The reaction between X and Y is first order with respect to X and zero order with respect to Y.
| Experiment | ${{[X]} \over {mol\,{L^{ - 1}}}}$ | ${{[Y]} \over {mol\,{L^{ - 1}}}}$ | ${{Initial\,rate} \over {mol\,{L^{ - 1}}\,{{\min }^{ - 1}}}}$ |
|---|---|---|---|
| I | 0.1 | 0.1 | $2 \times {10^{ - 3}}$ |
| I | 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)
Answer (integer)
40
Solution
$r=k[X][Y]^{0}=k[X]$
<br/><br/>
Using I & II
<br/><br/>
$\frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\left(\frac{L}{0.1}\right) \quad \Rightarrow \quad \mathrm{L}=0.2$
<br/><br/>
Using I & III
<br/><br/>
$\frac{M \times 10^{-3}}{2 \times 10^{-3}}=\frac{0.4}{0.1} \quad \Rightarrow \quad \mathrm{M}=8$
<br/><br/>
$\frac{M}{L}=\frac{8}{0.2}=40$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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