Rate Equations

Section 1 of 28Rate of reaction

Definition

• Definition the change in amount or concentration of a reactant or product per unit time

Unit

• Unit mol dm^-3 s^-1

Measurement

• Measurement measure the decrease in reactant concentration or the increase in product concentration over time

Section 2 of 28The rate equation

1Origin

• Origin rate equations can only be found experimentally, not from the stoichiometric equation

2General form

• General form for P + Q → products, rate = k[P]^m[Q]ⁿ

  What the symbols mean: rate (mol dm⁻³ s⁻¹) · k = rate constant · [ ] = concentration (mol dm⁻³) · each power is the order with respect to that reactant
  • (units depend on the overall order)

3

Terms

• k is the rate constant (proportionality constant).
• [P] and [Q] are reactant concentrations.
• m and n are the orders, which can only be 0, 1 or 2.

4Products

• Products their concentrations are not included, as they do not affect the rate

5First-order example

• First-order example rate = k[D], because rate ÷ [D] is constant at 6.67 × 10^-4 s^-1 for [D] of 3.00, 2.00 and 1.00 mol dm^-3

Section 3 of 28Nitric oxide and hydrogen

Reaction

• 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g), with rate = k[NO]²[H₂]

Method

• Method keep one concentration constant to see the effect of the other

With [H₂] constant

• With [H₂] constant the rate is proportional to [NO]²

With [NO] constant

• With [NO] constant the rate is proportional to [H₂]

Key point

• H₂ is not second order even though its coefficient is 2, because orders come from experiment, not from the equation

Section 4 of 28Order of reaction

Order = the power to which the concentration of a reactant is raised in the rate equation

Zero order

• Changing the concentration has no effect on the rate.
• The reactant does not appear in the rate equation.

First order

• The rate is directly proportional to the concentration.
• Doubling the concentration doubles the rate.

Second order

• The rate is proportional to the square of the concentration.
• Doubling the concentration increases the rate fourfold.

Overall order

• Overall order the sum of the individual orders, so rate = k[NO]²[H₂] is second order in NO, first order in H₂ and third order overall

Section 5 of 28Using the rate equation

N₂O₅ decomposition

• N₂O₅ decomposition rate = k[N₂O₅]
• The order is 1.
• Tripling the concentration triples the rate.

Bromate and bromide

• Bromate and bromide rate = k[BrO₃^-][Br^-][H⁺]
• Each reactant is first order, so the overall order is 3.
• Doubling [BrO₃^-] doubles the rate and halving [Br^-] halves it, so there is no overall change.

Section 6 of 28Deducing orders from a table

• Approach
  • use a graph or a table of results with calculation
    • (its shape shows the order)

1

Find two experiments in which only one reactant concentration changes.

2

Calculate how that concentration changed.

3

Calculate how the rate changed.

4

Decide the order for that reactant.

5

Repeat for each reactant.

2-bromo-2-methylpropane and hydroxide

• Experiments 1 and 2: [(CH₃)₃CBr] doubles and the rate doubles, so first order.
• Experiments 1 and 3: [OH^-] doubles and the rate increases fourfold, so second order.
• Rate = k[(CH₃)₃CBr][OH^-]².

Writing the equation

• Writing the equation omit zero-order reactants, and omit the power 1 for first-order reactants

Caution

• Caution read standard form values carefully

Section 7 of 28Calculating the rate constant

Method

• Method rearrange the rate equation to k = rate ÷ , and use one set of initial values
  • (concentration terms)

Example

• Example rate = k[Na₂CO₃][Cl^-] with an initial rate of 4.38 × 10^-6 mol dm^-3 s^-1, [Na₂CO₃] = 0.0250 and [Cl^-] = 0.0125 mol dm^-3

Result

• Result k = 1.40 × 10^-2 dm³ mol^-1 s^-1

Consistency

• Consistency the other experiments give the same value of k

Units

• Units they depend on the overall order, as for first order (s^-1) and second order (dm³ mol^-1 s^-1)

Section 8 of 28The Arrhenius equation

Rate constant

• Rate constant k is constant only if concentration is the only factor changed, and it changes with temperature or a catalyst

Temperature effect

• Temperature effect at higher temperature a greater proportion of molecules have energy of at least the activation energy, so both rate and k increase

Equation

• Equation k = Ae^−E_a/RT
  What the symbols mean: k = rate constant · A = pre-exponential (frequency) factor · E_a = activation energy (J mol⁻¹) · R = 8.31 J mol⁻¹ K⁻¹ · T = temperature (K)

Terms

• E_a and A are characteristic constants of a reaction, and A varies only slightly with temperature.
• R is the gas constant.
• k and T are the only variables.

Scope

• Scope it applies to reactions in gases, in solution and on catalyst surfaces

Section 9 of 28Using the Arrhenius equation

Logarithmic form

• Logarithmic form ln k = ln A − E_a/RT, which is easier to use

Higher temperature

• Higher temperature ln k increases, so k and the rate increase

Higher activation energy

• Higher activation energy a smaller proportion of molecules has enough energy, so k and the rate decrease

Typical task

• Typical task use measured k and T values to calculate the activation energy

Rearranged

• E_a = (ln A − ln k) × RT

Example

• Example at 400 K with k = 6.25 × 10^-4 s^-1, A = 4.6 × 10¹³ and R = 8.31 J K^-1 mol^-1, E_a = × 3324 = 129 kJ mol^-1
  • (31.4597 + 7.3778)

Syllabus note

• Syllabus note the equations need not be learned, but must be rearranged

Section 10 of 28Arrhenius plot

• Plot
  • ln k (y) against 1/T (x) gives a straight line

Part of y = mx + c	Quantity
y	ln k
x	1/T
m (gradient)	−Ea/R
c (intercept)	ln A

Activation energy

• E_a = −gradient × R

Example

• Example a gradient of −3666.6 gives E_a = 3666.6 × 8.31 = 30.5 kJ mol^-1

Constant A

• Constant A substitute a point from the graph, such as (2.60 × 10^-3, −7.0), into ln k = (−E_a/R)(1/T) + ln A, giving ln A = 2.53 and A = 12.55

Meaning

• Meaning a higher T gives a greater ln k and so a faster reaction

Section 11 of 28Concentration–time graphs

Use

• Use they show the overall rate and the order for a reactant

Zero order

• The reactant concentration falls in a straight line.
• The rate is constant.

First order

• The concentration falls as a downward curve that eventually plateaus.

Second order

• The concentration falls as a steeper curve, decreasing more at first than in first order.

Section 12 of 28Initial rate from a concentration–time graph

• Initial rate = the rate at the very start of the reaction, when the reactant concentrations are known exactly

Initial rate from a concentration–time graph

Draw a tangent to the curve at t = 0 s.

Calculate the gradient of the tangent, which is the initial rate.

Section 13 of 28Initial-rate experiments

Conditions

• Conditions the temperature must stay constant and only one reactant concentration changes in each experiment

Procedure

• Procedure for each experiment draw a concentration–time graph and find the initial rate from the tangent at t = 0

General reaction

• General reaction for 2A + B + C, experiment 1 uses base concentrations, experiment 2 changes [A], experiment 3 changes [B] and experiment 4 changes [C]

Experiment	[A] / mol dm-3	[B] / mol dm-3	[C] / mol dm-3	Initial rate / mol dm-3 s-1
1	1.5 × 10-3	1.5 × 10-3	1.5 × 10-3	2.1 × 10-3
2	3.0 × 10-3	1.5 × 10-3	1.5 × 10-3	2.1 × 10-3
3	1.5 × 10-3	3.0 × 10-3	1.5 × 10-3	4.2 × 10-3
4	1.5 × 10-3	1.5 × 10-3	3.0 × 10-3	8.4 × 10-3

Deducing the orders

• Experiments 1 and 2: [A] doubles and the rate is unchanged, so zero order in A.
• Experiments 1 and 3: [B] doubles and the rate doubles, so first order in B.
• Experiments 1 and 4: [C] doubles and the rate quadruples, so second order in C.

Rate equation

• Rate equation rate = k[B][C]²

  What the symbols mean: rate (mol dm⁻³ s⁻¹) · k = rate constant · [ ] = concentration (mol dm⁻³) · each power is the order with respect to that reactant
  • (units depend on the overall order)

Section 14 of 28Rate–concentration graphs (start)

Method

• Method rates measured at various initial concentrations of each reactant are plotted against concentration

Zero order

• The rate does not depend on concentration, so the graph is a horizontal line.
• Rate = k.

Section 15 of 28Rate–concentration graphs

1Zero order

• Zero order rate = k
• The rate stays constant, so the graph is a horizontal line.

2First order

• First order rate = k[A]

  What the symbols mean: rate (mol dm⁻³ s⁻¹) · k = rate constant · [ ] = concentration (mol dm⁻³) · each power is the order with respect to that reactant
  • (units depend on the overall order)
• The rate is proportional to concentration, so the graph is a straight line.
• Its gradient is k.

3Second order

• Second order rate = k[A]²
• The rate is proportional to the square of the concentration, so the graph is a smooth curve.

4Exam note

• Exam note a question may give a graph for one reactant and a table for the others, so use both

Section 16 of 28Rate equations and mechanisms

Mechanism = a description of the steps in which bonds break and form in a reaction

Rate-determining step = the slowest step in a mechanism

Link

• Link every reactant in the rate equation appears in the rate-determining step

Exclusions

• Exclusions zero-order reactants and intermediates are not in the rate-determining step

Section 17 of 28Predicting a mechanism

Example

• NO₂(g) + CO(g) → NO(g) + CO₂(g) with rate = k[NO₂]²

Deduction

• Deduction it is second order in NO₂ and zero order in CO, so two NO₂ molecules are in the rate-determining step

Possible mechanism

• Step 1 (slow): 2NO₂ → NO + NO₃.
• Step 2 (fast): NO₃ + CO → NO₂ + CO₂.
• Adding the steps and cancelling gives NO₂ + CO → NO + CO₂.

Section 18 of 28Deducing the rate equation from a mechanism

Example

• 2NO + 2H₂ → N₂ + 2H₂O

Mechanism

• Step 1 (fast): NO + NO → N₂O₂.
• Step 2 (slow, rate-determining): N₂O₂ + H₂ → H₂O + N₂O.
• Step 3 (fast): N₂O + H₂ → N₂ + H₂O.

Deduction

• Deduction the slow step contains N₂O₂ (from two NO) and one H₂

Result

• Result rate = k[NO]²[H₂], third order overall

Section 19 of 28Identifying the rate-determining step and catalysts

1

Finding the rate-determining step from the rate equation

• Propane bromination in alkali has rate = k[CH₃CH₂CH₃][OH^-].
• Only the hydrocarbon and OH^- are involved, not bromine, so the rate-determining step is the step containing them, step 1.

2

Identifying a catalyst

• A species in the rate equation but not in the overall equation is a catalyst.
• Butanone halogenation has rate = k[CH₃CH₂COCH₃][H⁺], and H⁺ is not in the equation, so it is a catalyst.
• The butanone and H⁺ are in the rate-determining step as an intermediate formed in that step.

Section 20 of 28Required Practical 7: overview

Required Practical 7: overview

Parts

• Part A is an initial rate method and Part B is a continuous monitoring method

Part A aim

• Part A aim to use an iodine clock to find the order with respect to iodide ions in the reaction of iodide with hydrogen peroxide in acid

Apparatus

• 0.25 mol dm^-3 H₂SO₄, 0.10 mol dm^-3 KI, a burette of 0.05 mol dm^-3 sodium thiosulfate, 0.10 mol dm^-3 H₂O₂, starch, 50 cm³ burettes with funnel, stand and clamp, white tile, pipette, measuring cylinders, 100 and 250 cm³ beakers, stirrer, timer and deionised water

Section 21 of 28Required Practical 7 Part A: method

1

Rinse and fill the burette with potassium iodide solution.

2

In a clean dry 250 cm³ beaker put 25 cm³ sulfuric acid, 20 cm³ water, 1 cm³ starch, 5.0 cm³ KI and 5.0 cm³ sodium thiosulfate, adding the thiosulfate last.

3

Stir the mixture.

4

Transfer 10.0 cm³ hydrogen peroxide to the beaker and start the timer immediately.

5

Stir and stop the timer when the mixture turns blue-black, which may take several minutes.

6

Record the time to a suitable precision.

7

Empty, rinse with deionised water and dry the beaker.

Experiment	H2SO4 / cm3	Starch / cm3	Water / cm3	KI / cm3	Na2S2O3 / cm3
1	25	1	20	5.0	5.0
2	25	1	15	10.0	5.0
3	25	1	10	15.0	5.0
4	25	1	5	20.0	5.0
5	25	1	0	25.0	5.0

• Repeat
  • experiments 2 to 5 use the same steps with these volumes

Section 22 of 28Required Practical 7 Part A: practical tip and analysis

• Hydrogen peroxide strength
  • it is often labelled by volume, with "10 vol" meaning 1 cm³ produces 10 cm³ of oxygen on decomposition (2H₂O₂ → O₂ + 2H₂O),
    • and 10 vol or 3% is about 0.979 mol dm^-3

1

Calculate the concentration of KI in each mixture, such as 5 cm³ of 0.1 mol dm^-3 in 250 cm³ giving 0.0005 mol dm^-3.

2

Convert the time to a rate, rate = 1 ÷ time.

3

Plot concentration of KI (x) against rate (y).

4

Use the graph to find the order for iodide ions.

Required Practical 7 Part A: practical tip and analysis

Example data

• Example data concentrations 0.015 to 0.075 mol dm^-3 give times of 40, 20, 13, 10 and 8 s and rates of 0.025 to 0.120 s^-1

Conclusion

• Conclusion the rate is directly proportional to [KI], so the reaction is first order with respect to iodide ions

Section 23 of 28Required Practical 7 Part B: continuous monitoring

Required Practical 7 Part B: continuous monitoring

Aim

• Aim to monitor the reaction of magnesium with hydrochloric acid continuously by gas collection

Apparatus

• 6 cm strips of magnesium, 0.8 mol dm^-3 HCl, 50 cm³ measuring cylinder, 100 cm³ conical flask, stand, boss and clamp, a gas syringe with bung and delivery tube (or a water trough with inverted 100 cm³ measuring cylinder), timer and deionised water

Experiment 1

• Put 50 cm³ of 0.8 mol dm^-3 HCl in the flask.
• Set up the gas collection equipment.
• Add one 6 cm magnesium strip, fit the bung firmly and start the timer immediately.
• Record the hydrogen volume every 15 seconds for 2.5 minutes.

Experiment 2

• Mix 25 cm³ of 0.8 mol dm^-3 HCl with 25 cm³ of water to give 50 cm³ of 0.4 mol dm^-3 acid.
• Repeat the same steps with this acid.

Section 24 of 28Practical tip and results table

Magnesium surface

• Magnesium surface if the ribbon is not new and shiny, clean it with sandpaper

Results table

• Results table time every 15 s (0 to 150 s) with the volume of gas produced by 0.8 mol dm^-3 HCl and by 0.4 mol dm^-3 HCl

Section 25 of 28Evaluation of the results

1

Plot time (x-axis, seconds) against volume of gas (y-axis, cm³).

2

Add a smooth best-fit curve for the 0.8 mol dm^-3 results.

3

Draw the tangent at t = 0 s and calculate its gradient, which is the initial rate.

4

Add a smooth best-fit curve for the 0.4 mol dm^-3 results.

5

Draw the tangent at t = 0 s and calculate its gradient.

• Comparison
  • the two initial rates show the effect of acid concentration on rate

Section 26 of 28Example calculation

Reaction

• Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g), with acid at 0.5 and 1.0 mol dm^-3

Rate equation

• Rate equation rate = volume ÷ time in cm³ s^-1

Example

• 18 cm³ in 30 s is 0.6 cm³ s^-1, and 12 cm³ in 40 s is 0.3 cm³ s^-1

Conclusion

• Conclusion the higher concentration gives the larger initial rate

Section 27 of 28Key terms

Term	Meaning
Rate equation	An expression linking rate to reactant concentrations, rate = k[A]m[B]n.
Rate constant	The constant k in the rate equation.
Order of reaction	The power of a reactant concentration in the rate equation.
Overall order	The sum of the orders of all reactants in the rate equation.
Initial rate	The rate at the start of a reaction, at known concentrations.

Term	Meaning
Arrhenius equation	k = Ae−Ea/RT, linking rate constant, temperature and activation energy.
Arrhenius plot	A graph of ln k against 1/T, with gradient −Ea/R.
Initial rate	The reaction rate at the start, found from the tangent at t = 0.
Zero order	A reaction order in which the rate is independent of concentration.

Term	Meaning
Mechanism	The sequence of steps by which a reaction occurs.
Rate-determining step	The slowest step in a reaction mechanism.
Intermediate	A species formed in one step and used up in a later step.
Iodine clock	An experiment in which the time for a blue-black colour to appear measures a rate.
Continuous monitoring	Measuring a changing quantity throughout a reaction.

Term	Meaning
Initial rate	The gradient of the tangent to the volume–time graph at t = 0.
Tangent	A line touching a curve at a point, with a gradient equal to the rate at that point.
Continuous monitoring	Measuring the volume of gas at regular time intervals throughout a reaction.
Smooth curve of best fit	A single smooth line drawn through the plotted points.

Section 28 of 28Exam technique

1

Using the stoichiometric coefficients as orders — orders are found by experiment.

2

Including product concentrations in the rate equation — they do not appear unless they affect the rate.

3

Saying doubling a second-order reactant doubles the rate — it quadruples it.

4

Quoting units of k without checking the overall order — units depend on the order.

1

Reading the order from the wrong graph type — concentration–time and rate–concentration graphs differ in shape.

2

Using a chord instead of a tangent at t = 0 — the initial rate needs a tangent.

3

Taking ln k against T instead of 1/T — the Arrhenius plot uses 1/T.

4

Changing more than one concentration at a time — only one concentration should change.

1

Including zero-order reactants in the rate-determining step — they do not appear in it.

2

Adding the thiosulfate before the other reagents — it must be added last.

3

Taking the time as the rate — rate = 1 ÷ time.

4

Calling a catalyst an intermediate — a catalyst is in the rate equation but not the overall equation, while an intermediate is formed and used in the mechanism.

1

Using the average rate over the whole experiment as the initial rate — use the tangent at t = 0.

2

Using a chord rather than a tangent — it does not give the rate at that moment.

3

Not starting the timer as soon as the bung is fitted — the first readings are lost.

4

Using old, oxidised magnesium — the surface should be cleaned.