Question

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

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

Answer 1

Correct Answer: 40

Experiment	[X] (mol L-1)	[Y] (mol L-1)	Initial Rate (mol L-1 min-1)
I	0.1	0.1	2 × 10-3
II	L	0.2	4 × 10-3
III	0.4	0.4	M × 10-3
IV	0.1	0.2	2 × 10-3

The rate law for the reaction is given by: Rate = k[X], indicating first order with respect to X and zero order with respect to Y.

Using Experiments I and IV (same [X] and different [Y]):
Rate_I = Rate_IV = 2 × 10^-3 mol L^-1 min^-1

Using Experiments I and II (change in [X]): 
(Rate_II) / (Rate_I) = ([X]_II / [X]_I), assuming [Y] does not affect rate.
4 × 10^-3 / 2 × 10^-3 = L / 0.1
L = 0.2 mol L^-1

Using Experiments I and III:
(Rate_III) / (Rate_I) = ([X]_III / [X]_I)
M × 10^-3 / 2 × 10^-3 = 0.4 / 0.1
M = 8

The ratio M/L = 8 / 0.2 = 40.
Since 40 is within the range of 40 to 40, the solution is validated.

Answer 2

\(Rate [X]^1[Y]^0\)
\(Rate = k[X]\)
From Exp I and II,
\(\frac{4 \times 10^{-3}}{2 \times 10^{-3}} = \left(\frac{L}{0.1}\right)^1 \left(\frac{0.2}{0.1}\right)^0\)

\(2 = (10 L)^1.\)
Hence L = 0.2 mol/L
From Exp III and IV,
\(\frac{M \times 10^{-3}}{2 \times 10^{-3}} = \left(\frac{0.4}{0.1}\right) \left(\frac{0.4}{0.2}\right)^0\)

\(\frac{M}{2} = 4\)
\(M = 8\)
\(\frac{M}{L} = \frac{8}{0.2}\)

\(\frac{M}{L} = 40\)
So, the answer is 40.