Kinetics II
Notes for every section, eight new exam-style questions with full worked solutions, and online recall cards for your review days. Nothing here is copied from the book.
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16.1 The key terms
rate = k[A]m[B]n
- Rate equation: shows how the rate depends on the concentrations of reactants. It can only be found by experiment, not from the balanced equation.
- Order with respect to a reactant: the power of its concentration in the rate equation (usually 0, 1 or 2).
- Overall order: the sum of the individual orders, m + n.
- Rate constant, k: the constant of proportionality. It changes only with temperature (it increases as temperature rises).
Zero order: changing the concentration has no effect on the rate. First order: doubling the concentration doubles the rate. Second order: doubling it quadruples the rate.
Check yourself: If doubling [A] makes the rate four times faster, what is the order with respect to A?
Second order
16.2 Using rate equations and the units of k
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16.3 Choosing a technique to measure rate
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16.4 Initial rates and continuous monitoring
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16.5 Concentration–time graphs and half-life
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16.6 Rate–concentration graphs and initial-rate data
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16.7 Rate-determining steps and mechanisms
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16.8 Core Practical 13a: the iodination of propanone
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16.9 Core Practical 13b: a clock reaction
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16.10 Evidence for SN1 and SN2 mechanisms
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16.11 Activation energy from an Arrhenius plot
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16.12 Core Practical 14: finding the activation energy
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Extra worked questions
Cover the solution, answer on paper, then mark yourself against each step. Each step shows the mark it earns.
All 8 worked solutions are free for members. The first one is open to everyone.
Order and units
For rate = k[A][B]2, give the overall order and the units of k.
Try it first, then show the worked solution
| Order | 1 + 2 = 3, third order overall | 1 |
|---|---|---|
| Units | k = rate ÷ [A][B]2 = mol dm−3 s−1 ÷ (mol dm−3)3 = mol−2 dm6 s−1 | 1 |
Answer: Third order; mol−2 dm6 s−1
Examiner tip: Show the cancelling. It is quick and earns the mark even if you slip at the end.
Using a half-life
A reaction is first order in X with a half-life of 25 s. How long does [X] take to fall from 0.80 to 0.10 mol dm−3?
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Rate equation from initial rates
Initial-rate data at constant temperature: Exp 1 [A] 0.10, [B] 0.10, rate 2.0 × 10−4; Exp 2 [A] 0.20, [B] 0.10, rate 4.0 × 10−4; Exp 3 [A] 0.10, [B] 0.20, rate 8.0 × 10−4 (concentrations in mol dm−3, rates in mol dm−3 s−1). Find the rate equation and the value and units of k.
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Which step is slow?
2-bromo-2-methylpropane reacts with hydroxide ions. The rate equation is rate = k[(CH3)3CBr]. Suggest a two-step mechanism consistent with this, and identify the rate-determining step.
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Activation energy from a graph
A plot of ln k against 1/T for a reaction is a straight line with gradient −6000 K. Calculate the activation energy in kJ mol−1. R = 8.31 J K−1 mol−1.
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Which substitution mechanism?
For the reaction of 2-bromo-2-methylpropane with aqueous hydroxide ions, rate = k[(CH3)3CBr]. Name the mechanism and explain how the rate equation supports it.
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Arrhenius at two temperatures
A reaction has Ea = 50.0 kJ mol−1 and A = 1.0 × 1010 s−1. Calculate k at 300 K and at 310 K, and the factor by which the rate increases.
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Activation energy from a clock reaction
In a clock reaction run at five temperatures, a graph of ln(1/t) against 1/T is a straight line with gradient −6.25 × 103 K. Explain why 1/t can be used, and calculate the activation energy in kJ mol−1 (R = 8.31 J K−1 mol−1).
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Video walkthroughs
Narrated walkthroughs of exam-style questions from this topic of the book, showing where each mark comes from. Try the question first, then watch. Captions are available in the player.
Orders and rate constant from initial rates
Half-lives and a first-order reaction
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Rate equations and mechanisms
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Activation energy from an Arrhenius plot
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Bromate and bromide: a three-reactant rate equation
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Recall cards
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