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First Order Reaction

H 2 O 2→ H 2 ...

Question

$H_{2} O_{2} \to H_{2} O + O_{2}$
The rate constant of this reaction is $2 \times 10^{- 2} m i n^{- 1}$ (at certain temperature) and reaction is started with $0.5 M$ concentration of $H_{2} O_{2}$ , after $23.03 m i n$ the reaction was suddenly stopped by lowering the temperature and remaining $H_{2} O_{2}$ was completely reacted with $N a O C l$ , if number of moles of $O_{2}$ produced in second reaction is $\frac{1}{x}$ . Find $x$ .
(Given : $N a O C l + H_{2} O_{2} \to O_{2} + N a C l + H_{2} O$ $l o g 1.5 = 0.2$ )

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Solution

$H_{2} O_{2} \to H_{2} O + O_{2}$
$N a O C l + H_{2} O_{2} \to O_{2} + N a C l + H_{2} O$
In 1 min $0.02 m o l / L H_{2} O_{2}$ decomposes so in $20 m i n$
$k = \frac{2.303}{t} l o g \frac{a}{a - x}$
$0.02 = \frac{2.303}{23.03} l o g \frac{0.5}{a - x}$
$0.02 = \frac{2.303}{23.03} l o g \frac{0.5}{a - x}$
$0.2 = l o g \frac{0.5}{a - x} (t a k e l o g 1.5 = 2)$
$1.5 = \frac{0.5}{a - x}$
$a - x = \frac{0.5}{1.5} = \frac{1}{3}$
$\frac{1}{3} m o l e H_{2} O_{2} is left so \frac{1}{3} m o l e O_{2} is produced in second reaction$

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