Chemistry · Chemical Kinetics · NEET
It means the rate of the reaction is proportional to the zero power of the reactant concentration. Since any number to the power zero is 1, Rate = k[R]^0 = k. So the rate stays the same (a constant) even if you add more reactant. Adding more reactant does NOT make it go faster. This matters for NEET because many students wrongly assume more reactant always means faster - here it does not.
Start from Rate = -d[R]/dt = k. Rearrange to d[R] = -k dt. Integrate both sides to get [R] = -kt + I, where I is the integration constant. At time t = 0 the concentration is [R]0, so [R]0 = I. Put I back in: [R] = [R]0 - kt. That is the integrated rate equation. From it, k = ([R]0 - [R]) / t.
If you plot concentration [R] on the y-axis against time t on the x-axis, you get a STRAIGHT LINE going down. The slope is -k and the y-intercept (starting point) is [R]0. Compare it with y = mx + c: here [R] = (-k)t + [R]0. A common NEET graph question shows this straight line with a constant negative slope - that tells you the reaction is zero order.
Half-life t(1/2) = [R]0 / 2k. It is DIRECTLY proportional to the initial concentration [R]0. So if you double [R]0, the half-life doubles. This is opposite to first order, where half-life does not depend on concentration at all. NEET has asked this exact point directly (2018).
For zero order, Rate = k, and rate has units of concentration per time = mol L^-1 s^-1. So k also has units mol L^-1 s^-1 (concentration/time). General rule: k units = (mol L^-1)^(1-n) s^-1, and for n = 0 this gives mol L^-1 s^-1. NEET tests this in matching questions.
They are uncommon but happen under special conditions. NCERT examples: some enzyme-catalysed reactions, and reactions on metal surfaces. The decomposition of gaseous ammonia on a hot platinum surface is zero order at high pressure. Also, the decomposition of phosphine on tungsten becomes zero order at high pressure when the surface is fully covered (saturated).
When the initial concentration of the reactant is doubled, the half-life period of a zero order reaction
For a certain reaction R -> Product, the plot of concentration [R] versus time is a straight line with a constant negative slope. The order of the reaction is:
2A -> B is a zero-order reaction with k = 1.0 mol L^-1 min^-1. If the initial concentration of A is 2 M, the time taken to complete 75% of the reaction is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
[R] = [R]0 - kt, where [R]0 is the initial concentration, [R] is the concentration at time t, and k is the rate constant. Rearranged, k = ([R]0 - [R]) / t.
t(1/2) = [R]0 / 2k. It depends on the initial concentration - it doubles when the initial concentration doubles.
mol L^-1 s^-1 (same units as the rate itself, because Rate = k for zero order).
Because Rate = k[R]^0, and any quantity to the power zero is 1. This happens in surface (metal-catalysed) or enzyme reactions where the surface or enzyme is saturated, so extra reactant cannot speed it up.
Zero order: rate is constant (rate = k), [R] vs t is a straight line, and half-life increases with concentration. First order: rate depends on [R], ln[R] vs t is a straight line, and half-life is constant (independent of concentration).
