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Reaction rates and gas-volume graphs
Change how quickly a reaction produces gas, then compare the curve, the measurements and the final gas amount.
Change the model
Change a value, then use Update model or leave the field to redraw. The table gives the same quantities as the graph. Your entries are not saved to progress; changing the model resets the check question.
Try these comparisons
- Keep the gas limit at 60 cm³. Compare rate factors 1 and 2 at 20 s and 60 s.
- Keep the rate factor fixed and change the gas limit. Explain which control changed the total product available.
- Compare mean rates over 0–20 s and 20–60 s. Which interval has the greater mean rate?
Read the curve
The vertical axis shows the total gas collected so far. A rising curve means more gas has been collected. Its gradient shows how quickly gas is being collected at that time. A steeper curve means a greater rate.
The curve becomes less steep as the reaction proceeds. In many reactions, reactants are being used up, so fewer successful collisions occur each second. A curve levelling off does not mean the gas has vanished.
Separate speed from amount
The rate factor changes how quickly this model approaches its gas limit. It leaves the limit unchanged. Compare a factor of 1 with a factor of 2: at the same early time, more gas has been collected in the faster run, but both approach the same total.
A catalyst or smaller solid pieces can increase rate without changing the final gas amount, when the reacting amounts and other conditions are suitable. Changing the amount of a limiting reactant can change the final amount instead. The gas-limit control represents that separate choice.
Calculate a mean rate
Choose two times. Subtract their gas volumes, then divide by the time difference. Between 20 s and 60 s, the interval is 40 s. Dividing the final gas volume by 60 s would describe a different interval.
The table rounds volumes to two decimal places. Keep extra digits in your working where available. A mean rate describes the whole interval; it is not the exact rate at either endpoint.
Worked example
If collected volume rises from 24 cm³ at 20 s to 40 cm³ at 60 s, the mean rate is (40 − 24) ÷ (60 − 20) = 0.40 cm³/s. The 40 cm³ reading alone is not the volume change.
What the model assumes
This is an illustrative mathematical curve, not a prediction for a particular reaction. Collection is leak-free at constant temperature and pressure. The curve approaches its gas limit smoothly; real measurements can vary.