Chemical Kinetics - Result Question 17
####20. Decomposition of $H _2 O _2$ follows a first order reaction. In 50 min, the concentration of $H _2 O _2$ decreases from 0.5 to $0.125 M$ in one such decomposition. When the concentration of $H _2 O _2$ reaches $0.05 M$, the rate of formation of $O _2$ will be (2016 Main)
(a) $6.93 \times 10^{-4} mol min^{-1}$
(b) $2.66 L min^{-1}$ at STP
(c) $1.34 \times 10^{-2} mol min^{-1}$
(d) $6.93 \times 10^{-2} mol min^{-1}$
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Answer:
Correct Answer: 20. (b)
Solution:
- For first order reaction, $k=\frac{2.303}{t} \log \frac{a}{a-x}$
Given, $\quad t=50 min, a=0.5 M, a-x=0.125 M$
$\therefore \quad k=\frac{2.303}{50} \log \frac{0.5}{0.125}=0.0277 min^{-1}$
Now, as per reaction
$$ \begin{gathered} 2 H _2 O _2 \longrightarrow 2 H _2 O+O _2 \ -\frac{1}{2} \frac{d\left[H _2 O _2\right]}{d t}=\frac{1}{2} \frac{d\left[H _2 O\right]}{d t}=\frac{d\left[O _2\right]}{d t} \end{gathered} $$
Rate of reaction, $-\frac{d\left[H _2 O _2\right]}{d t}=k\left[H _2 O _2\right]$
$\therefore \quad \frac{d\left[O _2\right]}{d t}=-\frac{1}{2} \frac{d\left[H _2 O _2\right]}{d t}=\frac{1}{2} k\left[H _2 O _2\right]$
When the concentration of $H _2 O _2$ reaches $0.05 M$,
$$ \begin{aligned} \frac{d\left[O _2\right]}{d t} & =\frac{1}{2} \times 0.0277 \times 0.05 \ \text { or } \quad \frac{d\left[O _2\right]}{d t} & =6.93 \times 10^{-4} mol min^{-1} \end{aligned} $$
$$ \text { [from Eq. (i)] } $$
Alternative Method
In fifty minutes, the concentration of $H _2 O _2$ decreases from 0.5 to $0.125 M$ or in one half-life, concentration of $H _2 O _2$ decreases from 0.5 to $0.25 M$. In two half-lives, concentration of $H _2 O _2$ decreases from 0.5 to $0.125 M$ or $2 t _{1 / 2}=50 min$
$$ \begin{array}{cc} & t _{1 / 2}=25 min \ \therefore & k=\left(\frac{0.693}{25}\right) min^{-1} \ \text { or } \frac{d\left[O _2\right]}{d t}=-\frac{1}{2} \frac{d\left[H _2 O _2\right]}{d t}=\frac{k\left[H _2 O _2\right]}{2}=6.93 \times 10^{-4} mol min^{-1} \end{array} $$
