The reaction between X and Y is first order with respect to X and zero order with respect to Y . Experiment X molL - 1 Y molL - 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 Examine…

Question

The reaction between X and Y is first order with respect to X and zero order with respect to Y . Experiment X molL - 1 Y molL - 1 Initial &#160; rate mol L - 1 min - 1 I 0 . 1 0 . 1 2 &#215; 10 - 3 II L 0 . 2 4 &#215; 10 - 3 III 0 . 4 0 . 4 M &#215; 10 - 3 IV 0 . 1 0 . 2 2 &#215; 10 - 3 Examine the data of table and calculate ratio of numerical values of M and L .

Aicharya · Accepted Answer

Given that reaction is first order with respect to X and of zero order with respect to Y.
Means $r = k [X] {[Y]}^{0}$ Where k = rate constant.
Using data $I$ & $II$
$\frac{4 \times 10^{- 3}}{2 \times 10^{- 3}} = (\frac{L}{0.1}) {(\frac{0.2}{0.1})}^{0} \Rightarrow L = 0.2$
Using data from $I$ & $III$
$\frac{M \times 10^{- 3}}{2 \times 10^{- 3}} = \frac{0.4}{0.1} \Rightarrow M = 8$
Not the ratio which is asked in question
$\frac{M}{L} = \frac{8}{0.2} = 40$

Experiment	$\frac{[X]}{{molL}^{- 1}}$	$\frac{[Y]}{{molL}^{- 1}}$	$\frac{Initial rate}{mol L^{- 1} \min^{- 1}}$
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	$0.1$	$0.2$	$2 \times 10^{- 3}$

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

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