reaction rates

REACTION RATESTopic 6

6.1.1 Define the term rate of reaction.

IB Core Objective

Different chemical reactions occur at different rates (ie. speeds) Rapid reactions, eg. The neutralisation of

a strong acid by a strong base in aqueous solution

Slow reactions, eg. Rusting of iron

Rate of a chemical reaction is a measure of the speed at which products are formed, measured as the change in concentration divided by the change in time (units = mol dm-3s-1)

6.1.1 Define the term rate of reaction.

6.1.2 Describe suitable experimental procedures for measuring rates of reactions.

IB Core Objective


Different changes can be measured vs. time to give a rate: Rate = Change/time Rate = Δ[Concentration]/ Δ[time] Rate = Δ[Pressure]/ Δ[time] Rate = Δ[Absorbance/ transmittance]/ Δ[time] Rate = Δ[pH]/ Δ[time]

Reaction rate describes how fast a reaction will take place.


What factors will determine rate? Surface area (usually refers to solid) Concentration or pressure (if gas) Temperature Catalyst

How is it found? Through experimentation only Rate of decreasing reactants

−Δ[Concentration]/ Δ[time] Rate of increasing products

Δ[Concentration]/ Δ[time]

Notice the negative

sign


pH changes (pH meter or acid/base titration)

Volume, pressure or mass changes Conductivity (aq) (Conductivity meter

or titration) Spectrometer or colorimeter for

colour change

6.1.2 Describe suitable experimental procedures for measuring rates of reactions. Data loggers and

probes Conductivity Acid/base Temperature

Titrations Acid/ base Concentration

TITRATION Removal of small samples

from the reaction mixture at different times and then titrating the sample to determine the concentration of either one of the reactants or one of the products at this time.

Results can then be used directly to generate a graph of concentration against time.

Best for quite slow reactions


6.1.2 Describe suitable experimental procedures

for measuring rates of reactions. Spectrometer

Absorbance and transmittance

Measures concentration

Collecting gas

COLLECTION OF EVOLVED GAS Gas produced in the reaction is collected either in

a gas syringe or in a graduated vessel over water. The volume of gas collected at different times can

be recorded. This technique is limited to reactions that produce

a gas (obviously!) PLUS if the gas is to be collected over water, the gas must not be water soluble.

An alternative technique is to carry out the reaction in a vessel of fixed volume and monitor the increase in the gas pressure


COLLECTION OF EVOLVED GAS

Example: Measuring the rate of reaction between a moderately reactive metal (such as zinc) and an acid (such as hydrochloric acid).

Zn (s) + 2 H+ (aq) → Zn2+ (aq) + H2 (g)


MEASUREMENT OF THE MASS OF REACTION MIXTURE The total mass of the reaction mixture will only

vary if a gas is evolved. The gas should have a high molar mass (ie. not

hydrogen) so there is a significant change in mass Example: measuring the rate of reaction

between a metal carbonate (such as calcium carbonate, marble chips) and an acid (such as hydrochloric acid) by measuring the rate of mass loss resulting from the evolution of carbon dioxide:

CaCO3 (s) + 2H+ (aq) → Ca2+ (aq) + H2O (l) + CO2 (g)


IB Core Objective

6.1.3 Analyse data from rate experiments.

Students should be familiar with graphs of changes in concentration, volume, and mass against time.

Numerical value will vary according to amount of the substance involved in the stoichiometric equation

MnO4- (aq) + 8 H+ (aq) + 5 Fe2+ (aq) → Mn2+

(aq) 4 H2O (l) + 5 Fe3+ (aq)

The rate of appearance of Fe3+ is five times as great as the rate at which MnO4

- is consumed

Rate = - ∆ [MnO4-] = 1 ∆ [Fe3+]

∆ t 5 ∆ t


Units: mol dm-3 s-1

Or more simply for a reaction: a A → b B, then

Rate = - 1 ∆ [A] = 1 ∆ [B] a ∆ t b ∆ t

Any property that differs between the reactants and products can be used to measure the rate of the reaction


General Formula


To find rate at an instant in time (instantaneous rate) we calculate the slope of a tangent on the experimentally obtained graph.

Here, Δx is the change in time and Δy is the change in concentration.

IB Core Objective

6.2.1 Describe the kinetic theory in terms of the movement of particles whose average energy is proportional to temperature in kelvins.

As heat is supplied to a substance, the velocity (kinetic energy) of the particles will increase.

When the velocity increases, so does the temperature.

Therefore, the absolute temperature in kelvin is proportional to the average kinetic energy of all the particles.

How does kinetic energy

and contact differ for solids

when compared to liquids and

gases?

Solids are fixed and

only vibrate, so kinetic energy is limited

IB Core Objective

6.2.2 Define the term activation energy, Ea.

The minimum amount of energy required for reaction is the activation energy, Ea.

6.2.2 Define the term activation energy, Ea.• Just as a ball cannot get over a hill

if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

IB Core Objective

6.2.3 Describe the collision theory Just because chemicals collide

/interact does not mean they will react!

Reaction rate depends on: Collision frequency Number of particles with E ≥ Ea. Appropriate collision geometry or

orientation.

IB Core Objective

6.2.4 Predict and explain, using collision theory, the qualitative effects of particle size, temperature, concentration and pressure on the rate of a reaction.

6.2.4 Surface Area By increasing the surface area

we increase the contact area. Collision rate will increase.

MarbleMarble

chip

6.2.4 Concentration & Pressure Increasing the concentration or

pressure will increase opportunity for molecules to react Collision rate will increase Pressure only affects gases

6.2.1 and 6.2.4 Temperature and Rate

• Temperature is directly related to kinetic energy or particle speed.

• Faster particles increases probability of molecule interaction

• Will increase # of particles with sufficient energy to over come activation energy.

Orientation

IB Core Objective

6.2.6 Describe the effect of a catalyst on a chemical reaction.

What do you know about catalysts? How would you define one in your own terms?

Have you heard the expression “catalyst for change”?

Catalysts 6.2.6 Catalysts provide an alternate reaction

pathway. One that takes less energy. Similar to a tunnel

through a mountain side, whether you go up and over, or go through you still end up at the same place.

Catalysts are just a facilitator They are not used up They lower the activation energy allowing

for more particles to have the correct energy requirement

Analogy Catalysts are like a dating service

They bring compatible people together They are not involved in what happens

after the pairs are together They can be reused again and again.

This is not to say that people will not get together on their own, however it lowers the energy required to find a match

6.2.6 Describe the effect of a catalyst on a chemical reaction.

One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

Draw an Energy Diagram for a catalyzed reaction

IB Core Objectives

6.2.5 Sketch and explain qualitatively the Maxwell-Boltzmann energy distribution curve for a fixed amount of gas at different temperatures and its consequences for changes in reaction rate.

Students should be able to explain why the area under the curve is constant and does not change with temperature.

Maxwell–Boltzmann Distributions 6.2.5

Activation Energy

Maxwell–Boltzmann Distributions 6.2.5

As the temperature increases, the curve flattens and broadens.

At higher temperatures, a larger number of molecules has higher energy.

This system has a fixed number of particles

Maxwell–Boltzmann Distributions

If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier.

• As a result, the reaction rate increases.

Be sure to draw the higher temp. Curve shorter and wider than the original.

IB Core Objective

6.2.7 Sketch and explain Maxwell-Boltzmann curves for reactions with and without catalysts.

Catalysts and Activation Energy 6.2.7

Is this a Maxwell-Boltzmann distribution curve?

Why or why not?

Sketch what a Maxwell-Boltzmann curve would look like.

HLt1/ 2

[A]0

2kt1/ 2

ln 2k

0.693

k1

[ A]t

k t 1

[A]0

1[A]t

t1/ 2 1

k[A]0

Rate Law

Order in [A]

RateLaw

Integrated Form,y = mx + b

StraightLine Plot

Half-Lifet1/2

zerothorder

(n = 0)rate = k [A]o= k [A]t = - k t +[A]o

[A]t vs. t(slope = - k)

t½ = [A]0/2k

firstorder

(n = 1)rate = k [A]1 ln[A]t = - k t + ln[A]o

ln[A]t vs. t

(slope = - k)t½ = ln 2/k

secondorder

(n = 2)rate = k [A]2 1/[A]t = kt + 1/[A]o

1/[A]t vs. t

(slope = k)t½ = 1/(k[A]o)

Return

HL Objective

16.1.1 Distinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant.

‘k’ represents a constant and

does not change EXCEPT • Temperature!!• Particle size

(solids)

16.1.1 Distinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant. Rate Rxn = Δ[A]/ Δt

Rate = k[A]m [B]n k = rate constant A and B are reactants. m and n represent the order of

reaction for each reactant. m + n = overall order of reaction

HL Objective

16.1.2 Deduce the rate expression for a reaction from experimental data.


1) If concentration of [A] is doubled yet no change in rate, than order = zero. [A]0

2) If doubling conc. of [A] = a doubling of rate, than order = one. [A]1

3) If doubling conc. of [A] = a quadrupling of rate, than order = 2. [A]2


Expt. Number

[A] mol dm-3 [B] mol dm-3 Rate of formation of C mol dm-3 s-1

1 0.10 0.05 0.02 x 102

2 0.10 0.10 0.04 x 102

3 0.05 0.10 0.01 x 102

4 0.10 0.20 0.08 x 102

For the reaction A + 2B → C

1. Deduce the order of reaction in respect to reactant A2. Deduce the order of reaction in respect to reactant B3. State what the overall order of the reaction is.4. Deduce the rate expression for this reactionA: 1. 2nd order. 2. 1st order 3. 2+1=3 3rd order 4. rate = k[A]2[B]

IB HL Objective

16.1.3 Solve problems involving the rate expression.


From the previous problem, rate = k[A]2[B].

To find the rate constant, plug in numbers from the results:Expt.

Number[A] mol dm-3 [B] mol dm-3 Rate of

formation of C mol dm-3 s-1

1 0.10 0.05 0.02 x 102

2 0.10 0.10 0.04 x 102

3 0.05 0.10 0.01 x 102

4 0.10 0.20 0.08 x 102

A: k= 4 x 103 mol-2 dm6 s-1Notice the

units!


Units for rate constantZero order overall: mol dm-3 s-1

First order overall: s-1

Second order overall: mol-1 dm3 s-1

Third order overall: mol-2 dm6 s-1

IB HL Objective

16.1.4 Sketch, identify and analyse graphical representation for zero-, first- and second-order reactions.

Note: Students should be familiar with both concentration-time and rate-concentration graphs.


Zero order reaction 1st order reaction 2nd order reaction

Conc

entr

atio

n

Time

Rat

e

Time

Order of Reaction

Concentration doubled. The rate:

Concentration tripled. The rate:

Zero Does not change

Does not change

1st Doubles (x2) Triples (x3)

2nd Quadruples (x4) x9


Zero order = linear. Decrease in concentration does not affect the rate of reaction.

First order = exponential decay (same as radioactive decay). (t1/2)

Second order = parabolic, because it depends on the square of the concentration.

Other Useful Information FOR 1ST ORDER REACTIONS ONLY Half life: The time it takes for half a

substance to decay/ disappear t1/2 = ln(2)/k

Half life

Conc

entr

atio

n /

Pres

sure

Calculate the half life for the

previous question

IB HL Objective

16.2.1 Explain that reactions can occur by more than one step and that the slowest step determines the rate of reaction (rate-determining step).


Several reactions may have simple equations, but sometimes there are several intermediate steps which occur to get to the final product(s).

These various intermediate steps can occur at different rates.

The slowest step is the rate-determining step.


Unimolecular: One species breaks up or rearranges to form products.

Bimolecular: Two species collide and interact to form the product. (Doubling the concentration of either will double the collision rate. What would be the overall order for this?)

Molecularity

Reaction Rate Law

Uni A Product Rate = k[A]Bi A +B

ProductRate = k[A][B]

Bi A + A Product

Rate = k[A]2

IB HL Objective

16.2.2 Describe the relationship between reaction mechanism, order of reaction and rate-determining step.

Mechanism Must accurately represent the

original stoichiometric equation

2XY2 2XY + Y2

Possible MechanismXY2 + XY2 X2Y4

X2Y4 X2 + 2Y2

X2 + Y2 2XY

Determining the Slowest Step 1) Determine the overall reaction

equation

2) Check to see if adding the reactions = the overall stoichiometric equation

3) Check for consistency with regards to the rate equation and the rate determining step Does the proposed rate equation = the

overall reaction equation?

Question Through experimentation it was

found that overall rate equation for:NO2(g) + CO(g) NO(g) + CO2(g)

Rate = k[NO2]2

The proposed reaction mechanism is:(Slow) NO2 + NO2 N2O4

(Fast) N2O4 + CO NO + CO2 + NO2

NO2(g) + CO(g) NO(g) + CO2(g) Does the sum match the original stoichiometric equation?

Found to be zero order

Question Continued

The proposed reaction mechanism is:(Slow) NO2 + NO2 N2O4

(Fast) N2O4 + CO NO + CO2 + NO2

NO2(g) + CO(g) NO(g) + CO2(g)

Does the RDS rate eqn = the overall rate eqn?

RDS rate = k[NO2][NO2] k[NO2]2Appears to be consistent with the original rate equation

H2 + NO X fast stepX + NO Y + H2O slow stepY + H2 N2 + H2O fast step

Question 22NO(g) + 2H2(g) → N2(g) + 2H2O(g)

rate = k[H2][NO]2

State and explain whether this mechanism agrees with the experimental rate expression in

NO is part of two steps, therefore must be considered in the rate expression.

Question 32H2O2 2H2O + O2

Which of the following mechanisms are correct?

Mech. 1H2O2 2OH2OH + Br- BrO- + 2H2O2

2H2O2 BrO- 2H2O + O2 + Br-

Mech. 2H2O2 + Br- H2O + BrO-

BrO- + H2O2 H2O + O2 + Br-

Br- cat.

A: Mech 2

IB HL Objective

16.3.1 Describe qualitatively the relationship between the rate constant (k) and temperature (T).

16.3.2 Determine activation energy (Ea) values from the Arrhenius equation by a graphical method.

16.3.1 Describe qualitatively the relationship between the rate constant (k) and temperature (T).

The Arrhenius Equation is k = Ae(-Ea/RT). This equation can be found in your data booklet.

A is the Arrhenius constant. It is dependent on collision rate and steric factors (geometry of the colliding particles).

This expression indicates that the rate constant k depends exponentially on temperature. Temperature has a large effect on the reaction rate.

Arrhenius Equation (16.3) The relationship between rate

constant and temperature.k = Ae(-

Ea/RT) The equation of a straight line can be determined by taking the natural log.

lnk = lnA –(Ea/R)1/T

Activation Energy can be found by varying

the temperature of an experiment

ln k

1/T

Arrhenius Equation (16.3)

Graphing ln k VS 1/T will give a straight line

Slope will = -Ea/RExtrapolate to find:ln [A]

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