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. (a)
Solution:
- For first order reaction, $k=\dfrac{2.303}{t} \log \dfrac{a}{a-x}$
Given, $\quad t=50$ $ min, a=0.5$ $ M, a-x=0.125 $ $M$
$\therefore \quad k=\dfrac{2.303}{50} \log \dfrac{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 \\ -\dfrac{1}{2} \dfrac{d\left[H _2 O _2\right]}{d t}=\dfrac{1}{2} \dfrac{d\left[H _2 O\right]}{d t}=\dfrac{d\left[O _2\right]}{d t} \end{gathered} $
Rate of reaction, $-\dfrac{d\left[H _2 O _2\right]}{d t}=k\left[H _2 O _2\right]$
$\therefore \quad \dfrac{d\left[O _2\right]}{d t}=-\dfrac{1}{2} \dfrac{d\left[H _2 O _2\right]}{d t}=\dfrac{1}{2} k\left[H _2 O _2\right]$$\quad$ …….. (i)
When the concentration of $H _2 O _2$ reaches $0.05 $ $M$,
$ \begin{aligned} \dfrac{d\left[O _2\right]}{d t} & =\dfrac{1}{2} \times 0.0277 \times 0.05 \quad \text {[from eq. (i)]} \\ \text { or } \quad \dfrac{d\left[O _2\right]}{d t} & =6.93 \times 10^{-4} mol \hspace {1mm} min^{-1} \end{aligned} $
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$
$ t _{1 / 2}=25 min $
$\therefore k=\left(\dfrac{0.693}{25}\right) min^{-1} $
$\text { or } \dfrac{d\left[O _2\right]}{d t}=-\dfrac{1}{2} \dfrac{d\left[H _2 O _2\right]}{d t}=\dfrac{k\left[H _2 O _2\right]}{2}=6.93 \times 10^{-4} $ $mol $ $min^{-1}$
BETA
