Chemical Kinetics - Result Question 21
####21. 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 $5 g$ ?
(a) $138.6 sec$
(b) $346.5 sec$
(c) $693.0 sec$
(d) $238.6 sec$
[2017]
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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,
$t=2 \times \frac{0.693}{10^{-2}} sec=138.6 sec$.
$N=N_0(\frac{1}{2})^{n}$
where $N=$ mass after $n$ number of half life
$n=$ number of half life
$N_0=$ Initital mass
