A second-order reaction requires $70$ min to change the concentration of reactants from $0.08$ M to $0.01$ M. How much time will it require to become $0.04$ M?

10

min

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20

min

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30

min

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40

min

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Solution

The correct option is A $10$ min
For second order, when $(a - x)$ $= 0.01$ M

$k_{2} = \frac{1}{t} \cdot \frac{x}{a (a - x)}$

$k_{2} = \frac{1}{70} \cdot \frac{0.07}{0.08 \times 0.01}$ (i)

When $(a - x)$ $= 0.04$ M

$k_{2} = \frac{1}{t} \cdot \frac{0.04}{0.08 \times 0.04}$ (ii)

From equations (i) and (ii),

$\frac{0.07}{70 \times 0.08 \times 0.01} = \frac{0.04}{t \times 0.08 \times 0.04}$

$t = 10$ min

Hence, option $A$ is correct.

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A second-order reaction requires 70 min to change the concentration of reactants from 0.08 M to 0.01 M. How much time will it require to become 0.04 M?

The correct option is A 10 minFor second order, when (a−x) =0.01 Mk2=1t⋅xa(a−x)k2=170⋅0.070.08×0.01 (i)When (a−x) =0.04 Mk2=1t⋅0.040.08×0.04 (ii)From equations (i) and (ii),0.0770×0.08×0.01=0.04t×0.08×0.04t=10 minHence, option A is correct.

A second-order reaction requires $70$ min to change the concentration of reactants from $0.08$ M to $0.01$ M. How much time will it require to become $0.04$ M?