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First Order Reaction

The catalytic...

Question

The catalytic decomposition of $H_{2} O_{2}$ was studied by titrating it at different intervals with $K M n O_{4}$ and the following data were obtained:
$\begin{matrix} t (s e c o n d s) & 0 & 600 & 1200 V o f K M n O_{4} (m L) & 22.8 & 13.8 & 8.3 \end{matrix}$
Calculate the velocity constant for the reaction assuming it to be a first order reaction.

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Solution

For first order reaction
$k = \frac{1}{t} l o g \frac{[A_{o}]}{[A_{t}]}$ $[A_{o}] \to$ Initial case
$[A_{t}] \to$ Final case
For $t_{1} = 0$ to $t_{2} = 600 s e c \Rightarrow t = t_{2} - t_{1} = 600 s e c$
$[A_{o}] = 22.8 m o l / L^{- 1}$ & $[A_{t}] = 13.8 m o l L^{- 1}$
$k = \frac{1}{600} l o g \frac{22.8}{13.8} = \frac{1}{600} \times 0.2180 = 3.633 \times 10^{- 4} 5^{- 1}$
For $t_{2} = 600$ to $t_{3} = 1200 s e c \Rightarrow t = t_{3} - t_{2} = 600 s e c$
$[A_{c}] = 13.8 m o l^{- 1}$ & $[A_{t}] = 8.3 m o l L^{- 1}$
$K = \frac{1}{600} l o g \frac{13.8}{8.3} = \frac{1}{600} \times 0.2208 = 3.68 \times 10^{- 4} S^{- 1}$
Since k for both conditions are approximately equal
So, rate constant for the reaction is -
$3.65 \times 10^{- 4} s e c^{- 1}$

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