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Rate Equations (Oxford AQA International A Level (IAL) Chemistry): Revision Note

Exam code: 9622

Richard Boole

Written by: Richard Boole

Reviewed by: Stewart Hird

Updated on

Rates Equations & Rates Terms

  • The rate of reaction refers to the change in the amount or concentration of a reactant OR product per unit time

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

  • It can be found by:

    • Measuring the decrease in the concentration of a reactant over time

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

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)

Error converting from MathML to accessible text. (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 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

Worked Example

The following equation represents the oxidation of bromide ions in acidic solution

BrO3- (aq) + 5Br- (aq) + 6H+ (aq) → 3Br2 (l) + 3H2O (l)

The rate equation for this reaction is:

Rate = k[BrO3- (aq)][Br- (aq)][H+ (aq)]

  1. State the overall order of the reaction

  2. Deduce the effect on the rate of reaction if the concentration of bromate ions is doubled and the concentration of bromide ions is halved

Answers:

  1. The overall order of the reaction:

    • All three reactants feature in the rate equation but they are not raised to a power

    • This means that the order with respect to each reactant is order 1 / first order.

    • So, the overall order of the reaction is 1 + 1 + 1 = 3 or third order.

  2. The effect of changing concentrations:

    • Since each reactant is first order, the concentration of each reactant is directly proportional to the effect that it has on rate

    • If the concentration of the bromate ion is doubled, then the rate of reaction will also double

    • If the concentration of the bromide ion is halved then the rate will also halve

    • Therefore, there is no overall effect on the rate of reaction - one change doubles the rate and the other change halves it

Deducing Orders

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

  • 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

Examiner Tips and Tricks

Be careful when reading the values in standard form! It is easy to make a mistake.

Rate Calculations

  • The rate constant (k) of a reaction can be calculated using initial rates and the rate equation

Calculating the rate constant

  • The reaction of sodium carbonate (Na2CO3) with chloride (Cl–) ions to form sodium chloride (NaCl) will be used as an example to calculate the rate constant from the initial rate and initial concentrations

  • The reaction and rate equations are:

Na2CO3 (aq) + 2Cl– (aq) + 2H+ (aq) → 2NaCl (aq) + CO2 (g) + H2O (l)

Rate = k [Na2CO3] [Cl–]

  • The progress of the reaction can be followed by measuring the initial rates of the reaction using various initial concentrations of each reactant

Experimental results of concentrations & initial rates table

 Experiment

[Na2CO3]
(mol dm-3)

[Cl–]
(mol dm-3)

[H+]
(mol dm-3)

Initial rate of reaction
(mol dm-3 s-1)

1

0.0250

0.0125

0.0125

4.38 x 10-6

2

0.0375

0.0125

0.0125

6.63 x 10-6

3

0.00625

0.0250

0.0250

2.19 x 10-6

  • Rearrange the rate equation to find k:

    • Rate = k [Na2CO3] [Cl–] → k = fraction numerator rate over denominator open square brackets Na subscript 2 CO subscript 3 close square brackets space open square brackets Cl to the power of minus close square brackets end fraction

  • Substitute the values of one of the experiments to find k:

    • For example, using the measurements from experiment 1

    • k = fraction numerator 4.38 cross times 10 to the power of negative 6 end exponent over denominator open square brackets 0.0250 close square brackets space open square brackets 0.0125 close square brackets end fraction

    • k = 1.40 x 10-2 dm3 mol-1 s-1

  • The measurements from experiments 2 or 3 could also have been used to find k

    • They would also give the same result of 1.40 x 10-2 dm3 mol-1 s-1 

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    Author
    Reviewer
    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.

    Stewart Hird

    Reviewer: Stewart Hird

    Expertise: Chemistry Content Creator

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.