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 ^ - 1 I 0 . 1 0 . 1 2 × 10 ^ - 3 II L 0 . 2 4 × 10 ^ - 3

Question

Quantrex Academy · Accepted Answer

Correct answer: 40

Given that reaction is first order with respect to X and of zero order with respect to Y.
Means $r = k$ \left[X\right]{ $\left[Y\right]}^{0}$ Where k = rate constant.
Using data $I$ & $II$
$\dfrac{4 \times {10}^{- 3}}{2 \times {10}^{- 3}} = \left(\dfrac{L}{0 . 1}\right){\left(\dfrac{0 . 2}{0 . 1}\right)}^{0} ⇒ L = 0 . 2$
Using data from $I$ & $III$
$\dfrac{M \times {10}^{- 3}}{2 \times {10}^{- 3}} = \dfrac{0 . 4}{0 . 1} ⇒ M = 8$
Not the ratio which is asked in question
$\dfrac{M}{L} = \dfrac{8}{0 . 2} = 40$

Experiment	$\dfrac{\left[X\right]}{{molL}^{- 1}}$	$\dfrac{\left[Y\right]}{{molL}^{- 1}}$	$\dfrac{Initial rate}{mol{L}^{- 1}\min^{- 1}}$
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...

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