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Simple Rate Equations

Learn and apply Simple Rate Equations in the published Chemistry course sequence.

  • GCE A-Level H1 Chemistry 8873-2027
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H1 Simple Rate Equations and Orders: Orientation

A rate equation is an experimental model of how reactant concentrations control initial rate. H1 questions use simple single-step reactions and individual orders of zero, one or two.

H1 8873 scope
  • Use rate equations and calculate initial rates from supplied concentration data.
  • Do not infer a multi-step mechanism or rate-determining step, and do not calculate k from an initial-rates table.

Definitions (Must Know)

  • Rate of reaction is the change in concentration of a reactant or product per unit time.
  • A rate equation has the form rate = k[A]^m[B]ⁿ.
  • The order with respect to A is the experimentally determined power m.
  • The overall order is m + n.
  • The rate constant, k, is the proportionality constant for a stated reaction at a stated temperature.
  • Activation energy, Eₐ, is the minimum energy required for a successful collision.

Detailed Explanations

A. Reading an order

If only [A] changes by a factor f, its rate contribution changes by f^m. Match the observed rate factor to 1, f or f².

B. Predicting a rate change

For rate = k[A]²[B], doubling A and halving B changes rate by 2² × 1/2 = 2.

C. Calculating an initial rate

Substitute the initial concentrations and the supplied k into the equation. Preserve concentration powers and units. This calculates rate; it does not establish a mechanism.

D. The H1 boundary

The simple single-step restriction lets you use the stated rate equation without importing H2 tests of proposed multi-step mechanisms.

E. Why concentration can increase rate

Increasing a reactant concentration places more particles in the same volume. The particles collide more frequently, so the number of successful collisions per unit time can increase. The rate equation shows how strongly the measured rate responds: first order gives a proportional change, second order gives a squared change, and zero order gives no observed concentration effect under the stated conditions.

Keep concentration and k separate. At constant temperature, changing concentration changes the collision frequency and the concentration terms in the rate equation; it does not change k.

Worked Examples

Modelled example 1

Concentration factors

Core

Problem

For rate = k[A]²[B]⁰ at fixed temperature, A is doubled and B is tripled. Find the rate factor.
Study the worked solution

  1. Apply A's power

    Method

    Raise A’s concentration factor 2 to power 2.

    Reason

    The reaction is second order in A.

    Working

    2² = 4.

  2. Apply B's power

    Method

    Raise B’s concentration factor 3 to power 0.

    Reason

    A zero-order concentration term contributes factor 1.

    Working

    3⁰ = 1.

  3. Combine

    Method

    Multiply the independent factors.

    Reason

    The concentration terms multiply in the rate equation.

    Working

    4 × 1 = 4: the rate quadruples.

Guided practice 2

Initial-rate calculation

About 7 min

Problem

For rate = k[A][B]², k = 0.50 dm⁶ mol⁻²s⁻¹, [A] = 0.20 and [B] = 0.30 mol dm⁻³. Calculate the initial rate.

Try this before viewing the solution

Hints

Hint 1: preserve powers
Substitute A once and B twice.
Hint 2: evaluate B first
(0.30)² = 0.090 before multiplying by 0.20 and 0.50.
View solution step by step

  1. Substitute

    Method

    Insert k and both initial concentrations into the stated rate equation.

    Reason

    The equation directly predicts the initial rate for the supplied conditions.

    Working

    rate = 0.50(0.20)(0.30)².

  2. Evaluate

    Method

    Square B’s concentration and complete the product.

    Reason

    B is second order while A is first order.

    Working

    rate = 9.0 × 10⁻³ mol dm⁻³s⁻¹.

Common misconception 3

Correct a zero-order interpretation

Find and correct the mistake

Learner claim

For rate = k[X]⁰[Y], a learner says, “X is not a reactant because it does not affect the rate.” Identify and correct the error.

Interpret zero order

Zero order in X means

View solution step by step

  1. Read the mathematical effect

    Method

    Use [X]⁰ = 1.

    Reason

    Changing X’s concentration does not change the rate predicted under the stated conditions.

    Working

    rate = k[Y] numerically.

  2. Keep the chemical meaning

    Method

    State that X may still be consumed in the chemical reaction.

    Reason

    Reaction order describes kinetic dependence, not whether a species appears in the overall chemistry.

    Working

    Correct conclusion: rate is zero order in X, not “X is absent.”

Examiner practice 4

Explain a combined rate change

3 marks

Problem

For rate = k[A]²[B] at constant temperature, [A] doubles while [B] halves. Calculate the overall rate factor. [3 marks]

Try this before viewing the solution

View solution step by step

  1. A contribution

    1 mark

    Method

    Square A’s factor 2.

    Reason

    A is second order.

    Working

    2² = 4.

  2. B contribution

    1 mark

    Method

    Use B’s factor 1/2 to power 1.

    Reason

    B is first order.

    Working

    (1/2)¹ = 1/2.

  3. Overall factor

    1 mark

    Method

    Multiply the contributions.

    Reason

    The rate-law terms act together.

    Working

    4 × 1/2 = 2: the rate doubles.

Challenge 5

Keep the rate unchanged

Minimal support

Problem

For rate = k[A][B]² at constant temperature, [A] is increased by a factor of 4. By what factor must [B] change to keep the rate unchanged?

Try this before viewing the solution

Hints

Hint 1: set the total factor to one
If B changes by factor f, the total rate factor is 4f².
Hint 2: solve the square
Set 4f² = 1 and take the positive concentration factor.
View solution step by step

  1. Build the factor equation

    Method

    Multiply A’s first-order factor by B’s squared factor.

    Reason

    The rate must retain an overall factor of 1.

    Working

    4f² = 1.

  2. Solve for B

    Method

    Find the positive factor f.

    Reason

    Concentration factors are positive.

    Working

    f² = 1/4, so f = 1/2: [B] must be halved.

Mind Stretchers

Attempt each unfamiliar application before opening the hint, then compare your reasoning with the solution.

Mind stretcher 1: Separating individual and overall orderExtension

Question. A reaction is zero order in X and second order in Y. What happens to rate when both concentrations are halved?

Show Hint

Apply each concentration factor to its own power.

Show Answer

(1/2)⁰(1/2)² = 1/4. The overall order is 2, but X contributes no rate change.

Mind stretcher 2: Connect the rate equation to collisionsExtension

Question. A reaction is first order in A. Its temperature is kept constant while [A] is doubled. Explain the rate change using both the rate equation and collision theory.

Show Hint

State the mathematical factor first, then explain what doubling concentration changes inside a fixed volume.

Show Answer

The rate doubles because the factor from A is 2¹ = 2. Doubling [A] places twice as many A particles in each unit volume, so collisions involving A occur more frequently. Temperature is unchanged, so k is unchanged.

Mind stretcher 3: Reverse-engineering a concentrationExtension

Question. For rate = k[A]², what factor change in A makes rate nine times larger?

Show Hint

Solve f² = 9.

Show Answer

[A] must increase by a factor of 3.