IBChemistryHLRevision NotesHow Much, How Fast & How Far?How Fast? The Rate of Chemical ChangeRate Equation (HL)

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Rate Equation (HL) (HL IB Chemistry)

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Rate Equation

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

Rate of Reaction

  • 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 and a graph of rate against [D] is drawn
  • The gradient at each concentration is calculated by fraction numerator rate over denominator open square brackets straight D close square brackets end fraction

Rate of reactions table

[D] (mol dm-3) Rate (mol dm-3 s-1) gradient = fraction numerator bold rate over denominator stretchy left square bracket D stretchy right square bracket end fraction bold space stretchy left parenthesis s to the power of negative 1 end exponent stretchy right parenthesis
3.00 2.000 x 10-3 6.67 x 10-4
2.00 1.334 x 10-3 6.67 x 10-4
1.00 6.670 x 10-4 6.67 x 10-4

  • A directly proportional relationship between the rate of reaction and concentration of D is observed when the results are plotted on the graph
  • The value of the gradient is constant as it is a straight line

Graph to show rate against [D]

A graph of rate against the concentration of D gives a straight line showing a directly proportional relationship

Rate of reaction over various concentrations of D

  • This leads to a very common rate expression:

Rate ∝ [D]       or       Rate = k[D]

  • This rate expression means that if the concentration of D is doubled, then the rate doubles
    • Equally, if the concentration of D halves, then the rate halves
  • However, the rate of a reaction does not always show this directly proportional relationship with the concentration of a reactant and we can use the rate equation and orders of reaction to explain this

What is the rate equation?

  • The following reaction will be used to discuss rate equations:

A (aq) + B (aq) → C (aq) + D (g) 

  • The rate equation for this reaction is:

Rate of reaction = k [A]m [B]n

  • Rate equations depend on the mechanism of the reaction and can only be determined experimentally, they cannot be found from the stoichiometric equations
  • In the above rate equation:
    • [A] and [B] are the concentrations of the reactants
    • m and n are orders with respect to each reactant involved in the reaction
  • Products and catalysts may feature in rate equations
  • Intermediates do not feature in rate equations

What is the order of reaction?

  • The order of a reactant shows how the concentration of a chemical, typically 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 a positive, negative or fractional value
    • Orders that are a fraction suggest that the reaction involves multiple steps

Zero order

  • 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

First order

  • 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

Second order

  • 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)

Overall order

  • The overall order of reaction is the sum of the powers of the reactants in a rate equation (m + n)

Exam Tip

  • In an exam, you may be presented with information about a reaction that uses orders that are fractions
  • Examples include:
    • The decomposition of ethanal: Rate = k[CH3CHO]3/2
    • The reaction of hydrogen and bromine: Rate = k[H2][Br2]1/2
    • The reaction between carbon monoxide and chlorine: Rate = k[CO]2[Cl2]1/2
  • However, for calculations only values of 0, 1 or 2 need to be considered

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

 

Answer 1:

  • 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

Answer 2:

  • 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

 

Answer 1:

  • 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
  • The overall order of the reaction is 1 + 1 + 1 = 3 or third order

Answer 2:

  • 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

How can the rate equation be deduced from experimental data?

  • The following reaction will be used to deduce the rate equation, using experimental data

(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
4 1.5 x 10-3 4.0 x 10-3 4.5 x 10-3

  • To derive the rate equation for a reaction, you must first determine all of the orders with respect to each of the reactants
  • This can be done using the tabulated data provided
  • Take the reactants one at a time and find the order with respect to each reactant individually:
    1. Identify two experiments where the concentration of one reactant changes, but the concentrations of all other reactants are constant
    2. Calculate what happens to the concentration
    3. Calculate what happens to the rate of reaction
    4. Deduce the order of reaction with respect to that chemical
    5. Repeat this for all of the reactants, one at a time, until you have determined the order with respect to all reactants

Order with respect to [(CH3)3CBr]

  1. In experiments 1 and 2, the concentration of (CH3)3CBr changes while the concentration of OH- remains constant
  2. The [(CH3)3CBr] has doubled
  3. The rate of the reaction has also doubled
  4. Therefore, the order with respect to [(CH3)3CBr] is 1 (first order)
    • [Change in concentration]order = change in rate
    • [2]order = 2
    • [2]1 = 2

Order with respect to [OH–]

  1. In experiments 1 and 3, the concentration of OH– changes while the concentration of (CH3)3CBr remains constant
  2. The [OH–] has doubled
  3. The rate of the reaction has increased by a factor of 4
  4. Therefore, the order with respect to [OH–] is 2 / second order
    • [Change in concentration]order = change in rate
    • [2]order = 4
    • [2]2 = 2

Building the rate equation

  • Once  the order with respect to all of the reactants is known the rate equation can be constructed
    • Zero order reactants are not included in the rate equation
    • First order reactants are included in the rate equation - they do not require a power
    • Second order reactants are included in the rate equation - they are raised to the power of 2
  • So, for this reaction the rate equation will be:

Rate = k [(CH3)3CBr] [OH–]2

Exam Tip

  • Examiners will often give concentration and rate data in standard form to test your mathematical skills!
  • Take your time because it is easy to make a mistake
    • The most common mistake is failing to notice a factor of ten, e.g. one rate value is x10-4 while the rest are x 10-3
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Caroline

Author: Caroline

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.