Rate equations and orders of reactionEdexcel A-Level Chemistry: Flashcards
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Define the rate of reaction.
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- Define the rate of reaction.
- The change in concentration of a reactant or product per unit time (mol dm⁻³ s⁻¹).
- What does the order with respect to a reactant mean?
- The power to which that reactant's concentration is raised in the rate equation.
- Can the rate equation be written from the balanced equation?
- No. It must be found by experiment.
- What does doubling a concentration do to the rate for zero, first and second order?
- Zero: no change. First: rate doubles. Second: rate increases by a factor of 4.
- What are the units of k for an overall first-order reaction?
- s⁻¹
- What are the units of k for an overall second-order reaction?
- dm³ mol⁻¹ s⁻¹
- What are the units of k for an overall third-order reaction?
- dm⁶ mol⁻² s⁻¹
- Define half-life.
- The time taken for the concentration of a reactant to halve.
- How does half-life show that a reaction is first order?
- Successive half-lives are constant, whatever the concentration.
- What is the relationship between k and half-life for a first-order reaction?
- k = ln 2 ÷ t½ = 0.693 ÷ t½
- What shape is a rate–concentration graph for first order?
- A straight line through the origin; the gradient equals k.
- What shape is a concentration–time graph for zero order?
- A straight line with a negative gradient (constant rate).
- How is the rate found from a concentration–time graph?
- Draw a tangent at that time; the magnitude of the gradient is the rate.
- What does k depend on?
- Temperature (and the presence of a catalyst); it does not depend on concentration.
Exam questions on Rate equations and orders of reaction
- The reaction between peroxodisulfate ions and iodide ions is S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq). Initial rates were measured at constant temperature. Experiment 1: [S₂O₈²⁻] = 0.020 mol dm⁻³, [I⁻] = 0.040 mol dm⁻³, initial rate = 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 2: [S₂O₈²⁻] = 0.040 mol dm⁻³, [I⁻] = 0.040 mol dm⁻³, initial rate = 4.8 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 3: [S₂O₈²⁻] = 0.040 mol dm⁻³, [I⁻] = 0.080 mol dm⁻³, initial rate = 9.6 × 10⁻⁶ mol dm⁻³ s⁻¹.Calculate the value of the rate constant, k, for this reaction, including its units.2 marks
- Dinitrogen pentoxide, N₂O₅, decomposes in solution at constant temperature. The initial concentration of N₂O₅ is 0.80 mol dm⁻³. It falls to 0.40 mol dm⁻³ after 120 s and to 0.20 mol dm⁻³ after a further 120 s. The rate constant for the reaction at this temperature is 5.78 × 10⁻³ s⁻¹.Calculate the initial rate of reaction, including units.2 marks
- Nitrogen monoxide reacts with hydrogen: 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g). Initial rates were measured at constant temperature. Experiment 1: [NO] = 0.010 mol dm⁻³, [H₂] = 0.010 mol dm⁻³, initial rate = 2.0 × 10⁻⁷ mol dm⁻³ s⁻¹. Experiment 2: [NO] = 0.020 mol dm⁻³, [H₂] = 0.010 mol dm⁻³, initial rate = 8.0 × 10⁻⁷ mol dm⁻³ s⁻¹. Experiment 3: [NO] = 0.010 mol dm⁻³, [H₂] = 0.030 mol dm⁻³, initial rate = 6.0 × 10⁻⁷ mol dm⁻³ s⁻¹.Use the data to deduce the order of reaction with respect to NO and with respect to H₂, and write the rate equation.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).