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ChemistryRates and equilibrium

The rate of reaction formula

The rate of reaction formula is mean rate = quantity of reactant used ÷ time taken, or mean rate = quantity of product formed ÷ time taken. The quantity can be a mass in grams, a gas volume in cm³ or, at GCSE Higher and A-level, an amount in moles, giving rates in g/s, cm³/s or mol/s. For example, if a reaction gives off 48 cm³ of gas in 30 seconds, its mean rate is 48 ÷ 30 = 1.6 cm³/s. The rate at one particular moment is the gradient of a tangent to the graph of quantity against time.

The rate of reaction formula

\text{mean rate of reaction} = \frac{\text{quantity of reactant used}}{\text{time taken}}
\text{mean rate of reaction} = \frac{\text{quantity of product formed}}{\text{time taken}}

AQA's GCSE Chemistry specification gives both forms, and says the quantity can be measured as a mass in grams or a volume in cm³. Higher tier students also use amounts in moles. You can follow whichever substance is easiest to measure: a gas given off, a solid used up, or a change in mass as gas escapes.

Rate of reaction units

You measureRate unitExample
Mass, in gg/s (or g s⁻¹)mass lost as CO₂ escapes
Gas volume, in cm³cm³/s (or cm³ s⁻¹)hydrogen collected in a gas syringe
Amount, in molmol/s (or mol s⁻¹)moles of product made
Concentration, in mol/dm³mol dm⁻³ s⁻¹A-level rate equations

If time is in minutes, the unit becomes g/min or cm³/min. Always give the unit: a rate of "1.6" means nothing on its own.

How to calculate rate of reaction

Example 1: from a gas volume

Magnesium ribbon reacts with hydrochloric acid and 48 cm³ of hydrogen is collected in 30 s. Find the mean rate.

\text{mean rate} = \frac{48}{30} = 1.6\ \text{cm}^3\text{/s}

Example 2: from a mass change

Some marble chips fizz in acid while a balance records the total mass. It drops by 2.4 g in 2 minutes as carbon dioxide escapes. Find the mean rate in g/s.

Convert the time: 2 minutes = 120 s.

\text{mean rate} = \frac{2.4}{120} = 0.02\ \text{g/s}

Example 3: in moles (Higher)

A reaction produces 0.060 mol of product in 40 s. Find the mean rate.

\text{mean rate} = \frac{0.060}{40} = 1.5 \times 10^{-3}\ \text{mol/s}

Now try one in the course. Work out the mean rate, then press Mark to check it there.

Read a table of gas volumes or mass lost and type the mean rate between two times.

Press Mark to have your answer marked in the Reaction Kinetics course.

Rate of reaction graphs

A graph of the quantity of product against time shows the whole reaction at once. The graph below shows the volume of gas from a reaction. Here x is the time in seconds and y is the total volume of gas in cm³.

y = 60\cdot\left(1-{ e}^{\left(\frac{- x}{20}\right)}\right)
102030405060708090100-1010203040506070fastest at the start20 s: 37.9 cm³levelling off near 60 cm³
Steep at first, then flatter as the reactants run outDrag to pan, scroll to zoom. x from 0 to 100, y from -11.9 to 71.5.
  • The steeper the curve, the faster the reaction.
  • The reaction is fastest at the start, when there are the most reactant particles to collide.
  • The curve flattens as the reactants are used up, and goes level when the reaction stops. The final height shows the total amount of product.

Mean rate over a time interval

The mean rate between two times is the change in quantity divided by the change in time. On this graph, the volume is 23.6 cm³ at 10 s and 46.6 cm³ at 30 s:

\text{mean rate} = \frac{46.6 - 23.6}{30 - 10} = \frac{23.0}{20} = 1.15\ \text{cm}^3\text{/s}

Rate at a particular time: drawing a tangent

To find the rate at one instant, draw a tangent: a straight line that just touches the curve at that time, with the same slope as the curve there. Its gradient is the rate at that moment. AQA asks all students to draw tangents and use their slope as a measure of rate, and Higher tier students to calculate the gradient.

Example 4: rate at 20 s from a tangent

The tangent to this curve at t = 20 s goes through (0, 15.9) and (40, 60.0). Find the rate at 20 s.

\text{rate} = \text{gradient} = \frac{60.0 - 15.9}{40 - 0} = \frac{44.1}{40} = 1.10\ \text{cm}^3\text{/s}

Choose two points far apart on your tangent line, not on the curve, to keep the reading error small. The rate at 20 s, 1.10 cm³/s, is lower than the starting rate, because some reactants have already been used up.

Rate from a reaction time: 1/t

In some experiments you do not measure a quantity; you time how long something takes, such as a cross under a flask disappearing as a precipitate forms. A faster reaction takes less time, so the rate goes up as the time comes down, in proportion to 1 ÷ time:

\text{rate} \propto \frac{1}{t}

If the cross disappears after 50 s in one experiment and 80 s in another, the relative rates are 1 ÷ 50 = 0.020 s⁻¹ and 1 ÷ 80 = 0.0125 s⁻¹. The first is 1.6 times faster. AQA's required practical on rates asks students to investigate how changing concentration affects the rate, using one method that measures the volume of gas produced and one that involves a change in colour or turbidity.

What changes the rate

Collision theory explains it: a collision only leads to a reaction if it has at least the activation energy. The rate increases if you raise:

  • concentration of a solution, or pressure of a gas, so collisions are more frequent;
  • surface area of a solid, by breaking it into smaller pieces, so more particles are exposed;
  • temperature, so collisions are more frequent and more of them have enough energy;
  • or by adding a catalyst, which gives a route with a lower activation energy.

The Reaction Rate Simulator lets you heat, concentrate and catalyse particles and watch the collision rate change, and the Reaction Kinetics course builds up from measuring rate to rate equations. At A-level, the rate is linked to concentrations by a rate equation, rate = k[A]ᵐ[B]ⁿ, and the rate constant k depends on temperature through the Arrhenius equation.

Try it

Reaction Rate Simulator

Heat, concentrate and catalyse particles to see collision theory

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Common mistakes

  • Time in minutes. If the question wants g/s or cm³/s, convert minutes to seconds first.
  • Using the total volume at the end for a mean rate over part of the reaction. Use the change in volume over the interval you are asked about.
  • Gradient from points on the curve instead of the tangent. Read the two points from your straight tangent line.
  • A tangent that crosses the curve. It should touch the curve at one point only, with the curve on one side.
  • Forgetting units, or writing cm³ instead of cm³/s.

Rate of reaction questions with answers

1. A reaction gives off 90 cm³ of gas in 45 s. Find the mean rate.
90 ÷ 45 = 2 cm³/s
2. A flask loses 1.8 g in 3 minutes as gas escapes. Find the mean rate in g/s.
3 minutes = 180 s.
1.8 ÷ 180 = 0.01 g/s
3. 0.025 mol of a reactant is used up in 50 s. Find the mean rate in mol/s.
0.025 ÷ 50 = 5.0 × 10⁻⁴ mol/s
4. A tangent to a gas volume curve passes through (10, 12) and (50, 36). Find the rate at the point where it touches.
Gradient = (36 − 12) ÷ (50 − 10) = 24 ÷ 40 = 0.6 cm³/s
5. A cross disappears after 25 s in a hot solution and 100 s in a cold one. How many times faster is the hot reaction?
Rates are 1 ÷ 25 = 0.04 and 1 ÷ 100 = 0.01, so 4 times faster.

Practise chemistry in a short timed paper

GCSE Chemistry Higher and A-level Chemistry: Year 1 each give you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so rates of reaction turn up regularly but not in every paper.

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Common questions

What is the formula for rate of reaction?
Mean rate = quantity of reactant used ÷ time taken, or quantity of product formed ÷ time taken.
What are the units for rate of reaction?
Usually g/s for mass, cm³/s for gas volume, or mol/s for amount. At A-level, rates in rate equations are in mol dm⁻³ s⁻¹.
How do you find the rate of reaction from a graph?
For a mean rate, divide the change in quantity by the change in time. For the rate at one moment, draw a tangent at that time and find its gradient.
Why does the rate of reaction slow down over time?
The reactants are used up, so their particles become less concentrated and collide less often.
What is the difference between mean rate and instantaneous rate?
Mean rate is averaged over a time interval. Instantaneous rate is the rate at one moment, found from the gradient of a tangent.
Why is rate proportional to 1/time?
If a fixed amount of change, such as a cross disappearing, happens faster, the time is shorter. Twice the rate means half the time, which is exactly what 1 ÷ time captures.

Revision cards

Say the answer first, then flip the card to check it.

Where to go next

  • Solve it

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  • Learn it

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    Rates, catalysts and collision theory: what makes reactions fast or slow

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    GCSE Chemistry Higher

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