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A question gives concentrations at different times and asks for the order. With no initial-rate data, find which plot of the data is a straight line. This post shows which axes to choose, how to read k from the slope, a quick test that needs no graph paper, and the graph traps that cost marks.
Which graphs does the syllabus expect?
The rationalised NCERT Class 12 chapter on Chemical Kinetics (Reprint 2026-27) derives integrated rate equations for zero- and first-order reactions and draws each as a straight line: [R] against t for zero order (Fig. 3.3), ln[R] against t for first order (Fig. 3.4), and log([R]₀/[R]) against t (Fig. 3.5). Its Table 3.4 collects them.
The JEE (Main) 2026 syllabus lists "differential and integral forms of zero and first-order reactions, their characteristics and half-lives", and the NEET (UG) 2026 syllabus uses the same words. The JEE (Advanced) 2026 brochure also stops at zero and first order. Check the document for your own exam year.
Which plot is straight for which order?
Each integrated equation can be rearranged into the form y = mx + c:
Zero order: [R] = [R]₀ − kt. Plot [R] against t. Slope = −k, intercept = [R]₀.
First order: ln[R] = ln[R]₀ − kt. Plot ln[R] against t. Slope = −k, intercept = ln[R]₀.
First order, base-10 form: log([R]₀/[R]) = kt/2.303. Plot log([R]₀/[R]) against t. Slope = k/2.303, and the line passes through the origin.
The units of the slope are the units of k: mol L⁻¹ s⁻¹ for zero order, s⁻¹ for first order. Our guide to reading order from the units of the rate constant runs that logic in reverse.
Worked example 1: is this reaction first order?
Sample numbers for a reaction A → products:
t = 0 min: [A] = 0.600 M
t = 10 min: 0.450 M
t = 20 min: 0.338 M
t = 30 min: 0.253 M
t = 40 min: 0.190 M
Differences: 0.150, 0.112, 0.085, 0.063. Not constant, so not zero order.
Ratios: 0.450/0.600 = 0.750, then 0.751, 0.749, 0.751. Constant, so first order.
Plot ln[A]: −0.511, −0.799, −1.085, −1.374, −1.661. A best-fit line through these points has slope −0.0288 min⁻¹, so k = 0.0288 min⁻¹.
Half-life: t½ = 0.693/k = 0.693/0.0288 ≈ 24 min. Check against the table: [A] should reach 0.300 M between 20 min (0.338) and 30 min (0.253), and it does.
A warning about lines that look straight. Plotted directly against t, these five [A] values give r² = 0.973, which sounds convincing. The ln[A] plot gives r² = 1.000. Over a short stretch a gentle curve can pass for a line, so trust the pattern of differences and ratios, or use data covering at least two half-lives.
Worked example 2: a zero-order line
NCERT notes that zero-order reactions are relatively uncommon; its example is the decomposition of ammonia on a hot platinum surface at high pressure. Take these sample numbers: [A] = 0.50 M at t = 0, 0.42 M at 20 s, 0.34 M at 40 s and 0.26 M at 60 s.
The drop is 0.08 M every 20 s, every time, so the reaction is zero order. Slope = −0.08/20 = −0.0040 mol L⁻¹ s⁻¹, so k = 4.0 × 10⁻³ mol L⁻¹ s⁻¹.
t½ = [A]₀/2k = 0.50/(2 × 0.0040) = 62.5 s. The line reaches zero at [A]₀/k = 125 s. A zero-order reaction is complete after exactly two of its starting half-lives; a first-order reaction never quite finishes.
Worked example 3: reading k from a log plot
A graph of log([R]₀/[R]) against t (in minutes) has slope 0.0150 min⁻¹. Then k = 2.303 × 0.0150 = 0.0345 min⁻¹. On an ln[R] plot the slope would be −k directly. Mixing the two forms is the most common slip in this chapter.
The same reading works for temperature. NCERT's Fig. 3.10 plots ln k against 1/T with slope −Eₐ/R, so a slope of −6000 K gives Eₐ = 6000 × 8.314 ≈ 49.9 kJ mol⁻¹. Our post on the Arrhenius equation and activation energy covers that graph in detail.
What about second order?
The open textbook OpenStax Chemistry 2e gives 1/[A] = kt + 1/[A]₀ for a second-order reaction: plot 1/[A] against t, and the slope is +k. Useful to know, but not required by the syllabi above.
Why do students misread these graphs?
Rodriguez and colleagues (2019) analysed written answers from 70 first-year students at a Swedish university to a chemical kinetics graph question. They looked at how far students joined the mathematics of the graph's shape to the chemistry it describes. The habit to build is simple: before calculating, say what each axis is and what the slope means for the molecules.
Which mistakes cost marks?
Calling any falling curve first order. Every reaction's [R] falls with time. Only a straight ln[R] plot shows first order.
Mixing ln and log. The 2.303 belongs with log, and only once.
Sign of the slope. A slope of −k means k is positive. Do not report a negative rate constant.
Deciding order from two points. Two points always make a straight line. Use at least three.
Treating the zero-order half-life as constant. It halves each time [R]₀ halves.
A practice routine
Write the three linear forms from memory, with what goes on each axis and what the slope is.
Make up two sample tables, one with equal differences and one with equal ratios, and swap them with a friend to identify.
For every data table in NCERT's exercises, run the difference and ratio test before plotting.
When your slope comes out with the wrong sign or units and you cannot see why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can compare each step with yours.
Tonight, take Worked example 1, change every concentration to a ratio of 0.80 per 10 minutes, and find the new k and half-life.
