Nuclear Chemistry Ques 1
1. The half life of a substance in a certain enzyme catalysed reaction is $138 s$. The time required fo the concentration of the substance to fall fron $1.28 mg L^{-1}$ to $0.04 mg L^{-1}$, is :
[2011]
(a) $414 s$
(b) $552 s$
(c) $690 s$
(d) $276 s$
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Solution:
- (c) For a first order reaction
Total time $T=$ no. of half lives $(n) \times$ half life $(t _{1 / 2})$
$ \frac{N}{N_0}=(\frac{1}{2})^{n} $
where $n=$ no. of half lives
Give $N_0$ (original amount) $=1.28 mg / l$
$N$ (amount of substance left after time $T$ )
$=0.04 mgL^{-1}$
$ \begin{aligned} \therefore \frac{0.04}{1.28} & =(\frac{1}{2})^{n} ; \frac{1}{32}=(\frac{1}{2})^{n} ;(\frac{1}{2})^{5}=(\frac{1}{2})^{n} \\ n & =5 \\ T & =5 \times 138=690 \end{aligned} $
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