A reaction-progress graph answers two different questions: how fast is the reaction proceeding, and how much change has occurred? The slope gives a rate; the final height gives an amount. A steep curve can reach the same final level as a shallow one, so “more gas at an early time” does not necessarily mean “more gas in total”.
These worked questions on reaction-rate graphs use original constructed data. They are supplied-data exercises, with no chemical preparation or gas-collection instructions. The smooth curves are idealised models chosen to make interpretation and checking clear; they are not experimental measurements.
Read the axes before comparing curves
On a gas-volume-versus-time graph, average gas-production rate = change in gas volume ÷ elapsed time. If volume is in cm³ and time in seconds, the unit is cm³ s⁻¹. Use the interval length, not simply the final time written on the axis.
A product curve commonly rises as product accumulates. A reactant-remaining curve falls as reactant is consumed. The negative slope of a reactant curve means the amount is decreasing; the rate of disappearance is usually quoted as a positive magnitude.
For an instantaneous rate, use the gradient of a tangent at the specified time. The initial rate is the tangent gradient at time zero. A line joining two points on a curved progress graph gives an average over an interval, even if the first point is the origin.
The supplied chemical model and its limits
Runs A and B model decomposition according to 2H₂O₂ → 2H₂O + O₂. Each begins with 4.00 mmol of hydrogen peroxide, which is completely consumed. The supplied gas-volume conversion is 25.0 cm³ of oxygen per mmol at the same measurement temperature and pressure for both runs. Assume no gas loss, dissolution correction or competing reaction.
Run B represents a more effective catalysed pathway under otherwise matched model conditions. No catalyst identity, preparation, dose or operating method is needed. The chemical equation supplies a two-to-one reactant-to-oxygen ratio; the chosen rate curves supply how quickly progress occurs.
| Time (s) | Run A volume (cm³) | Run B volume (cm³) |
|---|---|---|
| 0 | 0 | 0 |
| 10 | 18 | 32 |
| 20 | 32 | 48 |
| 30 | 42 | 50 |
| 40 | 48 | 50 |
| 50 | 50 | 50 |
| 60 | 50 | 50 |
Complete graph alternative: Both solid curves start at zero and level off at 50 cm³. Run A follows V = 50[1 − (1 − t/50)²] from 0 to 50 s, then stays at 50. Run B follows V = 50[1 − (1 − t/25)²] from 0 to 25 s, then stays at 50. Here V is in cm³ and t in seconds. These formulas fully specify the plotted models; calculus is not needed for the questions. The dashed initial tangents pass through (0 s, 0 cm³) and (20 s, 40 cm³) for A, and through (0 s, 0 cm³) and (10 s, 40 cm³) for B.
Worked example 1: average rates over different intervals
Find Run A’s average gas-production rate from 10 to 30 s, then compare it with the average from 30 to 50 s.
First interval: (42 − 18) ÷ (30 − 10) = 24 ÷ 20 = 1.20 cm³ s⁻¹. Second interval: (50 − 42) ÷ (50 − 30) = 8 ÷ 20 = 0.40 cm³ s⁻¹. The earlier average is three times the later average.
Dividing 42 by 30 gives 1.40 cm³ s⁻¹, which is the average from zero to 30 s. It does not answer the requested 10-to-30-second question. Mark the two endpoint readings and subtract both coordinates before dividing.
The gas volume increases during both intervals, but it increases more slowly later. A high total volume is compatible with a low current rate because the volume includes product accumulated earlier.
Worked example 2: use the supplied initial tangents
Calculate the initial rate in each run from the dashed tangent coordinates. For A, gradient = (40 − 0) ÷ (20 − 0) = 2.00 cm³ s⁻¹. For B, gradient = 40 ÷ 10 = 4.00 cm³ s⁻¹. The initial model rate in B is twice that in A.
The tangent’s point at 20 s and 40 cm³ is not a point on Run A’s actual progress curve. At 20 s the curve gives 32 cm³. Points used to calculate tangent gradient must lie on the tangent line, not be replaced with nearby points on the curve.
Using the first ten seconds gives averages of 18/10 = 1.80 cm³ s⁻¹ for A and 32/10 = 3.20 cm³ s⁻¹ for B. Both are lower than the respective initial rates because these model curves flatten as the reaction proceeds.
Worked example 3: compare speed with total yield
Why do both runs finish at 50 cm³ even though B is faster? The equation requires two mmol of hydrogen peroxide for each mmol of oxygen. Thus 4.00 mmol of reactant yields 4.00/2 = 2.00 mmol of oxygen. At the supplied gas conditions, volume = 2.00 × 25.0 = 50.0 cm³.
The plateau is consistent with the same initial limiting amount and complete conversion in both cases. A reaches the plateau at 50 s and B at 25 s in the fully specified curves. The table alone brackets B’s completion between 20 and 30 s; only the additional curve model locates it at 25 s.
A catalyst supplies an alternative reaction pathway and can increase the rate. It does not create extra reactant. For a reversible reaction, a catalyst helps the system approach equilibrium faster without changing the equilibrium composition at a given temperature. Neither statement means that every real process with added catalyst must have an identical observed yield: side reactions or incomplete observations would need consideration.
Increasing concentration is not automatically a fair rate-only comparison. If the solution volume is unchanged, increasing reactant concentration also increases its amount and may raise the final gas volume. Check what is held constant before explaining either the slope or the plateau.
Worked example 4: a falling reactant curve
| Time (s) | Hydrogen peroxide remaining (mmol) |
|---|---|
| 0 | 4.00 |
| 10 | 2.56 |
| 20 | 1.44 |
| 30 | 0.64 |
| 40 | 0.16 |
| 50 | 0 |
| 60 | 0 |
The full curve is n = 4.00 − VA/12.5, where n is remaining mmol and VA is Run A’s cm³ value. From 10 to 30 s, the signed gradient is (0.64 − 2.56)/20 = −0.096 mmol s⁻¹. The average disappearance rate is therefore 0.096 mmol s⁻¹.
Over that interval, oxygen production is 1.20 cm³ s⁻¹, equivalent to 1.20/25.0 = 0.048 mmol s⁻¹. Hydrogen peroxide disappears twice as fast in mmol terms, matching the equation’s coefficient ratio. A species-production rate and a stoichiometrically normalised reaction rate are not interchangeable without this factor.
Explain the pattern without overclaiming
A decreasing reactant concentration can reduce the frequency of productive encounters and cause a curve to flatten. Higher temperature can increase the proportion of particles able to react, while greater exposed surface area can affect a reaction involving a solid. These are possible mechanisms, not information that can be read from an unlabelled curve alone.
A flat gas trace means no further measured gas accumulation during the observed interval. Complete reaction is one explanation under this lesson’s assumptions. In real evidence, a leak, an equilibrium state, a very slow reaction or a faulty measurement could also matter. State the supplied assumptions before treating a plateau as proof that every reactant has been consumed.
Three practice tasks with solved answers
A fresh constructed run begins with 3.60 mmol of hydrogen peroxide under the same ideal conversion assumptions. Its supplied times 0, 15, 30, 45 and 60 s correspond to oxygen volumes 0, 24, 39, 45 and 45 cm³. Its separately supplied initial tangent passes through (0 s, 0 cm³) and (15 s, 30 cm³).
Practice 1: an interval average
Find the average gas-production rate from 15 to 30 s.
Answer: (39 − 24)/(30 − 15) = 15/15 = 1.00 cm³ s⁻¹. The final reading alone is not the amount produced in this interval.
Practice 2: tangent versus chord
Find the initial rate and compare it with the zero-to-15-second average.
Answer: Initial tangent rate = 30/15 = 2.00 cm³ s⁻¹. The interval average is 24/15 = 1.60 cm³ s⁻¹. They describe different quantities.
Practice 3: check the plateau
Verify the 45 cm³ plateau. If the initial reactant amount doubled with complete conversion at the same gas conditions, what total volume would be expected?
Answer: 3.60/2 × 25.0 = 45.0 cm³. Doubling the limiting amount gives 90.0 cm³. The amount calculation alone does not establish how long the second run would take.
Connect rate curves with reversible reactions
The curves above describe how quickly measured quantities change. For a reversible reaction, also ask what mixture is reached at equilibrium. Continue with the ammonia equilibrium and recycling model, which distinguishes equal forward and reverse rates from equal concentrations. Its original condition comparisons explain pressure, temperature and catalyst effects, while the worked stream balance keeps recycling separate from single-pass conversion. A faster approach to equilibrium is not automatically a higher equilibrium yield.
Frequently asked questions
Is gas volume the same as reaction rate?
No. Volume is an accumulated amount. Its change per unit time gives a gas-volume production rate under the stated measurement conditions.
Can I find initial rate from the first two readings?
That gives an early interval average. It may approximate initial rate over a short interval, but the supplied initial tangent defines it more precisely here.
Which points should I use for tangent gradient?
Use two well-separated points on the tangent line. They need not lie on the reaction-progress curve.
Why is a reactant curve gradient negative?
Reactant amount decreases. The disappearance rate is usually quoted as the positive magnitude of that negative change per unit time.
Does a catalyst increase the final amount of gas?
Not in the complete-conversion model with the same limiting amount. A catalyst changes the pathway and rate, not the available amount of reactant.
Does a catalyst change equilibrium yield?
At a given temperature it does not change equilibrium composition. It can help the system reach that composition faster.
Does a plateau always prove completion?
No. It shows no further measured accumulation over the observed interval. Completion follows here from explicit assumptions; real data may need other explanations checked.
Sources and further reading
These references support the scientific background. Examples, datasets, diagrams and practice answers on this page are original. Reference links are for background reading, not instructions to perform a practical activity.
- OpenStax Chemistry 2e: chemical reaction rates
- OpenStax Chemistry 2e: collision theory
- OpenStax Chemistry 2e: catalysis
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