1.

The half life of a substance in a certain enzyme catalyzed reaction is 138s. The time required for the concentration of the substance to fall from `1.28 mg L^(-1) to 0.04 mg L^(-1)` :A. 267sB. 414sC. 552sD. 690s

Answer» Correct Answer - A
Enzyme catalyzed reactions follow first order Kinetics, where `t_(1//2)` is constant. At the end of every half life, half of the reactant is consumed. Thus, `1.28mgL^(-1)overset(t_(1//2))(rarr)0.64mgL_(-1)overset(t_(1//2))(rarr)0.32mgL^(-1)`overset(t_(1//2))(rarr)0.16mgL_(-1)overset(t_(1//2))(rarr)0.08mgL_(-1)overset(t_(1//2))(rarr)0.04mgL_(-1)`
Since five half-lives are involved to reduce `1.28mhL^(-1)` to `0.04mgL^(-1)` , the total times is
`5xx138s=690s`
Alternatively
`0.04mgL^(-1)=(1.28mgL^(-1))/((2)^(n))`
or `2^(n)=(1.28mgL^(-1))/(0.04mgL^(-1))=32=(2)^(5)`
Hence, `n=5` , i.e., five half lives are involved.


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