First Order Kinetics

Question

A first order reaction has a rate constant of 2.303 × 10^-3 s^-1; how long will it take for the reactant concentration to fall to one-tenth of its initial value?

RankGuru · Accepted Answer

For a first order reaction, the integrated rate law is t = (2.303/k) log([A_0]/[A]), which links the elapsed time to the ratio of initial and final concentrations through the rate constant. When the concentration drops to one-tenth, [A_0]/[A] = 10, so log 10 = 1. Substituting k = 2.303 × 10^-3 s^-1 gives t = (2.303/(2.303 × 10^-3)) × 1 = 1000 s. The clean cancellation of 2.303 leaves a simple reciprocal of the rate constant. Option 100 s underestimates by a factor of ten through a logarithm error. Option 2303 s mistakenly multiplies by 2.303 again. Option 230.3 s misplaces a decimal. The exponential decay characteristic of first order kinetics is central to NCERT kinetics. A common JEE pitfall is to ignore the role of first order reaction, yet it is exactly this factor that distinguishes the correct answer from the tempting alternatives. Working through the logic step by step, rather than memorising the result, makes it clear why rate constant governs the behaviour seen here. Plausibility check: one-tenth corresponds to log 10 = 1 decade of decay, and dividing 2.303 by a rate constant of the same value scaled by 10^-3 logically yields 1000 s.

First Order Kinetics

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