1.

The half life of a substance in a certain enzyme-catalysed reaction is 138 s . The time required for the concentration of the substance to fall from 1.28 mg L^(-1) to 0.04 mg L^(-1) is

Answer»

690s
276 s
414 s
552s

Solution :Enzyme catalysed reactions are initially FOLLOW FIRST ORDER kinetics when concentration decreases `1.28mg l^(-1)` to 0.04 mg `l^(-1)` . Then five half-life completed
No. of half-lives =5
So , TIMES required = `5 XX 138 = 690 `s


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