NEET UG 2024 — Chemistry Physical Chemistry

Let the rate equation is Rate $=\mathrm{k}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$=k[A]x[B]y Therefore, we can write $2 \times 10^{-3}=k[0.1] \times\left[0.1 \rrbracket^y\right.$2×10−3=k[0.1]×[0.1]]y...(i) $4 \times 10^{-3}=k[0.2]^x[0.1]^y$4×10−3=k[0.2]x[0.1]y...(ii) $1.6 \times 10^{-2}=\mathrm{k}[0.2]^{\times}[0.2]^{\mathrm{y}}$1.6×10−2=k[0.2]×[0.2]y...(iii) (ii) $\div$÷ (i); $\frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\frac{k[0.2]^x[0.1]^y}{k[0.1]^x[0.1]^y}$2×10−34×10−3​=k[0.1]x[0.1]yk[0.2]x[0.1]y​ $\Rightarrow \quad \frac{2}{1}=\frac{(0.2)^x}{(0.1)^x}=\left(\frac{2}{1}\right)^x$⇒12​=(0.1)x(0.2)x​=(12​)x $\therefore \quad x=1$∴x=1 (ii) $\div$÷ (iii); $\frac{4 \times 10^{-3}}{1.6 \times 10^{-2}}=\frac{k[0.2]^{\mathrm{x}}[0.1]^y}{\mathrm{k}[0.2]^{\mathrm{x}}[0.2]^{\mathrm{y}}}$1.6×10−24×10−3​=k[0.2]x[0.2]yk[0.2]x[0.1]y​ $\Rightarrow \quad \frac{1}{4}=\frac{(0.1)^y}{(0.2)^y}=\left(\frac{1}{2}\right)^y$⇒41​=(0.2)y(0.1)y​=(21​)y $\therefore \quad y=2$∴y=2 $\therefore \quad$∴ Rate $=k[A]^1[B]^2$=k[A]1[B]2 First order with respect to A while second order with respect to B .