Cambridge AS & A Level9701

Rate of reaction

Chemistry 9701 Chapter Notes

What this chapter covers

Rate of reactionEffect of temperature on reaction rates and the concept of activation energyHomogeneous and heterogeneous catalysts
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In this chapter

  1. 1Measuring the Rate of Reaction
  2. 2Rate Equations and Order of Reaction
  3. 3The Initial Rates Method
  4. 4Half-Life and Reaction Order
  5. 5Reaction Mechanisms and the Rate-Determining Step
  6. 6Homogeneous and Heterogeneous Catalysis

1. Measuring the Rate of Reaction

The rate of a chemical reaction is a measure of how fast reactants are used up or products are formed. It is formally defined as the change in concentration of a reactant or product per unit of time. The standard units for reaction rate are moles per decimetre cubed per second (mol dm⁻³ s⁻¹). As a reaction proceeds, the concentration of reactants decreases, so the rate of reaction typically slows down over time. To find the rate at a specific moment (the instantaneous rate), we plot a graph of concentration against time. The rate at any given time 't' is equal to the gradient of the tangent to the curve at that point. Since reactant concentration decreases, the gradient will be negative. The rate is a positive value, so we take the absolute value of the gradient, i.e., rate = -gradient.

Rate of reaction = (Change in concentration) / (Time taken)

Instantaneous rate = Gradient of tangent to concentration-time graph

Key term

Rate of Reaction: The change in concentration of a reactant or a product per unit of time.

Examiner insight

Examiners award marks for correctly drawing a tangent on the graph and for showing the calculation of its gradient using coordinates from the tangent line, not from the curve itself.

Common pitfall

Forgetting that the rate of reaction is always a positive value. When calculating rate from the gradient of a reactant concentration-time graph, the gradient will be negative, so you must take its positive value.

Worked example 13 marks

The decomposition of hydrogen peroxide, 2H₂O₂(aq) → 2H₂O(l) + O₂(g), was followed by measuring the concentration of H₂O₂ over time. The graph below shows the results. Calculate the initial rate of reaction.

  1. 1

    Step 1: To find the initial rate, draw a tangent to the curve at time t = 0.

  2. 2

    Step 2: Choose two points on the tangent to calculate the gradient. For example, the tangent passes through (0, 2.0) and (50, 0.0).

  3. 3

    Step 3: Calculate the gradient (Δy / Δx). Gradient = (0.0 - 2.0) mol dm⁻³ / (50 - 0) s = -2.0 / 50 = -0.040 mol dm⁻³ s⁻¹.

  4. 4

    Step 4: The rate of reaction is the positive value of the gradient. Rate = 0.040 mol dm⁻³ s⁻¹.

Recap

  • Reaction rate is the change in concentration of a substance over time.
  • The standard units for reaction rate are mol dm⁻³ s⁻¹.
  • The rate of reaction is fastest at the start and decreases as reactants are used up.
  • The instantaneous rate at a specific time is found by calculating the gradient of the tangent to the concentration-time graph at that point.

Quick check

  1. If the concentration of a product increases from 0.10 mol dm⁻³ to 0.25 mol dm⁻³ in 30 seconds, what is the average rate of reaction?2 marks

2. Rate Equations and Order of Reaction

A rate equation is a mathematical expression that shows how the rate of reaction depends on the concentrations of the reactants. For a general reaction A + B → Products, the rate equation takes the form: rate = k[A]ᵐ[B]ⁿ. Here, [A] and [B] are the concentrations of the reactants. 'k' is the rate constant, a proportionality constant that is specific to the reaction at a particular temperature. The powers 'm' and 'n' are the orders of reaction with respect to reactants A and B, respectively. The order of reaction tells you how the concentration of a reactant affects the rate. If m=0 (zero order), changing [A] has no effect on the rate. If m=1 (first order), doubling [A] doubles the rate. If m=2 (second order), doubling [A] quadruples the rate. The overall order of reaction is the sum of the individual orders (m + n). Importantly, the orders 'm' and 'n' can only be determined by experiment; they cannot be deduced from the stoichiometry of the balanced chemical equation.

rate = k[A]ᵐ[B]ⁿ

Key term

Order of Reaction: The power to which the concentration of a reactant is raised in the rate equation, which defines how the rate is affected by the concentration of that reactant.

Examiner insight

You must be able to determine the units of the rate constant for different overall orders. A common question asks you to calculate k and give its units, with one mark often reserved specifically for the correct units.

Common pitfall

Confusing the order of reaction (e.g., 'm' and 'n') with the stoichiometric coefficients in the balanced chemical equation. They are not related and must be found experimentally.

Worked example 13 marks

For the reaction X + Y → Z, the rate equation is found to be rate = k[X]²[Y]. The rate of reaction is 3.0 x 10⁻⁴ mol dm⁻³ s⁻¹ when [X] = 0.10 mol dm⁻³ and [Y] = 0.20 mol dm⁻³. Calculate the value of the rate constant, k, and state its units.

  1. 1

    Step 1: Rearrange the rate equation to solve for k. k = rate / ([X]²[Y]).

  2. 2

    Step 2: Substitute the given values into the equation. k = (3.0 x 10⁻⁴) / ((0.10)² x 0.20).

  3. 3

    Step 3: Calculate the numerical value. k = (3.0 x 10⁻⁴) / (0.01 x 0.20) = (3.0 x 10⁻⁴) / (0.002) = 0.15.

  4. 4

    Step 4: Determine the units of k by substituting units into the rearranged equation. Units = (mol dm⁻³ s⁻¹) / ((mol dm⁻³)² x (mol dm⁻³)) = (mol dm⁻³ s⁻¹) / (mol³ dm⁻⁹).

  5. 5

    Step 5: Simplify the units. Units = mol⁻² dm⁶ s⁻¹. So, k = 0.15 mol⁻² dm⁶ s⁻¹.

Recap

  • The rate equation links reaction rate to reactant concentrations.
  • The rate constant, k, is constant for a reaction at a specific temperature.
  • Order of reaction with respect to a reactant must be determined experimentally.
  • Zero order means concentration has no effect on rate.
  • First order means rate is directly proportional to concentration.
  • Second order means rate is proportional to the square of the concentration.
  • The units of k depend on the overall order of the reaction.

Quick check

  1. A reaction has the rate equation: rate = k[P][Q]². What is the overall order of the reaction?1 mark
  2. For the reaction in the previous question, what are the units of the rate constant k?1 mark

3. The Initial Rates Method

The initial rates method is a common experimental technique used to determine the orders of reaction and the rate equation. It involves carrying out a series of experiments where the initial concentration of one reactant is changed while the concentrations of all other reactants are kept constant. The initial rate of reaction is measured for each experiment (often by taking the gradient of the concentration-time graph at t=0). By comparing the change in concentration of a reactant with the resulting change in the initial rate, we can deduce the order of reaction with respect to that reactant. For example, if doubling the concentration of reactant A (while keeping others constant) causes the initial rate to double, the reaction is first order with respect to A. If the rate quadruples, it is second order. If the rate does not change, it is zero order. This process is repeated for each reactant to determine the full rate equation.

Key term

Initial Rate: The instantaneous rate of reaction at the very beginning of the reaction (at time t=0).

Worked example 16 marks

The reaction between propanone and iodine is acid-catalysed: CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + H⁺(aq) + I⁻(aq). The initial rates of reaction were measured in a series of experiments at a constant temperature.

Experiment[CH₃COCH₃] / mol dm⁻³[I₂] / mol dm⁻³[H⁺] / mol dm⁻³Initial Rate / mol dm⁻³ s⁻¹
10.500.200.501.4 x 10⁻⁵
21.000.200.502.8 x 10⁻⁵
31.000.400.502.8 x 10⁻⁵
41.000.201.005.6 x 10⁻⁵

a) Deduce the order of reaction with respect to each reactant.b) Write the rate equation for the reaction.c) Calculate the rate constant, k, using data from Experiment 4, and state its units.

  1. 1

    a) To find the order for CH₃COCH₃: Compare Experiments 1 and 2. [CH₃COCH₃] doubles (0.50 → 1.00), [I₂] and [H⁺] are constant. The rate doubles (1.4x10⁻⁵ → 2.8x10⁻⁵). Therefore, the reaction is first order with respect to CH₃COCH₃.

  2. 2

    a) To find the order for I₂: Compare Experiments 2 and 3. [I₂] doubles (0.20 → 0.40), [CH₃COCH₃] and [H⁺] are constant. The rate does not change (2.8x10⁻⁵ → 2.8x10⁻⁵). Therefore, the reaction is zero order with respect to I₂.

  3. 3

    a) To find the order for H⁺: Compare Experiments 2 and 4. [H⁺] doubles (0.50 → 1.00), [CH₃COCH₃] and [I₂] are constant. The rate quadruples (2.8x10⁻⁵ → 5.6x10⁻⁵ is a doubling, but the question has 2.8 to 5.6 - this is a doubling, not quadrupling. Let's assume the question meant 1.12x10⁻⁴ for quadrupling, or the order is 1. Let's re-read the data. Exp 2 rate is 2.8, Exp 4 rate is 5.6. Rate doubles. Therefore, the reaction is first order with respect to H⁺.

  4. 4

    b) The rate equation is rate = k[CH₃COCH₃]¹[I₂]⁰[H⁺]¹. This simplifies to rate = k[CH₃COCH₃][H⁺].

  5. 5

    c) Rearrange the rate equation: k = rate / ([CH₃COCH₃][H⁺]). Using data from Experiment 4: k = (5.6 x 10⁻⁵) / (1.00 x 1.00) = 5.6 x 10⁻⁵.

  6. 6

    c) To find the units: units = (mol dm⁻³ s⁻¹) / (mol dm⁻³ x mol dm⁻³) = (mol dm⁻³ s⁻¹) / (mol² dm⁻⁶) = mol⁻¹ dm³ s⁻¹. So, k = 5.6 x 10⁻⁵ mol⁻¹ dm³ s⁻¹.

Recap

  • The initial rates method determines reaction orders by changing one reactant's concentration at a time.
  • If doubling concentration doubles the rate, the order is 1.
  • If doubling concentration quadruples the rate, the order is 2.
  • If doubling concentration has no effect on the rate, the order is 0.
  • Once all orders are known, the rate equation can be written and the rate constant k can be calculated.

Quick check

  1. In an initial rates experiment, if the concentration of a reactant is tripled and the initial rate increases by a factor of nine, what is the order of reaction with respect to that reactant?1 mark

4. Half-Life and Reaction Order

The half-life (t½) of a reaction is the time taken for the concentration of a reactant to fall to half its initial value. The relationship between half-life and reaction order is a powerful tool for identifying the order from a single experiment. For a first-order reaction, the half-life is constant and independent of the initial concentration. This means it takes the same amount of time for the concentration to go from 1.0 to 0.5 mol dm⁻³, as it does to go from 0.5 to 0.25 mol dm⁻³, and so on. For a zero-order reaction, the half-life decreases as the reaction proceeds. For a second-order reaction, the half-life increases as the reaction proceeds. By measuring successive half-lives from a concentration-time graph, you can determine the order. The constant half-life of first-order reactions allows for a simple calculation of the rate constant, k, using the formula k = ln(2) / t½.

t½ = time for [Reactant] to halve

For first-order reactions: k = ln(2) / t½ ≈ 0.693 / t½

Key term

Half-life (t½): The time required for the concentration of a reactant to decrease to half of its initial value.

Fun fact

Carbon-14 dating, used to determine the age of ancient organic materials like fossils and artifacts, relies on the constant half-life (5730 years) of the first-order radioactive decay of carbon-14.

Worked example 14 marks

The radioactive decay of strontium-90 is a first-order process with a half-life of 28 years. A sample of strontium-90 has an initial concentration of 0.080 mol dm⁻³.a) Calculate the rate constant, k, for this decay, including units.b) Calculate the concentration of strontium-90 remaining after 84 years.

  1. 1

    a) For a first-order reaction, k = ln(2) / t½. k = ln(2) / 28 = 0.693 / 28 = 0.02475. The units are time⁻¹, so k = 0.0248 year⁻¹ (to 3 s.f.).

  2. 2

    b) 84 years is the duration. The number of half-lives is Time / t½ = 84 years / 28 years = 3 half-lives.

  3. 3

    b) After 1 half-life (28 years): concentration = 0.080 / 2 = 0.040 mol dm⁻³.

  4. 4

    b) After 2 half-lives (56 years): concentration = 0.040 / 2 = 0.020 mol dm⁻³.

  5. 5

    b) After 3 half-lives (84 years): concentration = 0.020 / 2 = 0.010 mol dm⁻³. The concentration remaining is 0.010 mol dm⁻³.

Recap

  • Half-life (t½) is the time for reactant concentration to halve.
  • A constant half-life is the defining characteristic of a first-order reaction.
  • For a zero-order reaction, half-life decreases over time.
  • For a second-order reaction, half-life increases over time.
  • The rate constant for a first-order reaction can be calculated using k = ln(2) / t½.

Quick check

  1. From a concentration-time graph, the time taken for the concentration to fall from 2.0 to 1.0 mol dm⁻³ is 50s. The time taken to fall from 1.0 to 0.5 mol dm⁻³ is also 50s. What is the order of reaction and the value of the half-life?2 marks

5. Reaction Mechanisms and the Rate-Determining Step

Most reactions do not occur in a single collision but proceed through a series of simpler, elementary steps. This sequence of steps is called the reaction mechanism. Within this sequence, one step will be significantly slower than all the others. This is the rate-determining step (RDS). The RDS acts as a bottleneck for the entire reaction, so the overall rate of the reaction is governed by the rate of this slow step. A crucial principle is that the rate equation for the overall reaction is determined by the reactants involved in the rate-determining step. The order of reaction with respect to each reactant is equal to the number of molecules of that reactant taking part in the RDS. For a proposed mechanism to be valid, two conditions must be met: 1) The elementary steps must add up to give the overall balanced equation for the reaction. 2) The rate equation derived from the RDS must match the experimentally determined rate equation.

Key term

Rate-Determining Step (RDS): The slowest step in a multi-step reaction mechanism that governs the overall rate of reaction.

Examiner insight

When asked to suggest a mechanism, you must ensure your proposed steps add up to the overall equation and that any intermediates are cancelled out. You must also explicitly state which step is the slow (rate-determining) step.

Worked example 14 marks

The reaction 2NO₂(g) + F₂(g) → 2NO₂F(g) has the experimentally determined rate equation: rate = k[NO₂][F₂]. A scientist proposes a two-step mechanism: Step 1: NO₂ + F₂ → NO₂F + F (slow) Step 2: NO₂ + F → NO₂F (fast) Is this proposed mechanism consistent with the experimental data? Justify your answer.

  1. 1

    Step 1: Check if the steps sum to the overall equation. (NO₂ + F₂ → NO₂F + F) + (NO₂ + F → NO₂F) Adding them gives: 2NO₂ + F₂ + F → 2NO₂F + F. The 'F' atom is an intermediate and cancels out, leaving 2NO₂ + F₂ → 2NO₂F. This matches the overall equation. The first condition is met.

  2. 2

    Step 2: Determine the rate equation from the proposed mechanism. The rate is determined by the slow step (Step 1). The reactants in Step 1 are one molecule of NO₂ and one molecule of F₂.

  3. 3

    Step 3: Write the rate equation based on the RDS. The rate equation would be rate = k[NO₂]¹[F₂]¹, or rate = k[NO₂][F₂].

  4. 4

    Step 4: Compare the derived rate equation with the experimental one. The derived rate equation (rate = k[NO₂][F₂]) matches the experimentally determined rate equation. The second condition is met.

  5. 5

    Conclusion: Yes, the proposed mechanism is consistent with the experimental data because the elementary steps sum to the overall equation and the rate equation derived from the slow step matches the observed rate equation.

Recap

  • A reaction mechanism is the series of elementary steps that make up an overall reaction.
  • The rate-determining step (RDS) is the slowest step in the mechanism.
  • The overall reaction rate is equal to the rate of the RDS.
  • The rate equation is determined by the species present in the RDS.
  • A valid mechanism must sum to the overall equation and be consistent with the experimental rate law.

Quick check

  1. A reaction's rate-determining step involves the collision of two molecules of reactant A. What is the order of reaction with respect to A?1 mark

6. Homogeneous and Heterogeneous Catalysis

A catalyst is a substance that increases the rate of a chemical reaction without being chemically changed itself at the end of the reaction. Catalysts work by providing an alternative reaction pathway with a lower activation energy (Ea), meaning more colliding particles have sufficient energy to react, thus increasing the rate. There are two main types of catalysts. Homogeneous catalysts are in the same physical state (phase) as the reactants, for example, two gases or two aqueous solutions. They work by forming an intermediate compound which then breaks down to form the product and regenerate the catalyst. Heterogeneous catalysts are in a different physical state from the reactants, for example, a solid catalyst with gaseous reactants. They typically work by providing a solid surface onto which reactant molecules can adsorb (bind), weakening their bonds and bringing them closer together in the correct orientation for reaction.

Key term

Catalyst: A substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy, while remaining chemically unchanged at the end of the reaction.

Worked example 13 marks

Explain the role of the iron catalyst in the Haber process (N₂(g) + 3H₂(g) ⇌ 2NH₃(g)). State the type of catalysis.

  1. 1

    This is an example of heterogeneous catalysis because the iron catalyst is a solid while the reactants (N₂ and H₂) are gases.

  2. 2

    The N₂ and H₂ molecules diffuse to the surface of the solid iron catalyst.

  3. 3

    The molecules are adsorbed onto the iron surface. This process weakens the strong covalent bonds within the N₂ and H₂ molecules, lowering the activation energy.

  4. 4

    The adsorbed atoms react on the surface to form NH₃ molecules.

  5. 5

    The NH₃ product molecules then desorb from the surface, freeing up the active sites for more reactants.

Worked example 23 marks

Atmospheric oxides of nitrogen, such as NO₂, catalyse the oxidation of sulfur dioxide to sulfur trioxide, a key step in acid rain formation. Explain this process, stating the type of catalysis.

  1. 1

    This is an example of homogeneous catalysis as both the catalyst (NO₂) and the reactant (SO₂) are gases.

  2. 2

    The catalyst works via a two-step mechanism, forming an intermediate.

  3. 3

    Step 1: The catalyst, NO₂, oxidises the sulfur dioxide: SO₂(g) + NO₂(g) → SO₃(g) + NO(g).

  4. 4

    Step 2: The catalyst is regenerated by the oxidation of NO by atmospheric oxygen: 2NO(g) + O₂(g) → 2NO₂(g).

  5. 5

    The overall reaction is 2SO₂(g) + O₂(g) → 2SO₃(g), with NO₂ acting as the catalyst as it is used in step 1 and regenerated in step 2.

Recap

  • Catalysts increase reaction rate by lowering the activation energy.
  • Homogeneous catalysts are in the same phase as the reactants.
  • Heterogeneous catalysts are in a different phase from the reactants.
  • The Haber process uses a solid iron (heterogeneous) catalyst.
  • Catalytic converters use solid Pt, Pd, and Rh (heterogeneous) catalysts to remove pollutants from exhaust fumes.
  • NO₂ acts as a homogeneous catalyst in the formation of acid rain.

Quick check

  1. The reaction between aqueous iodide ions (I⁻) and peroxodisulfate ions (S₂O₈²⁻) is catalysed by aqueous Fe³⁺ ions. What type of catalysis is this?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define 'rate constant' and state two factors that can affect its value.3 marks
  2. A reaction is first order with respect to reactant A and second order with respect to reactant B. (i) Write the rate equation for this reaction. (ii) State the overall order. (iii) By what factor will the initial rate of reaction increase if the concentrations of both A and B are doubled?4 marks
  3. The graph shows the change in concentration of a reactant X over time. Use the graph to show that the reaction is first order and calculate the rate constant, k, including its units.4 marks
  4. For the reaction P + 2Q → R, the following initial rates data were obtained. | Experiment | [P] / mol dm⁻³ | [Q] / mol dm⁻³ | Initial Rate / mol dm⁻³ s⁻¹ | |---|---|---|---| | 1 | 0.10 | 0.10 | 2.0 x 10⁻⁴ | | 2 | 0.20 | 0.10 | 4.0 x 10⁻⁴ | | 3 | 0.20 | 0.20 | 1.6 x 10⁻³ | Determine the rate equation and calculate the rate constant with units.5 marks
  5. Describe an experimental technique, other than measuring concentration, that could be used to follow the rate of the reaction: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g).2 marks
  6. The overall reaction for the oxidation of iodide ions by hydrogen peroxide in acidic solution is: H₂O₂(aq) + 2I⁻(aq) + 2H⁺(aq) → I₂(aq) + 2H₂O(l). The rate equation is found to be rate = k[H₂O₂][I⁻]. Suggest a two-step mechanism for this reaction, identifying the rate-determining step.3 marks
  7. A first-order reaction has a rate constant of 4.5 x 10⁻³ s⁻¹. Calculate the half-life of the reaction.2 marks
  8. Explain, with the aid of equations, the role of rhodium in a catalytic converter in the removal of nitrogen monoxide from car exhaust fumes.3 marks
  9. A reaction has the rate equation rate = k[A]². When the initial concentration of A is 0.50 mol dm⁻³, the initial rate is 8.0 x 10⁻⁵ mol dm⁻³ s⁻¹. Calculate the rate of reaction when the concentration of A has fallen to 0.10 mol dm⁻³.4 marks
  10. Distinguish between homogeneous and heterogeneous catalysis, giving one specific, named industrial example of each.4 marks

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