Effect of Concentration on Rate

This lesson covers how changing the concentration of a reactant in solution affects the rate of reaction, with full explanations using collision theory, graphs, and worked examples as required by the Edexcel GCSE Combined Science specification (1SC0).

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The Key Idea

When the concentration of a reactant in solution is increased, there are more particles of that reactant per unit volume. This means the particles are closer together and collisions between reactant particles happen more frequently.

More frequent collisions means more successful collisions per second, so the rate of reaction increases.

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Collision Theory Explanation

graph LR
    subgraph "Low concentration"
        L1["Few particles<br/>per unit volume"]
        L2["Collisions are<br/>less frequent"]
        L3["Fewer successful<br/>collisions/s"]
        L4["Slower rate"]
        L1 --> L2 --> L3 --> L4
    end
    subgraph "High concentration"
        H1["Many particles<br/>per unit volume"]
        H2["Collisions are<br/>more frequent"]
        H3["More successful<br/>collisions/s"]
        H4["Faster rate"]
        H1 --> H2 --> H3 --> H4
    end

    style L4 fill:#c0392b,color:#fff
    style H4 fill:#27ae60,color:#fff

Step-by-step explanation

1. More particles per unit volume — increasing concentration puts more reactant particles into the same volume.
2. More frequent collisions — particles are closer together so they bump into each other more often.
3. More successful collisions per second — the number of collisions with energy ≥ Eₐ increases.
4. Rate increases — products form more quickly.

Exam Tip: Increasing concentration does not give particles more energy. It only increases the frequency of collisions. The energy of individual collisions stays the same — that only changes with temperature.

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Graphical Evidence

When we carry out the same reaction at different concentrations and plot the results, we see clear differences.

Graph: Volume of Gas vs Time at Different Concentrations

Feature	Higher concentration	Lower concentration
Initial gradient	Steeper (faster rate)	Shallower (slower rate)
Total product formed	Same if limiting reactant amount is unchanged	Same
Time to complete	Shorter	Longer

Key points to remember when reading these graphs:

• A steeper curve at the start means a faster rate.
• Both curves level off at the same total volume if the same total amount of limiting reactant was used.
• The higher concentration experiment reaches the plateau first.

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Pressure and Gases

For reactions involving gases, increasing the pressure has the same effect as increasing concentration in solution.

Solution reactions	Gas reactions
Increase concentration	Increase pressure
More particles per unit volume	More particles per unit volume (gas molecules pushed closer together)
More frequent collisions	More frequent collisions
Faster rate	Faster rate

Exam Tip: If the question asks about gases, use the word pressure instead of concentration — the explanation is identical.

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Worked Example

In an experiment, magnesium ribbon reacts with hydrochloric acid. The volume of hydrogen gas is measured every 10 seconds.

Time (s)	Volume (cm³) — 1.0 mol/dm³ HCl	Volume (cm³) — 2.0 mol/dm³ HCl
0	0	0
10	12	24
20	22	40
30	30	48
40	36	48
50	40	48
60	44	48
70	48	48

Observations:

• With 2.0 mol/dm³ HCl the reaction finishes by about 30 s — the curve levels off early.
• With 1.0 mol/dm³ HCl the reaction takes about 70 s to finish.
• Both reach the same total volume (48 cm³) because the same amount of magnesium was used each time.
• The initial rate is faster with the higher concentration.

Calculating Mean Rate

For 2.0 mol/dm³:

$$\text{Mean rate} = \frac{48}{30} = 1.6 \text{ cm}^3\text{/s}$$Mean rate=3048​=1.6 cm3/s

For 1.0 mol/dm³:

$$\text{Mean rate} = \frac{48}{70} = 0.69 \text{ cm}^3\text{/s}$$Mean rate=7048​=0.69 cm3/s

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Concentration and 1/Time

In the disappearing cross experiment, we often plot 1/time (s⁻¹) against concentration. Since 1/time is proportional to rate, this gives a straight line through the origin — showing that rate is directly proportional to concentration.

Concentration (mol/dm³)	Time (s)	1/Time (s⁻¹)
0.02	200	0.005
0.04	100	0.010
0.06	67	0.015
0.08	50	0.020
0.10	40	0.025

A graph of 1/time vs concentration is a straight line through the origin, confirming that rate is directly proportional to concentration.

Exam Tip: If asked to "use the data to show that rate is proportional to concentration", calculate 1/time for each data point and show that the ratio of concentration to 1/time is constant (or that the graph is a straight line through the origin).

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Summary

• Increasing concentration means more particles per unit volume.
• This leads to more frequent collisions and therefore more successful collisions per second.
• The rate of reaction increases.
• On graphs, higher concentration gives a steeper initial curve but the same total product if the limiting reactant amount is unchanged.
• For gases, increasing pressure has the same effect as increasing concentration.
• Rate is directly proportional to concentration (shown by a straight-line graph of 1/time vs concentration).

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Extended Worked Example: Doubling Concentration

Question: When the concentration of HCl is doubled from 1.0 mol/dm³ to 2.0 mol/dm³, the rate of reaction with marble chips also doubles. Explain this using collision theory and state what would happen to the rate if the concentration were halved.

Answer: