Chemical Kinetics Ques 14

14. A first order reaction has a specific reaction rate of $10^{-2} s^{-1}$. How much time will it take for $20 g$ of the reactant to reduce to $5 g$ ?

[2017]

(a) $138.6 sec$

(b) $346.5 sec$

(c) $693.0 sec$

(d) $238.6 sec$

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Answer:

Correct Answer: 14.(a)

Solution:

(a) Half life for a first order reaction,

$ \begin{aligned} & t _{1 / 2}=\frac{0.693}{K} \\ & \text{ So, } t _{1 / 2}=\frac{0.693}{10^{-2}} sec \end{aligned} $

Also, for the reduction of $20 g$ of reactant to $5 g$, two half lives will be required.

$\therefore$ For $20 g$ of the reactant to reduce to $5 g$, timetaken,

$t=2 \times \frac{0.693}{10^{-2}} sec=138.6 sec$.