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 $\left(\frac{k _1}{k _0}\right)$ 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. (d)
Solution:
- For first order reaction $t _{1 / 2}=\frac{\ln 2}{k _1}=40 s$
For zero order reaction $t _{1 / 2}=\frac{[A] _0}{2 k _0}=20 s$
$$ \begin{aligned} & \Rightarrow \text { Eq. (ii)/(i) } \quad=\frac{1}{2}=\frac{[A] _0}{2 k _0} \times \frac{k _1}{\ln 2} \ & \Rightarrow \quad \frac{k _1}{k _0}=\frac{\ln 2}{[A] _0}=\frac{0.693}{1.386}=0.5 \end{aligned} $$
