JEE Main · 2019 · Shift-IhardCK-113

In the following reaction: xA yB _{10}[-{d[{A}]}{dt}] = _{10}[-{d[{B}]}{dt}] + 0.3010 'A' and 'B' respectively can be:

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

In the following reaction: $\mathrm{xA \rightarrow yB}$ $\log_{10}\left[-\frac{d[\mathrm{A}]}{dt}\right] = \log_{10}\left[-\frac{d[\mathrm{B}]}{dt}\right] + 0.3010$

'A' and 'B' respectively can be:

Options

a
$\mathrm{C_2H_4}$ and $\mathrm{C_4H_8}$
✓
b
$\mathrm{C_2H_2}$ and $\mathrm{C_6H_6}$
c
n-Butane and Iso-butane
d
$\mathrm{N_2O_4}$ and $\mathrm{NO_2}$

Correct Answera

$\mathrm{C_2H_4}$ and $\mathrm{C_4H_8}$

Detailed Solution

From the equation: $\log_{10}\left[-\frac{d[\mathrm{A}]}{dt}\right] - \log_{10}\left[-\frac{d[\mathrm{B}]}{dt}\right] = 0.3010 = \log_{10} 2$ $\frac{-d[\mathrm{A}]/dt}{d[\mathrm{B}]/dt} = 2$ For $\mathrm{C_2H_4 \rightarrow C_4H_8}$ , $2\mathrm{C_2H_4} \rightarrow \mathrm{C_4H_8}$ , so $-\frac{d[\mathrm{A}]}{dt} = 2 \frac{d[\mathrm{B}]}{dt}$ .

Answer: Option (1)

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