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Standard XII

Chemistry

Derivation of Kp and Kc

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 M $K M n O_{4}$ solution under acidic conditions. Calculate the rate constant for decomposition of $H_{2} O_{2}$ .

k = 0.022 h r^{- 1}

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k = 0.044 h r^{- 1}

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k = 0.011 h r^{- 1}

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None of these

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Solution

The correct option is A $k = 0.022 h r^{- 1}$
$2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$
ml. eq of $H_{2} O_{2}$ in 10 ml diluted
$= M_{1} e q . o f K M n O_{4} t i t r a t e d = 25 \times .025 \times 5 = 3.125$
ml. eq in 100 ml solu. = 31.25
but 22400 ml $O_{2} = 68 g m H_{2} O_{2}$

$1 m l = \frac{68}{22400}$

$20 m l = \frac{68 \times 20}{22400} i n 1 m l H_{2} O_{2}$

$= \frac{68 \times 20 \times 10}{22400} i n 10 m l H_{2} O_{2}$

No. of Meq. in 10 ml of 20%kg ml $H_{2} O_{2}$

$I n i t i a l l y = \frac{68 \times 20 \times 10 \times 1000}{17 \times 22400} = 35.71$

$\Rightarrow$ $k = \frac{2.303}{6} l o g (\frac{35.71}{31.25})$
$k = 0.022 h r^{- 1}$

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273

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