Chemical Kinetics - Result Question 18
####18. A first order reaction has a rate constant of $2.303 \times 10^{-3} s^{-1}$. The time required for $40 g$ of this reactant to reduce to $10 g$ will be
[Given that $\log _{10} 2=0.3010$ ]
[NEET Odisha 2019]
(a) $602 s$
(b) $230.3 s$
(c) $301 s$
(d) $2000 s$
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Solution:
- (a) For a first order reaction
Half life period, $t _{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{2.303 \times 10^{-3}} s^{-1}$ $=300.91 s$
Now, $40 g \xrightarrow{t _{1 / 2}} 20 g \xrightarrow{t _{1 / 2}} 10 g$
So, $40 g$ substance requires 2 half life periods to reduce upto $10 g$
$\therefore \quad$ Time taken in reduction $=2 \times 300.91 s$
$ =601.82 \simeq 602 s $
