Reading a graph in four steps
Many multiple-choice questions and FRQs give a graph or table that already holds the answer. Reading it in order avoids careless mistakes:
- Title and legend. What does the graph show? How many curves are there, and what condition does each represent?
- Axes, units and scale. What quantity is on each axis? Do the units include a factor like 10⁻³ or 10⁻⁵? Is the y-axis concentration or ln(concentration)? Does the scale start at 0?
- Trend. Rising or falling? Straight or curved? Is there a plateau or a sharp jump?
- Key points. Intercepts, slopes, peaks, inflection points, crossings, the half-equivalence point, the equivalence point: the quantity asked for usually corresponds to one of these.
Key points of common graphs
- Weak acid titrated with strong base. Look at the start, the half-equivalence point and the equivalence point. It tells you that pH = pKa at half-equivalence, and the equivalence volume gives the moles of acid.
- [A], ln[A] or 1/[A] vs time. Look for the one that is straight. It tells you the reaction order and how k relates to the slope.
- Reaction energy profile. Look at the heights of the peaks and the start vs the end. It tells you the activation energy, ΔH, the number of steps and the slowest step.
- Molecular speed distribution. Look at the position and width of the peak. It tells you which temperature is higher and how many molecules exceed Ea.
- Photoelectron spectrum. Look at the peak positions and relative heights. It tells you the binding energies and the electrons in each subshell. | Beer’s law calibration curve | Slope | Concentration from absorbance |
Data tables: control variables
Data-table questions often give several trials in which one condition changes at a time. Find two trials where only one quantity differs and see how the result changes; that shows the effect of that quantity, as in finding reaction orders from initial rates (Example 1).
Pick a graph, then pick an answer to see the matching spot on the graph:
25.00 mL of 0.100 M weak acid HA is titrated with 0.100 M NaOH, giving the curve below. At which point does the pH equal the pKa of HA?
At A there’s only the weak acid HA; the pH comes from its partial ionization, about 2.9, which is below the pKa.
At B exactly half the NaOH needed for equivalence has been added, so the HA left equals the A⁻ formed, and pH = pKa ≈ 4.7.
C is the equivalence point. The solution is mostly A⁻, with a pH of about 8.7; it’s above 7 because A⁻ is a weak base.
At D there’s excess NaOH, and the excess OH⁻ sets the pH at about 12.2.
A plot of ln[A] against time for a reaction gives a straight line. What can you conclude from the graph?
For a zero-order reaction it’s the [A]–t plot that’s straight, not the ln[A]–t plot.
For a second-order reaction the 1/[A]–t plot is straight, with slope +k.
A straight ln[A]–t plot means first order. The slope is −1.40 ÷ 40 min = −0.035 min⁻¹, so k = −slope = 0.035 min⁻¹.
The intercept is ln[A]₀ = ln 0.80 = −0.22. It tells you the starting concentration, not the rate constant.
The graph shows the distribution of molecular speeds for one gas sample at two temperatures. Which curve is at the higher temperature?
A taller peak means the speeds are bunched into a narrower range; lower temperatures give narrower, taller peaks, so curve 1 is the colder one.
The same sample has the same number of molecules, so the areas under the two curves are equal; the shapes differ because the temperatures differ.
The areas under the two curves are equal (each represents all the molecules), so area can’t tell you the temperature.
At higher temperature the average kinetic energy is greater, so the most probable speed shifts right and the distribution becomes broader and lower. Curve 2 is hotter.
Solutions of known concentration give the calibration curve shown. An unknown solution has an absorbance of 0.495 at the same wavelength. About what is its concentration?
Go across from 0.495 on the y-axis to the line, then down to the x-axis: 4.5 (in units of 10⁻⁵ M). Or use the slope: 0.495 ÷ 0.110 = 4.5.
That reading is too high: the absorbance at 5.0 on the x-axis is 0.55.
0.495 is the absorbance (y-axis), not the concentration; you have to go through the line to the x-axis.
That multiplies the absorbance by the slope (0.495 × 0.110 = 0.054, i.e. 5.4 × 10⁻⁷ M); the concentration is the absorbance divided by the slope.
Example 1: a rate law from initial-rate data
Three trials of A + B → C at the same temperature gave these data:
| Trial | [A]₀ (M) | [B]₀ (M) | Initial rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.20 | 0.30 | 2.4 × 10⁻³ |
Find the rate law and the rate constant k.
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Write the general rate law. rate = k[A]m[B]n; we need m and n.
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Find m from two trials with the same [B]. Trials 1 and 2 both have [B] = 0.10; only [A] changes:
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Find n from two trials with the same [A]. Trials 2 and 3 both have [A] = 0.20:
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Write the rate law.
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Find k from any trial. Using trial 1:
The units come from rate ÷ concentration³.
Example 2: reading data from an energy profile
On a reaction’s energy profile, the key points have these energies (kJ/mol, left to right): reactants 40, first peak 120, intermediate 60, second peak 95, products 10. How many steps are there, which step is slowest, and what is ΔH?
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Count the peaks. There are two peaks (two transition states) with a dip between them (an intermediate), so the reaction has 2 steps.
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Activation energy of each step. For each step, Ea = energy of the peak − energy at the start of that step:
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Find the slowest step. Step 1 has a much larger activation energy, so it is slowest: it is the rate-determining step决速步.
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Find ΔH. Only the start and end of the whole reaction matter:
The products are lower in energy than the reactants, so the reaction is exothermic. The intermediate and the peaks don’t affect ΔH.
- k from the ln[A]–t plot of a first-order reaction
k = −slope.
- The forward activation energy from an energy profile
Energy of the peak minus energy of the reactants.
- pKa from a weak acid titration curve
The pH at the half-equivalence point.
- The starting concentration from an ln[A]–t plot
The intercept is ln[A]₀.
- εb from a Beer’s law calibration curve
A = εbc, so the slope is εb.
- Moles of acid titrated, from a titration curve
Read the equivalence volume and multiply by the titrant’s molarity.
- The binding energy of a subshell’s electrons from a photoelectron spectrum
Look at the position of its peak.
- k from the [A]–t plot of a zero-order reaction
For zero order the [A]–t plot is straight, and k = −slope.
Common mistakes
Mistake 1: “Start calculating as soon as you see the graph, without checking the axes.” A 10⁻⁵ in the axis units, an axis of ln[A] instead of [A], or a scale that doesn’t start at 0 all change the reading completely. Check the axes before reading values.
Mistake 2: “The pH at the equivalence point is always 7.” Only for a strong acid with a strong base. For a weak acid titrated with a strong base, the solution at equivalence contains the weak base A⁻, so the pH is above 7; for a weak base titrated with a strong acid, it’s below 7.
Mistake 3: “The slope of the line is k.” First decide which plot is straight, then check the sign: the slope is −k for zero- and first-order reactions, but +k for the 1/[A]–t plot of a second-order reaction.
Quick check
A weak acid is titrated with a strong base, with the equivalence point at 30.0 mL. To read the pKa from the curve, at which volume do you read the pH?
15.0 mL is the half-equivalence point, where the HA left equals the A⁻ formed, so pH = pKa.
A plot of 1/[A] against time is a straight line with positive slope. What is the order of the reaction?
For a second-order reaction the 1/[A]–t plot is straight with slope k.
On an energy profile, the reactants are at 50 kJ/mol, the products at 20 kJ/mol and the peak at 130 kJ/mol. What is the forward activation energy?
Ea = 130 − 50 = 80 kJ/mol; ΔH = 20 − 50 = −30 kJ/mol.
Doubling [B] with [A] held constant leaves the initial rate unchanged. What is the order with respect to B?
The rate doesn’t depend on [B], so the reaction is zero order in B.
How does the speed distribution of a gas sample change when its temperature rises?
The distribution spreads toward higher speeds, so a larger fraction exceeds Ea; the number of molecules, and so the area, stays the same.
A graph’s y-axis is labeled “Concentration (10⁻³ M)” and you read 2.5. What is the actual concentration?
The label means concentration ÷ 10⁻³ M = 2.5, so the concentration is 2.5 × 10⁻³ M.
AP-style practice
At 300 K, the concentration of X in the decomposition X → Y + Z changes with time as follows:
| t (min) | [X] (M) |
|---|---|
| 0 | 0.500 |
| 10 | 0.370 |
| 20 | 0.274 |
| 30 | 0.203 |
| 40 | 0.151 |
(a) Explain how a graph can show that the reaction is first order in X.
(b) Find the rate constant k, with units.
(c) Find the half-life of the reaction.
(d) Predict [X] at 60 min.
(e) How would the ln[X]–t graph change if the experiment were repeated at a higher temperature?
Model answer and scoring
(a) Plot ln[X] against t. The values of ln[X] are −0.693, −0.994, −1.295, −1.595 and −1.890, which lie on a straight line, so the reaction is first order (neither the [X]–t nor the 1/[X]–t plot is straight).
(b) Slope = (−1.890 − (−0.693)) ÷ 40 min ≈ −0.030 min⁻¹, so k = −slope = 0.030 min⁻¹.
(c) t½ = 0.693 ÷ k = 0.693 ÷ 0.030 min⁻¹ ≈ 23 min.
(d) ln[X] = ln 0.500 − 0.030 × 60 = −0.693 − 1.80 = −2.49, so [X] = e−2.49 ≈ 0.083 M.
(e) At higher temperature k is larger, so the line is steeper (a more negative slope); the intercept is still ln 0.500.
Scoring: (a) a straight ln[X]–t plot shows first order; (b) k with units; (c) half-life; (d) [X] at 60 min; (e) steeper slope with the same intercept.
Summary
- Four steps for graphs: title and legend, axes and units, trend, key points.
- Key points and what they give: pH = pKa at half-equivalence, the slope of a straight plot and k, peaks and activation energy, start and end points and ΔH, the slope of a calibration curve and εb.
- For data tables, find two trials where only one quantity differs and use the ratio to find that quantity’s effect.
AP tip: When you read values off a graph, mark the points you used (on scratch paper, or on the FRQ graph) and write down each value with its units. FRQ scoring guidelines usually accept a range for values read from a graph, but the units and order of magnitude must be right.
Sources: the titration curve is calculated exactly from charge balance, using Ka = 1.8 × 10⁻⁵ for acetic acid from OpenStax Chemistry 2e Appendix H; the other graphs and tables use data designed by this site for practice.