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A LevelChemistryCambridge (CIE)Revision Notes26. Reaction KineticsSimple Rate Equations, Orders of Reaction & Rate ConstantsChemical Kinetics Terminology

Chemical Kinetics Terminology (Cambridge (CIE) A Level Chemistry) : Revision Note

Richard Boole

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Chemical Kinetics Terminology

  • The rate of reaction refers to the change in the amount or concentration of a reactant or product per unit time and can be found by:

  • Measuring the decrease in the concentration of a reactant OR

  • Measuring the increase in the concentration of a product over time

    • The units of rate of reaction are mol dm-3 s-1 

begin mathsize 16px style bold Rate bold space bold of bold space bold reaction bold equals fraction numerator bold change bold space bold in bold space bold concentration over denominator bold time end fraction end style

Rate equation

  • Rate equations can only be determined experimentally

    • They cannot be found from the stoichiometric equations

  • For the general reaction of P and Q reacting together to form products:

P (aq) + Q (aq) → R (aq) + S (g)

  • The rate equation will include:

    • A rate / proportionality constant, k

      • This can be calculated from the gradient of the graph

    • The concentration of the reactants

      • They are shown in square brackets for concentration, e.g. [P] and [Q]

    • The order to which each reactant is raised

      • They are shown as powers, e.g. m and n

      • The order with respect to any reactant can only be 0, 1 or 2

Rate of reaction = k [P]m [Q]n

  • The rate equation does not include the concentration of the products

    • This is because they do not affect the rate of reaction

Example reactions for rate equations

  • The following general reaction will be used as an example to study the rate of reaction

D (aq) → E (aq) + F (g) 

  • The rate of reaction at different concentrations of D is measured and tabulated

Rate of reactions table

[D] (mol dm-3)

Rate (mol dm-3 s-1)

fraction numerator bold rate over denominator stretchy left square bracket D stretchy right square bracket end fraction (s-1)

3.00

2.00 x 10-3

6.67 x 10-4

2.00

1.33 x 10-3

6.67 x 10-4

1.00

6.60 x 10-4

6.67 x 10-4

  • A directly proportional relationship between the rate of the reaction and the concentration of D is observed when a graph is plotted

Rates [D] graph, downloadable AS & A Level Chemistry revision notes

Rate of reaction over various concentrations of D

  • For the above reaction, the rate equation is:

Rate = k [D]

  • The value of the rate / proportionality constant, k, can be calculated from the results or from two points on the graph

    • For this example, the value is is 6.67 x 10-4 s-1

Nitric oxide and hydrogen

  • The reaction between nitric oxide and hydrogen is:

2NO (g) + 2H2 (g) → N2 (g) + 2H2O (g)

  • The rate equation for this reaction is:

rate = k [NO]2 [H2]

  • By keeping the concentration of one reactant constant, the rate equation can show the effect of each reactants

  • Keeping [H2] constant:

    • This means that [H2] will not affect the rate of reaction

    • Any change in the rate of reaction is caused by [NO]

    • The change in the rate of reaction is proportional to the square of [NO]:

    Rate = k1 [NO]2

  • Keeping [NO] constant:

    • This means that [NO] will not affect the rate of reaction

    • Any change in the rate of reaction is caused by [H2]

    • The change in the rate of reaction is proportional to [H2]:

    Rate = k2 [H2]

  • The individual equations can be combined to give the overall rate equation

    • k = k1 + k2

Rate = k [NO]2 [H2]

  • Notice that the [H2] does not have an order of 2

    • This is because the order must be determined experimentally, not from the equation

Order of reaction

  • The order of a reactant shows how the concentration of a reactant affects the rate of reaction

  • It is the power to which the concentration of that reactant is raised in the rate equation

    • The order can be 0, 1 or 2

  • When the order of reaction with respect to a chemical is 0

    • Changing the concentration of the chemical has no effect on the rate of the reaction

    • Therefore, it is not included in the rate equation

  • When the order of reaction with respect to a chemical is 1

    • The concentration of the chemical is directly proportional to the rate of reaction, e.g. doubling the concentration of the chemical doubles the rate of reaction

    • The chemical is included in the rate equation

  • When the order of reaction with respect to a chemical is 2

    • The rate is directly proportional to the square of the concentration of that chemical, e.g. doubling the concentration of the chemical increases the rate of reaction by a factor of four

    • The chemical is included in the rate equation (appearing as a squared term)

  • The overall order of reaction is the sum of the powers of the reactants in a rate equation

Rate = k [NO]2 [H2]

  • For example, in the rate equation above, the reaction is:

    • Second-order with respect to NO

    • First-order with respect to H2

    • Third-order overall (2 + 1)

Worked Example

The chemical equation for the thermal decomposition of dinitrogen pentoxide is:

2N2O5 (g) → 4NO2 (g) + O2 (g)

The rate equation for this reaction is:

Rate = k[N2O5 (g)]

  1. State the order of the reaction with respect to dinitrogen pentoxide

  2. Deduce the effect on the rate of reaction if the concentration of dinitrogen pentoxide is tripled

Answers:

  1. The order with respect to dinitrogen pentoxide:

    • Dinitrogen pentoxide features in the rate equation, therefore, it cannot be order zero / 0

    • The dinitrogen pentoxide is not raised to a power, which means that it cannot be order 2 / second order

    • Therefore, the order with respect to dinitrogen pentoxide must be order 1 / first order

  2. The effect of tripling [N2O5]:

    • Since the reaction is first order, the concentration of dinitrogen pentoxide is directly proportional to the rate

    • This means that if the concentration of the dinitrogen pentoxide is tripled, then the rate of reaction will also triple

Deducing Orders

  • To derive the rate equation for a reaction, you can use a graph or a table of results

    • The type and shape of the graph indicates the order with respect to a reactant

    • A table or results requires calculation

  • Take the reactants one at a time and find the order with respect to each reactant individually

  • Steps to derive a rate equation:

    1. Identify two experiments where:

      • The concentration of one reactant changes and the concentrations of all other reactants remain constant

      • Calculate what has happened to the concentration of the reactant

      • Calculate what has happened to the rate of reaction

      • Determine the order with respect to that reactant

    2. Repeat this for all of the reactants

      • Work methodically through each reactant, one at a time

      • Determine the order with respect to all reactants

Worked Example

Use the information in the table to determine the rate equation for the nucleophilic substitution of 2-bromo-2-methylpropane by hydroxide ions:

(CH3)3CBr  +  OH-  →  (CH3)3COH  +  Br-

Table to show the experimental data of the above reaction

Experiment

Initial [(CH3)3CBr]
/ mol dm-3 

Initial [OH–]
/ mol dm-3 

Initial rate of reaction
/ mol dm-3 s-1

1

1.0 x 10-3

2.0 x 10-3

3.0 x 10-3

2

2.0 x 10-3

2.0 x 10-3

6.0 x 10-3

3

1.0 x 10-3

4.0 x 10-3

1.2 x 10-2

Answer:

Order with respect to [(CH3)3CBr]:

  • Using experiments 1 and 2:

    • The [OH-] has remained constant

    • The [(CH3)3CBr] has doubled

    • The rate of the reaction has also doubled

    • Therefore, the order with respect to [(CH3)3CBr] is 1 (first order)

Order with respect to [OH-]:

  • Using experiments 1 and 3:

    • The [(CH3)3CBr] has remained constant

    • The [OH-] has doubled

    • The rate of reaction has increased by a factor of 4 (i.e. increased by 22)

    • Therefore, the order with respect to [OH-] is 2 (second order)

Building the rate equation:

  • Once you know the order with respect to all of the reactants, you put them together to form the rate equation

    • If a reactant is order 0, it should not appear in the rate equation

    • If a reactant is order 1, then it features in the rate equation

      • There is no need to include the number 1 as a power

    • If a reactant is order 2, then it features in the rate equation with the number 2 as a power

  • For this reaction, the rate equation will be:

Rate = k [(CH3)3CBr] [OH-]2

Half-life

  • The half-life (t1/2) is the time taken for the concentration of a limiting reactant to become half of its initial value

Rate-determining step & intermediates

  • The rate-determining step is the slowest step in a reaction

  • If a reactant appears in the rate-determining step, then the concentration of that reactant will also appear in the rate equation

  • For example, the rate equation for the reaction below is rate = k [CH3Br] [OH–]

CH3Br + OH– → CH3OH + Br–

  • This suggests that both CH3Br and OH– take part in the slow rate-determining step

  • This reaction is, therefore, a bimolecular reaction

    • Unimolecular: one species involved in the rate-determining step

    • Bimolecular: two species involved in the rate-determining step

  • The intermediate is derived from substances that react together to form it in the rate-determining step

    • For example, for the reaction above the intermediate would consist of CH3Br and OH–

The intermediate formed during the reaction of CH3Br and hydroxide ions

Reaction Kinetics - Intermediate, downloadable AS & A Level Chemistry revision notes

The intermediate is formed from the species that are involved in the rate-determining step (and thus appear in the rate equation)

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    Richard Boole

    Author: Richard Boole

    Expertise: Chemistry Content Creator

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.