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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)` isA. 690sB. 276 sC. 414 sD. 552 s |
Answer» Correct Answer - a Enzyme catalysed reactions are initially follow first order kinetics when concentration decreases `1.28 mg 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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