Cambridge AS & A Level9701

Simple rate equations, orders of reaction and rate constants

Chemistry 9701 Chapter Notes

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Simple rate equations, orders of reaction and rate constantsHomogeneous and heterogeneous catalysts
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In this chapter

  1. 1Rate Equations, Order, and Rate Constant
  2. 2The Initial Rates Method
  3. 3Concentration-Time Graphs and Order
  4. 4Half-Life and First-Order Reactions
  5. 5Mechanisms and the Rate-Determining Step

1. Rate Equations, Order, and Rate Constant

The rate of a chemical reaction depends on the concentration of its reactants. A rate equation is a mathematical expression, found by experiment, that shows this relationship. 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 reactants A and B. The powers 'm' and 'n' are the 'orders of reaction' with respect to each reactant. They tell you how much the rate changes when you change the concentration of that reactant. The 'overall order' is the sum of the individual orders (m + n). The term 'k' is the rate constant, a value specific to a reaction at a certain temperature. Importantly, the orders 'm' and 'n' can only be found from experimental data; you cannot simply read them from the balancing numbers in the overall chemical equation.

rate = k[A]^m[B]^n

Key term

Rate Equation: An experimentally determined equation that links the rate of reaction to the concentrations of the reactants, each raised to a specific power.

Examiner insight

Examiners frequently test the calculation of units for the rate constant, k. Always show your working by rearranging the rate equation and cancelling units, as this is where marks are often awarded.

Common pitfall

Confusing the orders of reaction (the powers in the rate equation) with the stoichiometric coefficients (the balancing numbers) in the overall chemical equation. Orders must be found by experiment.

Worked example 12 marks

A reaction has the rate equation: rate = k[H₂][NO]². Determine the units of the rate constant, k.

  1. 1

    Step 1: Write down the rate equation: rate = k[H₂][NO]²

  2. 2

    Step 2: Rearrange the equation to make k the subject: k = rate / ([H₂][NO]²)

  3. 3

    Step 3: Substitute the standard units for rate (mol dm⁻³ s⁻¹) and concentration (mol dm⁻³): k = (mol dm⁻³ s⁻¹) / ((mol dm⁻³) × (mol dm⁻³)²)

  4. 4

    Step 4: Simplify the denominator: k = (mol dm⁻³ s⁻¹) / (mol³ dm⁻⁹)

  5. 5

    Step 5: Cancel and combine the units. For mol: 1 - 3 = -2. For dm: -3 - (-9) = 6. The s⁻¹ remains. So, the units are mol⁻² dm⁶ s⁻¹.

  6. 6

    Step 6: Write the final units, usually with positive indices first: dm⁶ mol⁻² s⁻¹.

Worked example 23 marks

For the reaction BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l), the experimental rate equation is found to be rate = k[BrO₃⁻][Br⁻][H⁺]². State the order of reaction with respect to each reactant and the overall order of the reaction.

  1. 1

    Step 1: Identify the power of [BrO₃⁻] in the rate equation. The power is 1. So, the reaction is first-order with respect to BrO₃⁻.

  2. 2

    Step 2: Identify the power of [Br⁻] in the rate equation. The power is 1. So, the reaction is first-order with respect to Br⁻.

  3. 3

    Step 3: Identify the power of [H⁺] in the rate equation. The power is 2. So, the reaction is second-order with respect to H⁺.

  4. 4

    Step 4: Calculate the overall order by summing the individual orders: Overall order = 1 (for BrO₃⁻) + 1 (for Br⁻) + 2 (for H⁺) = 4.

Recap

  • A rate equation shows how reactant concentration affects reaction rate.
  • The order of reaction with respect to a reactant is the power its concentration is raised to in the rate equation.
  • Overall order is the sum of the individual orders of reaction.
  • The rate constant, k, is constant for a reaction at a fixed temperature.
  • Orders of reaction must be determined experimentally and are not the same as stoichiometric coefficients.

Quick check

  1. A reaction is zero-order with respect to reactant Z. What happens to the rate if [Z] is doubled?1 mark
  2. What are the units of the rate constant for a simple first-order reaction, rate = k[A]?1 mark

2. The Initial Rates Method

The initial rates method is a common experimental approach to determine the orders of reaction. 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. By comparing the initial rate of reaction across these experiments, we can deduce the order with respect to the reactant that was changed. For example, if doubling the concentration of reactant A doubles the initial rate, the reaction is first-order with respect to A. If it quadruples the rate, it's second-order. If the rate is unchanged, it's zero-order. This process is repeated for each reactant to build the full rate equation.

Key term

Initial Rate: The instantaneous rate of a reaction at the very beginning (time t=0), before the concentrations of reactants have significantly decreased.

Examiner insight

When asked to deduce the order from data, you must explicitly state how the concentration and rate change between specific experiments to justify your conclusion. Simply stating the order is not enough for full marks.

Common pitfall

Trying to compare two experiments where more than one concentration has changed. You must isolate the effect of a single reactant by choosing experiments where all other concentrations are held constant.

Worked example 14 marks

The reaction A + B + C → Products was studied. Use the data below to deduce the order with respect to A, B, and C, and write the rate equation.

Exp[A]/mol dm⁻³[B]/mol dm⁻³[C]/mol dm⁻³Rate/mol dm⁻³s⁻¹
10.101.001.000.0078
20.201.001.000.0078
31.000.101.000.00008
41.000.201.000.00032
51.001.000.100.00078
61.001.000.200.00156
  1. 1

    Step 1: Find the order with respect to A. Compare Experiments 1 and 2, where [B] and [C] are constant. [A] doubles (0.10 → 0.20), but the rate is unchanged (0.0078 → 0.0078). Therefore, the reaction is zero-order with respect to A.

  2. 2

    Step 2: Find the order with respect to B. Compare Experiments 3 and 4, where [A] and [C] are constant. [B] doubles (0.10 → 0.20). The rate increases by a factor of 4 (0.00032 / 0.00008 = 4). Since 2² = 4, the reaction is second-order with respect to B.

  3. 3

    Step 3: Find the order with respect to C. Compare Experiments 5 and 6, where [A] and [B] are constant. [C] doubles (0.10 → 0.20). The rate also doubles (0.00156 / 0.00078 = 2). Since 2¹ = 2, the reaction is first-order with respect to C.

  4. 4

    Step 4: Combine the orders into the rate equation. rate = k[A]⁰[B]²[C]¹. Since [A]⁰ = 1, the final rate equation is rate = k[B]²[C].

Recap

  • The initial rates method helps find the orders of reaction experimentally.
  • To find the order for a reactant, compare two experiments where only its concentration changes.
  • Zero-order: changing concentration has no effect on rate.
  • First-order: doubling concentration doubles the rate.
  • Second-order: doubling concentration quadruples the rate.

Quick check

  1. In an experiment, tripling the concentration of reactant X causes the initial rate to increase by a factor of 9. What is the order with respect to X?1 mark

3. Concentration-Time Graphs and Order

Plotting the concentration of a reactant against time gives a curve whose shape reveals the order of reaction. A zero-order reaction has a constant rate, so its concentration-time graph is a straight line with a negative gradient. The gradient of this line is equal to -k. A first-order reaction slows down as the reactant is used up, producing a downward curve. Its key feature is a constant half-life. A second-order reaction also produces a downward curve, but it is steeper at the start and tails off more slowly than a first-order reaction. The rate at any point in time for any order can be found by drawing a tangent to the curve at that time and calculating its gradient.

Key term

Zero-Order Reaction: A reaction where the rate is independent of the concentration of the reactants.

Examiner insight

To prove a reaction is first-order from a graph, you must demonstrate the half-life is constant. It is not enough to just say the graph is a curve. You must take at least two half-life measurements from the graph and show they are the same.

Fun fact

The concept of half-life is crucial in pharmacology to determine drug dosage intervals, ensuring the concentration of a medicine in the body stays within a safe and effective range.

Worked example 14 marks

The graph shows the concentration of a reactant X versus time. By taking at least two measurements, show that the reaction is first-order and calculate the rate constant, k.

  1. 1

    Step 1: Identify the initial concentration at t=0. From the graph, [X] = 1.60 mol dm⁻³.

  2. 2

    Step 2: Find the time for the concentration to halve for the first time (the first half-life). Half of 1.60 is 0.80 mol dm⁻³. Reading from the graph, the time taken to reach 0.80 is 20 seconds. So, the first half-life is 20 s.

  3. 3

    Step 3: Find the time for the concentration to halve again (the second half-life). Starting from 0.80 mol dm⁻³ at t=20s, the next half-concentration is 0.40 mol dm⁻³. Reading from the graph, the time taken to reach 0.40 is 40 seconds. The time for this second interval is 40s - 20s = 20s.

  4. 4

    Step 4: Conclude the order. Since the first half-life (20s) is equal to the second half-life (20s), the half-life is constant. This is the characteristic of a first-order reaction.

  5. 5

    Step 5: Calculate the rate constant, k, using the formula k = ln(2) / t½. k = ln(2) / 20 s.

  6. 6

    Step 6: Calculate the final value. k ≈ 0.693 / 20 = 0.0347 s⁻¹.

Recap

  • A concentration-time graph for a zero-order reaction is a straight line with a negative gradient.
  • A concentration-time graph for a first-order reaction is a curve with a constant half-life.
  • A concentration-time graph for a second-order reaction is a steeper curve than first-order, with a non-constant half-life.
  • The rate at any time 't' is the gradient of the tangent to the curve at that point.

Quick check

  1. A concentration-time graph for a reactant is a straight line with a negative slope. What is the order of reaction with respect to this reactant?1 mark

4. Half-Life and First-Order Reactions

The half-life (t½) of a reaction is the time it takes for the concentration of a reactant to decrease to half its initial value. For zero and second-order reactions, the half-life changes as the reaction progresses. However, first-order reactions have a unique and important property: their half-life is constant. It does not depend on the initial concentration. This means it takes the same amount of time for the concentration to go from 100% to 50% as it does to go from 50% to 25%, or from 25% to 12.5%, and so on. This constant half-life has a direct mathematical relationship with the rate constant, k, given by the formula t½ = ln(2)/k.

t_1/2 = ln(2) / k

Key term

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

Fun fact

Radiocarbon dating works because the decay of carbon-14 is a first-order process with a constant half-life of about 5730 years, allowing scientists to date ancient organic objects.

Worked example 12 marks

The decomposition of hydrogen peroxide is first-order. If the rate constant, k, for the reaction is 8.3 x 10⁻⁴ s⁻¹, calculate the half-life of the reaction in seconds.

  1. 1

    Step 1: Identify the correct formula. For a first-order reaction, t½ = ln(2) / k.

  2. 2

    Step 2: Substitute the given values into the formula. t½ = ln(2) / (8.3 x 10⁻⁴ s⁻¹).

  3. 3

    Step 3: Calculate the value of ln(2), which is approximately 0.693.

  4. 4

    Step 4: Perform the division: t½ = 0.693 / (8.3 x 10⁻⁴) ≈ 835 seconds.

Worked example 22 marks

A first-order reaction has a half-life of 30 minutes. What percentage of the original reactant will remain after 90 minutes?

  1. 1

    Step 1: Determine how many half-lives have passed. Number of half-lives = Total time / Half-life = 90 mins / 30 mins = 3.

  2. 2

    Step 2: Calculate the remaining amount after each half-life. After 1st half-life (30 mins): 50% remains. After 2nd half-life (60 mins): 50% of 50% = 25% remains. After 3rd half-life (90 mins): 50% of 25% = 12.5% remains.

  3. 3

    Step 3: State the final answer. 12.5% of the original reactant will remain.

Recap

  • Half-life is the time for reactant concentration to halve.
  • For first-order reactions, the half-life is constant and independent of concentration.
  • The formula t½ = ln(2)/k links the half-life and rate constant for first-order reactions.
  • After 'n' half-lives, the fraction of reactant remaining is (1/2)ⁿ.

Quick check

  1. A first-order reaction has a half-life of 10 days. How long does it take for the concentration to fall from 0.8 mol dm⁻³ to 0.2 mol dm⁻³?2 marks

5. Mechanisms and the Rate-Determining Step

Many chemical reactions do not happen in a single collision but proceed through a sequence of simpler steps, known as the reaction mechanism. The slowest step in this sequence is called the rate-determining step (RDS). It acts as a bottleneck, controlling the overall speed of the reaction. The rate equation for the overall reaction is determined by the species involved in this slow step. Specifically, the order of reaction with respect to a reactant is the number of molecules of that reactant taking part in the rate-determining step. Therefore, by comparing the experimentally determined rate equation with a proposed mechanism, we can check for consistency. A valid mechanism must have a rate-determining step that matches the rate equation, and its individual steps must sum to give the overall balanced chemical equation.

Key term

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

Examiner insight

When suggesting a mechanism, you must clearly label the slow (rate-determining) and fast steps. The reactants of the slow step must match the species and powers in the given rate equation.

Common pitfall

Forgetting to check that the elementary steps of a proposed mechanism add up to the overall stoichiometric equation. A mechanism must be consistent with both the rate law and the overall reaction.

Worked example 13 marks

The overall reaction 2NO₂(g) + F₂(g) → 2NO₂F(g) has the experimental rate equation: rate = k[NO₂][F₂]. A two-step mechanism is proposed: Step 1: NO₂ + F₂ → NO₂F + F (slow) Step 2: NO₂ + F → NO₂F (fast) Show that this mechanism is consistent with the rate equation and the overall reaction.

  1. 1

    Step 1: Check consistency with the rate equation. The rate equation is determined by the slow step (RDS). The reactants in Step 1 are one molecule of NO₂ and one molecule of F₂.

  2. 2

    Step 2: This means the reaction should be first-order with respect to NO₂ and first-order with respect to F₂. This gives a predicted rate equation of rate = k[NO₂][F₂].

  3. 3

    Step 3: This predicted rate equation matches the experimental rate equation. So, the mechanism is consistent with the kinetics.

  4. 4

    Step 4: Check consistency with the overall reaction equation. Add the steps of the mechanism together: (NO₂ + F₂ → NO₂F + F) + (NO₂ + F → NO₂F).

  5. 5

    Step 5: Cancel any species that appear on both sides (intermediates). The 'F' atom is on the right in Step 1 and on the left in Step 2, so it cancels out.

  6. 6

    Step 6: The sum of the steps is NO₂ + F₂ + NO₂ → NO₂F + NO₂F, which simplifies to 2NO₂(g) + F₂(g) → 2NO₂F(g). This matches the overall equation. The mechanism is consistent.

Recap

  • A reaction mechanism is the series of steps by which a reaction occurs.
  • The slowest step is the rate-determining step (RDS) and it controls the overall rate.
  • The rate equation only includes species involved in or before the rate-determining step.
  • The order with respect to a reactant in the rate equation is the number of its molecules in the RDS.
  • A valid mechanism must match both the experimental rate equation and the overall stoichiometry.

Quick check

  1. A reaction has the rate equation rate = k[X]². Which of these could be the rate-determining step? (a) X + Y → P (b) 2X → Q1 mark

End-of-chapter exercise

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

  1. For a reaction with the rate equation rate = k[P]²[Q], what is the effect on the initial rate of reaction if the concentration of P is doubled and the concentration of Q is halved?2 marks
  2. The rate constant for a second-order reaction is 0.54 dm³ mol⁻¹ s⁻¹. What is the rate of reaction when the concentration of the reactant is 0.020 mol dm⁻³? The rate equation is rate = k[A]².2 marks
  3. Explain the difference between the order of a reaction and the overall order of a reaction.2 marks
  4. The radioactive isotope iodine-131 is used in medicine. It has a half-life of 8 days and its decay is a first-order process. A hospital receives a sample containing 20.0 mg of iodine-131. What mass of iodine-131 will remain after 24 days?3 marks
  5. For the reaction X + Y → Z, the following initial rates data were obtained. Deduce the rate equation and calculate the rate constant, k, including its units. | Exp | [X]/mol dm⁻³ | [Y]/mol dm⁻³ | Initial Rate/mol dm⁻³s⁻¹ | |---|---|---|---| | 1 | 0.1 | 0.2 | 4.0 x 10⁻⁴ | | 2 | 0.1 | 0.4 | 8.0 x 10⁻⁴ | | 3 | 0.3 | 0.4 | 7.2 x 10⁻³ |5 marks
  6. A concentration-time graph for a first-order reaction is a curve. How would you use this graph to confirm the reaction is first-order and not second-order?2 marks
  7. The reaction between nitrogen monoxide and hydrogen is: 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g). The experimental rate equation is rate = k[NO]²[H₂]. Suggest a two-step mechanism for this reaction, identifying the rate-determining step.3 marks
  8. Explain what is meant by the term 'rate-determining step' and why species that are only involved in steps after the rate-determining step do not appear in the rate equation.3 marks
  9. A reaction is third-order overall. Write down one possible rate equation for a reaction between two reactants, A and B, that fits this description, and determine the units of the rate constant, k.3 marks
  10. The decomposition of dinitrogen pentoxide, 2N₂O₅(g) → 4NO₂(g) + O₂(g), is a first-order reaction. In an experiment, the initial concentration of N₂O₅ is 1.24 x 10⁻² mol dm⁻³. If the rate constant is 6.2 x 10⁻⁴ s⁻¹, calculate the initial rate of reaction and the half-life of the decomposition.4 marks

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