) rate $=\mathrm{k}(0.3)^{\mathrm{x}}(0.2)^{\mathrm{y}}=7.2 \times 10^{-2}$=k(0.3)x(0.2)y=7.2×10−2 (III) rate $$=\mathrm{k}(0.3)^{\mathrm{x}}(0.40)^{\mathrm{y}}=2.88 \times 10^{-1}$$=k(0.3)x(0.40)y=2.88×10−1 (IV) rate $=\mathrm{k}(0.4)^{\mathrm{x}}(0.1)^{\mathrm{y}}=2.40 \times 10^{-2}$=k(0.4)x(0.1)y=2.40×10−2 On dividing Eq. (I) by (IV), we get $$\left(\frac{0.1}{0.4}\right)^x\left(\frac{0.1}{0.1}\right)^y=\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}$$(0.40.1​)x(0.10.1​)y=2.4×10−26.0×10−3​ or $$\left(\frac{1}{4}\right)^x=\left(\frac{1}{4}\right)^1$$(41​)x=(41​)1 $$\therefore \quad \mathrm{x}=1$$∴x=1 On dividing Eq. (II) by (III), we get $$\begin{aligned} & & \left(\frac{0.3}{0.3}\right)^x\left(\frac{0.2}{0.4}\right)^y & =\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}} \\ \text { or } & & \left(\frac{1}{2}\right)^y & =\frac{1}{4} \\ \text { or } & & \left(\frac{1}{2}\right)^y & =\left(\frac{1}{2}\right)^2 \\ \therefore & & y & =2 \end{aligned}$$ or  or ∴​​(0.30.3​)x(0.40.2​)y(21​)y(21​)yy​=2.88×10−17.2×10−2​=41​=(21​)2=2​ $\therefore$∴ Thus, rate law is, $$\text { rate }=\mathrm{k}[\mathrm{A}]^1[\mathrm{~B}]^2 \text { or }=\mathrm{k}[\mathrm{A}][\mathrm{B}]^2$$ rate =k[A]1[ B]2 or =k[A][B]2