The half-life of a substance in a certain enzyme catalysed 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 is:

Question

Accepted Answer

690 s. To find the time required for the concentration to fall from 1.28 mg L⁻¹ to 0.04 mg L⁻¹, we can use the formula for first-order kinetics: t = (ln(C₀/C)) / (ln(2) / t₁₂), where C₀ is the initial concentration, C is the final concentration, and t₁₂ is the half-life.

The half-life of a substance in a certain enzyme catalysed 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 is:

Solution

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