Measuring and Calculating Rates of Reaction

Knowing what changes the rate of a reaction is one thing; measuring the rate in the laboratory is another. Topic C5 of OCR Gateway Science A includes required practicals to follow a reaction over time, and you must be able to calculate a rate and interpret a rate–time graph. There are three standard ways to measure a rate, depending on what the reaction produces. This lesson sets out those methods, shows how to calculate a mean rate, and explains the shape of the graph you get.

By the end of this lesson you should be able to describe methods for measuring the rate of a reaction, calculate a mean rate with the correct units, and interpret a quantity-versus-time graph, including why it is steepest at the start and levels off.

Measuring a rate always comes down to the same idea: pick one thing that changes as the reaction proceeds, and measure how much it changes in a given time. The skill is choosing the right quantity to follow for a particular reaction, and then reading the resulting graph correctly.

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Methods for Measuring Rate

To measure a rate you follow one measurable quantity as it changes over time. The three standard methods are:

Method	What you measure	Suitable when…
Gas volume (gas syringe)	Volume of gas given off at set times	A gas is produced (e.g. Mg + acid)
Mass loss (balance)	Decrease in mass as gas escapes	A gas escapes from an open flask
Turbidity ("disappearing cross")	Time for a mark to disappear	The mixture turns cloudy (a precipitate forms)

Measuring gas volume

If a reaction gives off a gas, you can collect it in a gas syringe and record the volume at regular time intervals. For example, magnesium reacting with hydrochloric acid produces hydrogen, which pushes out the syringe plunger. A graph of gas volume against time shows how the rate changes.

Measuring mass loss

If the gas given off is allowed to escape from an open conical flask (often with a cotton-wool plug to stop spray), the flask is placed on a balance and the decreasing mass is recorded over time. The mass falls as gas leaves. This works well for reactions producing a dense gas such as carbon dioxide (e.g. a carbonate with acid).

The turbidity / "disappearing cross" method

When sodium thiosulfate reacts with hydrochloric acid, a fine precipitate of sulfur forms and gradually makes the solution cloudy:

$\text{Na}_2\text{S}_2\text{O}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{S} + \text{SO}_2 + \text{H}_2\text{O}$Na2​S2​O3​+2HCl→2NaCl+S+SO2​+H2​O

A cross is drawn on paper under the flask, and you time how long it takes for the cross to disappear from view as the sulfur clouds the mixture. A shorter time means a faster rate. This method gives the rate as $\dfrac{1}{\text{time}}$time1​ for the cross to disappear.

Exam Tip: Choose the method to match the reaction: a gas given off → gas syringe (volume) or a balance (mass loss); a precipitate that clouds the mixture → the disappearing-cross method. For the cross method, a shorter time = a faster rate.

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Calculating the Mean Rate

The mean (average) rate of a reaction over a period of time is:

$\text{mean rate} = \dfrac{\text{quantity of reactant used or product formed}}{\text{time}}$mean rate=timequantity of reactant used or product formed​

The units depend on what you measured: a gas volume gives cm³/s, a mass change gives g/s, and an amount in moles gives mol/s.

Worked Example 1 — Mean rate from a gas volume

A reaction produces $48 \text{ cm}^3$48 cm3 of gas in $60 \text{ s}$60 s. Calculate the mean rate of reaction.

Step 1 — write the equation: $\text{mean rate} = \dfrac{\text{volume of gas}}{\text{time}}$mean rate=timevolume of gas​.

Step 2 — substitute: $\text{mean rate} = \dfrac{48}{60}$mean rate=6048​.

Step 3 — calculate: $\text{mean rate} = 0.8 \text{ cm}^3/\text{s}$mean rate=0.8 cm3/s.

Answer: $0.8 \text{ cm}^3/\text{s}$0.8 cm3/s.

Worked Example 2 — Mean rate from a mass loss

In a reaction giving off carbon dioxide, the flask loses $1.2 \text{ g}$1.2 g of mass in $30 \text{ s}$30 s. Calculate the mean rate.

Step 1 — $\text{mean rate} = \dfrac{\text{mass lost}}{\text{time}}$mean rate=timemass lost​.

Step 2 — substitute: $\text{mean rate} = \dfrac{1.2}{30}$mean rate=301.2​.

Step 3 — calculate: $\text{mean rate} = 0.04 \text{ g/s}$mean rate=0.04 g/s.

Answer: $0.04 \text{ g/s}$0.04 g/s.

Exam Tip: Always give the rate's unit: cm³/s for a gas volume, g/s for a mass change, mol/s for moles. Divide the quantity changed by the time — a bare number with no unit will not score the final mark.

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Interpreting a Rate–Time Graph

A graph of quantity (gas volume or mass) against time has a characteristic shape:

• it is steepest at the start, because the concentration of the reactants is highest then, so collisions are most frequent and the reaction is fastest;
• it gradually becomes less steep as the reactants are used up and the rate falls;
• it levels off (becomes horizontal) when the reaction stops — usually because a reactant has run out.

The gradient (steepness) of the line at any point gives the rate at that moment. A steeper line means a faster rate.

Notice that the two curves in the diagram level off at the same final volume: changing the temperature changes how fast the gas is produced, but not the total amount of gas, which depends only on how much reactant there was. The hotter reaction is steeper at the start (faster) and finishes sooner.

Exam Tip: A rate–time curve is steepest at the start (highest concentration → fastest) and levels off when a reactant runs out. If two curves reach the same height, the same amount of product formed — the steeper one was just faster.

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Higher Tier: Instantaneous Rate from a Tangent

Higher tier only: The mean rate averages over a whole period, but the rate actually changes throughout the reaction. To find the rate at a particular instant, draw a tangent to the curve at that point and calculate its gradient:

$\text{rate at a point} = \text{gradient of the tangent} = \dfrac{\text{change in quantity}}{\text{change in time}}$rate at a point=gradient of the tangent=change in timechange in quantity​

The tangent at the start is the steepest, giving the highest (initial) rate; tangents drawn later are less steep, showing the rate falling as the reactants are used up.