The results given in the below table were obtained during kinetic studies of the following reaction: {2A + B C + D} |…

Step 1 — Determine orders:

From Exp I & II (A constant): Rate doubles when B doubles → order w.r.t. B = 2 (rate quadruples: $2.40/0.6 = 4$, so $n_B = 2$)

From Exp I & III (B constant): Rate doubles when A doubles → order w.r.t. A = 1

Rate law: $r = k[\mathrm{A}][\mathrm{B}]^2$

From Exp I: $k = \frac{6.00 \times 10^{-3}}{0.1 \times (0.1)^2} = 6.00\,\mathrm{L^2\,mol^{-2}\,min^{-1}}$

Step 2 — Find X (Exp IV): $7.20 \times 10^{-2} = 6.00 \times X \times (0.2)^2 = 6.00 \times X \times 0.04$ $X = \frac{7.20 \times 10^{-2}}{0.24} = 0.3$

Step 3 — Find Y (Exp V): $2.88 \times 10^{-1} = 6.00 \times 0.3 \times Y^2$ $Y^2 = \frac{0.288}{1.8} = 0.16 \Rightarrow Y = 0.4$

Answer: Option (3) — 0.3, 0.4