A reaction, which is second-order, has a rate constant of $0.002 L m o l^{- 1} s^{- 1}$ . If the initial conc. of the reactant is 0.2 M, how long will it take for the concentration to become 0.0400 M?

1000 sec

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400 sec

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200 sec

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10,000 sec

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Solution

The correct option is A 10,000 sec
$\frac{1}{a} = \frac{1}{a_{0}} + k t$
$\frac{1}{0.04} = \frac{1}{0.2} + 0.002 t$
$t = 10000 s e c$

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A reaction, which is second-order, has a rate constant of 0.002Lmol−1s−1. If the initial conc. of the reactant is 0.2 M, how long will it take for the concentration to become 0.0400 M?

The correct option is A 10,000 sec1a=1a0+kt10.04=10.2+0.002tt=10000sec

A reaction, which is second-order, has a rate constant of $0.002 L m o l^{- 1} s^{- 1}$ . If the initial conc. of the reactant is 0.2 M, how long will it take for the concentration to become 0.0400 M?

The correct option is A 10,000 sec
$\frac{1}{a} = \frac{1}{a_{0}} + k t$
$\frac{1}{0.04} = \frac{1}{0.2} + 0.002 t$
$t = 10000 s e c$