The rate of a first-order reaction is 0.04 mol L^-1s^-1 at 10 seconds and 0.03 mol L^-1s^-1 at 20 seconds after initiation of the reaction. The half-l - Chemistry MCQ Chemistry | ExamTest.live

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The rate of a first-order reaction is 0.04 mol L^{- 1} s^{- 1} at 10 seconds and 0.03 mol L^{- 1} s^{- 1} at 20 seconds after initiation of the reaction. The half-life period of the reaction is:

A $34.1 s$
B $44.1 s$
C $54.1 s$
D $24.1 s$

Solution & Step-by-step Explanation

For a first-order reaction, Rate = k [A] . So \frac{Rate _{1}}{Rate _{2}} = \frac{[ A ] _{1}}{[ A ] _{2}} . k = \frac{2.303}{t _{2} - t _{1}} lo g \frac{[ A ] _{1}}{[ A ] _{2}} = \frac{2.303}{20 - 10} lo g \frac{0.04}{0.03} = \frac{2.303}{10} lo g (1.33) k = 0.2303 \times 0.124 \approx 0.0287 s^{- 1} t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0287} \approx 24.1 s .

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The rate of a first-order reaction is 0.04 mol L^{- 1} s^{- 1} at 10 seconds and 0.03 mol L^{- 1} s^{- 1} at 20 seconds after initiation of the reaction. The half-life period of the reaction is:

34.1 s

44.1 s

54.1 s

24.1 s

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Chemical Kinetics First Order Reaction

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The rate of a first-order reaction is $0.04 mol L^{- 1} s^{- 1}$ at $10 seconds$ and $0.03 mol L^{- 1} s^{- 1}$ at $20 seconds$ after initiation of the reaction. The half-life period of the reaction is:

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The rate of a first-order reaction is 0.04mol L−1s−1 at 10seconds and 0.03mol L−1s−1 at 20seconds after initiation of the reaction. The half-life period of the reaction is:

Solution & Step-by-step Explanation

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Discussion

The rate of a first-order reaction is $0.04 mol L^{- 1} s^{- 1}$ at $10 seconds$ and $0.03 mol L^{- 1} s^{- 1}$ at $20 seconds$ after initiation of the reaction. The half-life period of the reaction is: