Chemical Kinetics - Result Question 28
28. Under the same reaction conditions, initial concentration of $1.386$ $ mol$ $ dm^{-3}$ of a substance becomes half in $40$ $ s$ and $20$ $ s$ through first order and zero order kinetics respectively. Ratio $\left(\dfrac{k _1}{k _0}\right)$ of the rate constants for first order $\left(k _1\right)$ and zero order $\left(k _0\right)$ of the reaction is
(a) $0.5 $ $mol^{-1}$ $ dm^{3}$
(b) $1.0$ $ mol $ $dm^{-3}$
(c) $1.5$ $ mol $ $dm^{-3}$
(d) $2.0$ $ mol^{-1}$ $ dm^{3}$
(2008, 3M)
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Answer:
Correct Answer: 28. (a)
Solution:
- For first order reaction $t _{1 / 2}=\dfrac{\ln 2}{k _1}=40 $ $s$$\quad$ …….. (i)
For zero order reaction $t _{1 / 2}=\dfrac{[A] _0}{2 k _0}=20 $ $s$$\quad$ …….. (ii)
$ \begin{aligned} & \Rightarrow \quad \text { Eq. (ii)/(i) } \quad=\dfrac{1}{2}=\dfrac{[A] _0}{2 k _0} \times \dfrac{k _1}{\ln 2} \\ & \Rightarrow \quad \dfrac{k _1}{k _0}=\dfrac{\ln 2}{[A] _0}=\dfrac{0.693}{1.386}=0.5 \end{aligned} $
BETA
