Chemistry · Chemical Kinetics · NEET
A first order reaction is a reaction whose rate depends on the concentration of only one reactant, and that concentration has power 1. So rate = k[A]. If you double [A], the rate doubles. The '1' in 'first order' is the sum of the powers in the rate law. This is important for NEET because most speed-based numericals in Chemical Kinetics use first order.
The equation is k = (2.303/t) log([A]₀/[A]). Here k is the rate constant, t is the time passed, [A]₀ is the starting concentration (or amount, or mass, or pressure), and [A] is the amount left after time t. You can also write it as ln([A]₀/[A]) = kt. NEET usually gives you three of the values and asks for the fourth, so just rearrange.
The real equation uses natural log (ln). But log tables and NEET data are in base-10 log. To change ln into log₁₀ you multiply by 2.303, because ln x = 2.303 × log₁₀ x. So 2.303 is just the conversion number from natural log to normal log. You do not derive it in the exam — you memorize that first order = 2.303.
Yes. Because the formula uses the RATIO [A]₀/[A], any unit cancels out. So you can put grams, moles, pressure, or number of particles — as long as both top and bottom use the same unit. That is why a NEET question can say '40 g reduces to 10 g' and you directly use log(40/10). This trick saves time.
For first order, k has units of time⁻¹, usually s⁻¹ (per second) or min⁻¹ (per minute). It does NOT depend on concentration units. A quick check: if a question gives k as 'per second' with no mol or L in it, the reaction is first order. This is a common NEET trap where they mix up units of different orders.
In zero order, rate does not depend on concentration at all (rate = k, units mol L⁻¹ s⁻¹), and a graph of concentration vs time is a straight sloping line. In first order, rate = k[A], k has units s⁻¹, and a graph of log[A] vs time (or ln[A] vs time) is a straight line. So the key marker of first order is: ln[A] vs t gives a straight line.
Plot ln[A] (or log[A]) on the y-axis against time t on the x-axis. You get a straight line going down. Its slope is −k (for ln) or −k/2.303 (for log). NEET sometimes asks 'which plot is linear for first order' — the answer is ln[A] vs t, not [A] vs t.
A first order reaction has a rate constant of 2.303 × 10⁻³ s⁻¹. The time required for 40 g of this reactant to reduce to 10 g will be (Given log₁₀2 = 0.3010):
For a first order reaction A → Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in min⁻¹ is:
A first order reaction has a specific reaction rate of 10⁻² s⁻¹. How much time will it take for 20 g of the reactant to reduce to 5 g?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. For first order, t½ = 0.693/k, and it does not depend on the starting concentration. So no matter how much you begin with, each half-life takes the same time. This is a favourite NEET point and is the reason radioactive decay is first order.
Rate = k[A], so rate changes as [A] falls, but k (the rate constant) stays fixed at a given temperature. Rate has units mol L⁻¹ s⁻¹, while k for first order has units s⁻¹. Do not confuse the two in numericals.
Yes, radioactive decay always follows first order kinetics, and carbon dating uses the same k = (2.303/t) log(N₀/N) idea. This is why NEET links first order reactions to radioactivity and dating problems.
Three quick signs: (1) k is given in s⁻¹ or min⁻¹ (time⁻¹), (2) the words 'first order' appear, or (3) a ln[A] vs t plot is a straight line. Then use k = (2.303/t) log([A]₀/[A]).
Both work. ln([A]₀/[A]) = kt, and k = (2.303/t) log([A]₀/[A]). The 2.303 appears only when you switch to base-10 log. Use whichever the data suits; NEET usually gives base-10 log values.
