The rate for a first order reaction is 0-639 - 10-2mol dm-1 min-1 and the initial concentration is 0-2 mol dm-3- The half-life period is

0.639× 10^ 2=K× 0.2 k=0693× 10^ 2/0.2 t_1/2=0.693/k=0.693× 0.2/0.693× 10^ 2 = 20 min = 1200 s ...

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Standard XII

Chemistry

First Order Reaction

The rate for ...

Question

The rate for a first order reaction is $0.639 \times 10^{- 2} m o l d m^{- 1} m i n^{- 1}$ and the initial concentration is 0.2 mol $d m^{- 3}$ . The half-life period is

1200 s

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0.33 s

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20 s

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600 s

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Solution

The correct option is A 1200 s
$0.639 \times 10^{- 2} = K \times 0.2 k = \frac{0693 \times 10^{- 2}}{0.2} t_{\frac{1}{2}} = \frac{0.693}{k} = \frac{0.693 \times 0.2}{0.693 \times 10^{- 2}}$
= 20 min = 1200 s

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The rate for a first order reaction is 0.639×10−2mol dm−1min−1 and the initial concentration is 0.2 mol dm−3. The half-life period is

The correct option is A 1200 s0.639×10−2=K×0.2k=0693×10−20.2t12=0.693k=0.693×0.20.693×10−2 = 20 min = 1200 s

The rate for a first order reaction is $0.639 \times 10^{- 2} m o l d m^{- 1} m i n^{- 1}$ and the initial concentration is 0.2 mol $d m^{- 3}$ . The half-life period is

The correct option is A 1200 s
$0.639 \times 10^{- 2} = K \times 0.2 k = \frac{0693 \times 10^{- 2}}{0.2} t_{\frac{1}{2}} = \frac{0.693}{k} = \frac{0.693 \times 0.2}{0.693 \times 10^{- 2}}$
= 20 min = 1200 s