Chemistry · Chemical Kinetics · NEET
For first order, t½ = 0.693/k. Here k is the rate constant. The number 0.693 comes from ln 2 (natural log of 2), because after one half-life the reactant drops to half. You do NOT need the starting concentration to find t½ for first order.
Look at the formula t½ = 0.693/k. There is no [A]₀ in it. This means whether you start with 10 mol or 100 mol, the time to fall to half is the same. This is a special mark of first order reactions. NEET loves testing this: for zero order t½ depends on [A]₀, for second order it depends on 1/[A]₀, but for first order it is constant.
99% completed means 99% is used up, so only 1% is left. So [A]₀/[A] = 100/1 = 100. Put this into t = (2.303/k) log([A]₀/[A]) = (2.303/k) × log 100 = (2.303/k) × 2 = 4.606/k. So the time for 99% is always 4.606/k, a fixed number for any first order reaction.
99.9% done means 0.1% is left, so [A]₀/[A] = 1000. Since 1000 ≈ 2¹⁰ = 1024, it takes about 10 half-lives. Quick rule: each half-life divides the reactant by 2. After 10 half-lives you have (1/2)¹⁰ ≈ 1/1000 left, which is 99.9% completion.
Always start from t = (2.303/k) log([A]₀/[A]). For n% completion, the fraction left is (100 − n)%, so [A]₀/[A] = 100/(100 − n). Example: 90% done → ratio 10 → t = (2.303/k)×1 = 2.303/k. 75% done → ratio 4 → t = (2.303/k)×0.602. Learn to convert 'percent completed' into the ratio first, then take the log.
No. For 90% completion the ratio is 10, so t = 2.303/k (one log). For 99% the ratio is 100, so t = 4.606/k (two logs), which is exactly twice as long. Do not mix these up. Also note: time for 99% = 2 × time for 90% for the same first order reaction.
If the rate constant for a first order reaction is k, the time t required for the completion of 99% of the reaction is given by:
If the half-life t½ for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:
The rate constant for a first order reaction is 4.606 × 10⁻³ s⁻¹. The time required to reduce 2.0 g of the reactant to 0.2 g is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. t½ = 0.693/k is ONLY for first order. For zero order t½ = [A]₀/2k, and for second order t½ = 1/(k[A]₀). Only first order has a half-life that does not depend on starting concentration.
For first order, k has units of time⁻¹, such as s⁻¹ or min⁻¹. If a question gives k in s⁻¹, that itself tells you the reaction is first order. This trick appears in NEET 2025.
It is the conversion factor from natural log to base-10 log: ln x = 2.303 log x. We use base-10 log because log tables and quick values (log 2 = 0.301, log 10 = 1) are easy to use in the exam.
Radioactive decay follows exactly the same first order maths. The half-life of a radioactive sample is also 0.693/λ, where λ is the decay constant. Carbon dating uses this idea, which is the next concept in this chapter.
