For the reaction A → B + C, the following data were obtained:

t in seconds

0

900

1800

Cone. of A

50.8

19.7

7.62

Prove that the reaction is of first order of A to decompose to one-half.

Solution

For the first order reaction can given by

$k = \frac{2.303}{t} \log \frac{{[A]}_{0}}{{[A]}_{t}}$

At t = 900 seconds,

${[A]}_{0} = 50.8 and {[A]}_{t} = 19.7 at t = 1800 seconds, {[A]}_{0} = 50.8 and {[A]}_{t} = 7.62$
$k = \frac{2.303}{t} \log \frac{{[A]}_{0}}{{[A]}_{t}}$
(i) $k = \frac{2.303}{900 s} \log \frac{50.8}{19.7} = \frac{2.303}{900 s} \times 0.4114 = 1.048 \times 10^{- 3} s^{- 1}$

(ii) $k = \frac{2.303}{1800 s} \log \frac{50.8}{7.62} = \frac{2.303}{1800 s} \times 0.839 = 1.052 \times 10^{- 3} s^{- 1}$

Since the value of k is both case is almost same thus it is first order reaction.

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Chemical Kinetics

For the reaction A → B + C, the following data were obtained:

t in seconds

0

900

1800

Cone. of A

50.8

19.7

7.62

Prove that the reaction is of first order of A to decompose to one-half.

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Chemical Kinetics

For the reaction A → B + C, the following data were obtained: t in seconds 0 900 1800 Cone. of A 50.8 19.7 7.62 Prove that the reaction is of first order of A to decompose to one-half.

Some More Questions From Chemical Kinetics Chapter

What will be the effect of temperature on rate constant?

A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction.

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Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

For the reaction A → B + C, the following data were obtained:

t in seconds

0

900

1800

Cone. of A

50.8

19.7

7.62

Prove that the reaction is of first order of A to decompose to one-half.

The rate law for the decomposition of N₂O₅ is rate = k[N₂O₅]. What is the significance of k in this equation?