JEE MAIN - Chemistry (2022 - 29th July Morning Shift - No. 21)
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
40
Explanation
$$r=k[X][Y]^{0}=k[X]$$
Using I & II
$\frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\left(\frac{L}{0.1}\right) \quad \Rightarrow \quad \mathrm{L}=0.2$
Using I & III
$\frac{M \times 10^{-3}}{2 \times 10^{-3}}=\frac{0.4}{0.1} \quad \Rightarrow \quad \mathrm{M}=8$
$\frac{M}{L}=\frac{8}{0.2}=40$
Using I & II
$\frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\left(\frac{L}{0.1}\right) \quad \Rightarrow \quad \mathrm{L}=0.2$
Using I & III
$\frac{M \times 10^{-3}}{2 \times 10^{-3}}=\frac{0.4}{0.1} \quad \Rightarrow \quad \mathrm{M}=8$
$\frac{M}{L}=\frac{8}{0.2}=40$
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