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Standard XII

Chemistry

First Order Reaction

Decomposition...

Question

Decomposition of $H_{2} O_{2}$ is a first order reaction. A solution of $H_{2} O_{2}$ labelled as 20 volumes was left open. Due to this, some $H_{2} O_{2}$ decomposed. To determine the new volume strength after 6 hours, 10 mL of this solution was diluted to 100 mL. 10 mL of this diluted solution was titrated against 25 mL of 0.025 N $K M n O_{4}$ solution under acidic conditions. The rate constant for decomposition of $H_{2} O_{2}$ is:

8.1 \times 10^{- 5} M s^{- 1}

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8.64 \times 10^{- 6} M s^{- 1}

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9.23 \times 10^{- 6} M s^{- 1}

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9.65 \times 10^{- 6} M s^{- 1}

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Solution

The correct option is A $8.1 \times 10^{- 5} M s^{- 1}$
Equivalent of $H_{2} O_{2}$ after dilution $= \frac{25}{1000} \times 0.025 = 6.25 \times 10^{- 4}$
Normality $= \frac{6.25 \times 10^{- 4}}{10} \times 1000 = 0.0625 N$
Normality before dilution $= 0.0625 \times 10 = 0.625 N$
Volume strength $= 5.6 \times$ normality $= 5.6 \times 0.625 = 3.5 N$
The integrated rate law for the first order reaction is $k = \frac{2.303}{t} l o g \frac{[A]_{0}}{[A]_{t}} = \frac{2.303}{6 \times 3600} l o g \frac{20}{3.5} = 8.1 \times 10^{- 5} M s^{- 1}$

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Similar questions

Q. Decomposition of

H_{2} O_{2}

is a first order reaction. A solution of

H_{2} O_{2}

labelled as

20

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H_{2} O_{2}

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6

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