The reaction between X and Y is first order with respect to X and zero order with respect to Y. | Experiment | [X] mol L$^{-1}$ | [Y] mol L$^{-1}$ | Initial rate mol L$^{-1}$ min$^{-1}$ | |------------|------------------|------------------|---------------------------------------| | I. | 0.1 | 0.1 | $2 \times 10^{-3}$ | | II. | L | 0.2 | $4 \times 10^{-3}$ | | III. | 0.4 | 0.4 | $M \times 10^{-3}$ | | IV. | 0.1 | 0.2 | $2 \times 10^{-3}$ | Examine the data of table and calculate ratio of numerical values of M and L. (Nearest Integer)

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Chemical Kinetics

The reaction between X and Y is first order with respect to X and zero order with respect to Y.

Experiment	[X] mol L $^{- 1}$	[Y] mol L $^{- 1}$	Initial rate mol L $^{- 1}$ min $^{- 1}$
I.	0.1	0.1	$2 \times 1 0^{- 3}$
II.	L	0.2	$4 \times 1 0^{- 3}$
III.	0.4	0.4	$M \times 1 0^{- 3}$
IV.	0.1	0.2	$2 \times 1 0^{- 3}$

Examine the data of table and calculate ratio of numerical values of M and L. (Nearest Integer)

Answer: 40

Rate law:

r = k [X]^{1} [Y]^{0} = k [X]

From experiments I and II:

\frac{4 \times 1 0 ^{- 3}}{2 \times 1 0 ^{- 3}} = \frac{L}{0.1}

2 = \frac{L}{0.1} \Rightarrow L = 0.2

From experiments I and III:

\frac{M \times 1 0 ^{- 3}}{2 \times 1 0 ^{- 3}} = \frac{0.4}{0.1}

\frac{M}{2} = 4 \Rightarrow M = 8

Therefore:

\frac{M}{L} = \frac{8}{0.2} = 40

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer