THE RATE EQUATION A guide for A level students KNOCKHARDY PUBLISHING.

THE RATE THE RATE EQUATIONEQUATIONA guide for A level studentsA guide for A level students

KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING

INTRODUCTION

This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards.

Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available.

Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm

Navigation is achieved by...

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THE RATE EQUATIONTHE RATE EQUATIONKNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING

THE RATE EQUATIONTHE RATE EQUATION

CONTENTS• Collision theory

• Methods for increasing rate

• Rate changes during a reaction

• The rate equation

• Worked examples

• Graphical methods for determining rate

• Half-life

• Rate determining step

• Check list

Before you start it would be helpful to…

• know how the energy changes during a chemical reaction

• know the basic ideas of Kinetic Theory

• know the importance of catalysts in industrial chemistry


COLLISION THEORYCOLLISION THEORY

Collision theory states that...

• particles must COLLIDE before a reaction can take place• not all collisions lead to a reaction• reactants must possess at least a minimum amount of energy - ACTIVATION ENERGY

plus• particles must approach each other in a certain relative way - the STERIC EFFECT

REVISIONREVISION

COLLISION THEORYCOLLISION THEORY

Collision theory states that...

• particles must COLLIDE before a reaction can take place• not all collisions lead to a reaction• reactants must possess at least a minimum amount of energy - ACTIVATION ENERGY

plus• particles must approach each other in a certain relative way - the STERIC EFFECT

According to collision theory, to increase the rate of reaction you therefore need...

more frequent collisions increase particle speed orhave more particles present

more successful collisions give particles more energy orlower the activation energy

REVISIONREVISION

INCREASING THE RATEINCREASING THE RATE

• INCREASE THE SURFACE AREA OF SOLIDS

• INCREASE TEMPERATURE

• SHINE LIGHT

• ADD A CATALYST

• INCREASE THE PRESSURE OF ANY GASES

• INCREASE THE CONCENTRATION OF REACTANTS

• INCREASE THE SURFACE AREA OF SOLIDS

• INCREASE TEMPERATURE

• SHINE LIGHT

• ADD A CATALYST

• INCREASE THE PRESSURE OF ANY GASES

• INCREASE THE CONCENTRATION OF REACTANTS

The following methods may be used to increase the rate of a chemical reaction

REVISIONREVISION

Reactions are fastest at the start and get slower as the reactants concentration drops.

In a reaction such as A + 2B ——> C the concentrations might change as shown

RATE CHANGE DURING A REACTIONRATE CHANGE DURING A REACTION

Reactants (A and B)Concentration decreases with time

Product (C)Concentration increases with time

• the steeper the curve the faster the rate of the reaction

• reactions start off quickly because of the greater likelihood of collisions

• reactions slow down with time as there are fewer reactants to collide

TIME

CO

NC

EN

TR

AT

ION

B A

C

REVISIONREVISION

Experimental Investigation

• the variation in concentration of a reactant or product is followed with time• the method depends on the reaction type and the properties of reactants/products

e.g. Extracting a sample from the reaction mixture and analysing it by titration. - this is often used if an acid is one of the reactants or products

Using a colorimeter or UV / visible spectrophotometer.

Measuring the volume of gas evolved.

Measuring the change in conductivity.

More details of these and other methods can be found in suitable text-books.

MEASURING THE RATEMEASURING THE RATE

RATE How much concentration changes with time. It is the equivalent of velocity.

MEASURING THE RATEMEASURING THE RATE

y

CO

NC

EN

TR

AT

ION

gradient = y x

x

TIME

• the rate of change of concentration is found from the slope (gradient) of the curve

• the slope at the start of the reaction will give the INITIAL RATE

• the slope gets less (showing the rate is slowing down) as the reaction proceeds

THE SLOPE OF THE GRADIENT OF THECURVE GETS LESS AS THEREACTION SLOWS DOWNWITH TIME


Format links the rate of reaction to the concentration of reactantsit can only be found by doing actual experimentsit cannot be found by just looking at the equation

the equation... A + B ——> C + D

might have a rate equation like this r = k [A] [B]2

r rate of reaction units of conc. / time usually mol dm-3 s-1

k rate constant units depend on the rate equation [ ] concentration units of mol dm-3

Interpretation

The above rate equation tells you that the rate of reaction is...proportional to the concentration of reactant A doubling [A] doubles rateproportional to the square of the concentration of B doubling [B] quadruples (22) rate

ORDER OF REACTION ORDER OF REACTION

Individual order The power to which a concentration is raised in the rate equation

Overall order The sum of all the individual orders in the rate equation.

Order tells you how much the concentration of a reactant affects the rate




e.g. in the rate equation r = k [A] [B]2

the order with respect to A is 1 1st Orderthe order with respect to B is 2 2nd Order

and the overall order is 3 3rd Order

Value(s) need not be whole numbers can be zero if the rate is unaffected by how much of a substance is present





e.g. in the rate equation r = k [A] [B]2

the order with respect to A is 1 1st Orderthe order with respect to B is 2 2nd Order

and the overall order is 3 3rd Order

Value(s) need not be whole numbers can be zero if the rate is unaffected by how much of a substance is present

NOTESThe rate equation is derived from experimental evidence not by looking at an equation.Species appearing in the stoichiometric equation sometimes aren’t in the rate equation.Substances not in the stoichiometric equation can appear in the rate equation - CATALYSTS



Experimental determination of order

Method 1Plot a concentration / time graph qnd calculate the rate (gradient) at points on the curvePlot another graph of the rate (y axis) versus the concentration at that point (x axis)

If it gives a straight line, the rate is directly proportional to concentration - 1st ORDER.

If the plot is a curve then it must have another order. Try plotting rate v. (conc.)2.A straight line would mean 2nd ORDER. This method is based on trial and error.


Experimental determination of order

Method 1Plot a concentration / time graph qnd calculate the rate (gradient) at points on the curvePlot another graph of the rate (y axis) versus the concentration at that point (x axis)

If it gives a straight line, the rate is directly proportional to concentration - 1st ORDER.

If the plot is a curve then it must have another order. Try plotting rate v. (conc.)2.A straight line would mean 2nd ORDER. This method is based on trial and error.

Method 2 - The initial rates method.Do a series of experiments (at the same temperature) at different concentrations of a reactant but keeping all others constant. Plot a series of concentration / time graphs and calculate the initial rate (slope of curve at start) for each reaction. From the results calculate the relationship between concentration and rate and hence deduce the rate equation. To find order directly, logarithmic plots are required.

THE RATE CONSTANT (k)THE RATE CONSTANT (k)

Units The units of k depend on the overall order of reaction.

e.g. if the rate equation is... rate = k [A]2 the units of k will be dm3 mol-1 sec-1

Divide the rate by as many concentrations as appear in the rate equation.

Overall Order 0 1 2 3units of k mol dm-3 sec-1 sec-1 dm3 mol-1 sec-1 dm6 mol-2 sec-1

example in the rate equation r = k [A] k will have units of sec-1

in the rate equation r = k [A] [B]2 k will have units of dm6 mol-2 sec-1

RATE EQUATION - SAMPLE CALCULATIONRATE EQUATION - SAMPLE CALCULATION

In an experiment between A and B the initial rate of reaction was found for various starting concentrations of A and B. Calculate...

• the individual orders for A and B• the overall order of reaction• the rate equation• the value of the rate constant (k)• the units of the rate constant

0.5 1 2

1.5 1 6

0.5 2 8

1

2

3

[A] [B] Initial rate (r)

r initial rate of reaction mol dm-3 s-1

[ ] concentration mol dm-3

Compare Experiments 1 & 2[B] same[A] 3 x biggerrate 3 x bigger rate [A]

FIRST ORDER with respect to (wrt) A

0.5 1 2

1.5 1 6

0.5 2 8

1

2

3


CALCULATING ORDER wrt A

Choose any two experiments where...

[A] is changed and, importantly,

[B] is KEPT THE SAME

See how the change in [A] affects the rate

As you can see, tripling [A] has exactly the same effect on the rate so...

THE ORDER WITH RESPECT TO A = 1(it is FIRST ORDER)


0.5 1 2

1.5 1 6

0.5 2 8

1

2

3


CALCULATING ORDER wrt B

Choose any two experiments where...

[B] is changed and, importantly,

[A] is KEPT THE SAME

See how a change in [B] affects the rate

As you can see, doubling [B] quadruples the rate so...

THE ORDER WITH RESPECT TO B = 2

It is SECOND ORDER


Compare Experiments 1 & 3[A] same[B] 2 x biggerrate 4 x bigger rate [B]2

SECOND ORDER wrt B



0.5 1 2

1.5 1 6

0.5 2 8

1

2

3


0.5 1 2

1.5 1 6

0.5 2 8

1

2

3



SECOND ORDER wrt B

OVERALL ORDER = THE SUM OF THE INDIVIDUAL ORDERS= 1 + 2

= 3


0.5 1 2

1.5 1 6

0.5 2 8

1

2

3


0.5 1 2

1.5 1 6

0.5 2 8

1

2

3


rate = k [A] [B]2


By combining the two proportionality relationships you can construct the overall rate equation


SECOND ORDER wrt B



0.5 1 2

1.5 1 6

0.5 2 8

1

2

3


0.5 1 2

1.5 1 6

0.5 2 8

1

2

3


Chose one experiment (e.g. Expt. 3) and substitute its values into the rate equation

k = 8 = 4 dm6 mol-2 sec-1

(0.5) (2)2

rate = k [A] [B]2

re-arranging k = rate [A] [B]2



SECOND ORDER wrt B



0.5 1 2

1.5 1 6

0.5 2 8

1

2

3


0.5 1 2

1.5 1 6

0.5 2 8

1

2

3


SUMMARYSUMMARYRATE EQUATION - SAMPLE CALCULATIONRATE EQUATION - SAMPLE CALCULATION


SECOND ORDER wrt B



Chose one experiment (e.g. Expt. 3) and substitute its values into the rate equation

k = 8 = 4 dm6 mol-2 sec-1

(0.5) (2)2

rate = k [A] [B]2

re-arranging k = rate [A] [B]2

RATE EQUATION QUESTIONSRATE EQUATION QUESTIONS

CALCULATE THE ORDER WITH RESPECT TO ATHE ORDER WITH RESPECT TO BTHE OVERALL ORDER OF REACTIONTHE FORMAT OF THE RATE EQUATIONTHE VALUE AND UNITS OF THE RATE CONSTANT

ANSWER ON NEXT PAGE

[A] / mol dm-3 [B] / mol dm-3 Rate / mol dm-3 s-1

Expt 1 0.25 0.25 4Expt 2 0.25 0.50 8Expt 3 0.50 0.25 8

No 1No 1


Expts 1&2 [A] is constant [B] is doubled Rate is doubledTherefore rate [B]1st order wrt B

Explanation: What was done to [B] had exactly the same effect on the rate


Expt 1 0.25 0.25 4Expt 2 0.25 0.50 8Expt 3 0.50 0.25 8

No 1No 1

ANSWERANSWER




Expts 1&3 [B] is constant [A] is doubled Rate is doubledTherefore rate [A] 1st order wrt A

Explanation: What was done to [A] had exactly the same effect on the rate


Expt 1 0.25 0.25 4Expt 2 0.25 0.50 8Expt 3 0.50 0.25 8

No 1No 1

ANSWERANSWER






Rate equation is r = k[A][B]r = k[A][B]


Expt 1 0.25 0.25 4Expt 2 0.25 0.50 8Expt 3 0.50 0.25 8

No 1No 1

ANSWERANSWER







Value of k Substitute numbers from Exp 1 to get value of kk = rate / [A][B] = 4 / 0.25 x 0.25 = 64


Expt 1 0.25 0.25 4Expt 2 0.25 0.50 8Expt 3 0.50 0.25 8

No 1No 1

ANSWERANSWER







Value of k Substitute numbers from Exp 1 to get value of kk = rate / [A][B] = 4 / 0.25 x 0.25 = 64

Units of k rate / conc x conc = dm3 mol-1 s-1


Expt 1 0.25 0.25 4Expt 2 0.25 0.50 8Expt 3 0.50 0.25 8

No 1No 1

ANSWERANSWER


CALCULATE THE ORDER WITH RESPECT TO CTHE ORDER WITH RESPECT TO DTHE OVERALL ORDER OF REACTIONTHE FORMAT OF THE RATE EQUATIONTHE VALUE AND UNITS OF THE RATE CONSTANT

ANSWER ON NEXT PAGE

[C] / mol dm-3 [D] / mol dm-3 Rate / mol dm-3 s-1

Expt 1 0.40 0.40 0.16Expt 2 0.20 0.40 0.04Expt 3 0.40 1.20 1.44

No 2No 2



Expt 1 0.40 0.40 0.16Expt 2 0.20 0.40 0.04Expt 3 0.40 1.20 1.44

No 2No 2

Expts 1&3 [C] is constant [D] is tripled Rate is 9 x biggerTherefore rate [D]2 2nd order wrt D

Explanation: Squaring what was done to D affected the rate (32 = 9)

ANSWERANSWER



Expt 1 0.40 0.40 0.16Expt 2 0.20 0.40 0.04Expt 3 0.40 1.20 1.44

No 2No 2



Expts 1&2 [D] is constant [A] is halved Rate is quarteredTherefore rate [C] 2 2nd order wrt C

Explanation: One half squared = one quarter

ANSWERANSWER



Expt 1 0.40 0.40 0.16Expt 2 0.20 0.40 0.04Expt 3 0.40 1.20 1.44

No 2No 2





Rate equation is r = k[C]r = k[C]22[D][D]22

ANSWERANSWER



Expt 1 0.40 0.40 0.16Expt 2 0.20 0.40 0.04Expt 3 0.40 1.20 1.44

No 2No 2





Rate equation is r = k[C]r = k[C]22[D][D]22

Value of k Substitute numbers from Exp 2 to get value of kk = rate / [C]2[D]2 = 0.04 / 0.22 x 0.42 = 6.25

Units of k rate / conc2 x conc2 = dm9 mol-3 s-1

ANSWERANSWER


CALCULATE THE ORDER WITH RESPECT TO ETHE ORDER WITH RESPECT TO FTHE OVERALL ORDER OF REACTIONTHE FORMAT OF THE RATE EQUATIONTHE VALUE AND UNITS OF THE RATE CONSTANT

No 3No 3

ANSWER ON NEXT PAGE

[E] / mol dm-3 [F] / mol dm-3 Rate / mol dm-3 s-1

Expt 1 0.40 0.40 0.16Expt 2 0.80 0.80 0.32Expt 3 0.80 1.20 0.32



Expt 1 0.40 0.40 0.16Expt 2 0.80 0.80 0.32Expt 3 0.80 1.20 0.32

No 3No 3

Expts 2&3 [E] is constant [F] is x 1.5 Rate unchangedRate is UNAFFECTED ZERO order wrt F

Explanation: Concentration of [F] has no effect on the rate

ANSWERANSWER



Expt 1 0.40 0.40 0.16Expt 2 0.80 0.80 0.32Expt 3 0.80 1.20 0.32

No 3No 3



Expts 1&2 [E] is doubled [F] is doubled Rate is doubledTherefore rate [E] 2 2nd order wrt E

Explanation: Although both concentrations have been doubled, we know [F]has no effect. The change must be all due to [E]

ANSWERANSWER



Expt 1 0.40 0.40 0.16Expt 2 0.80 0.80 0.32Expt 3 0.80 1.20 0.32

No 3No 3





Rate equation is r = k[E]r = k[E]

ANSWERANSWER



Expt 1 0.40 0.40 0.16Expt 2 0.80 0.80 0.32Expt 3 0.80 1.20 0.32

No 3No 3





Rate equation is r = k[E]r = k[E]

Value of k Substitute numbers from Exp 1 to get value of kk = rate / [E] = 0.16 / 0.4 = 0.40

Units of k rate / conc = s-1

ANSWERANSWER

GRAPHICAL DETERMINATION OF RATE GRAPHICAL DETERMINATION OF RATE

RATE CALCULATION

The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction

Place a rule on the outside of the curve and draw a line as shown on the graph.

y

x

gradient = y / x

In the reaction…

A(aq) + B(aq) ——> C(aq) + D(aq)

the concentration of B was measured every 200 minutes. The reaction is obviously very slow!

The variation in rate can be investigated by measuring the change in concentration of one of the reactants or products, plotting a graph and then finding the gradients of the curve at different concentrations.


RATE CALCULATION



y

x

gradient = y / x

concentration = 1.20 mol dm-3

gradient = - 1.60 mol dm-3

1520 min

rate = - 1.05 x 10-3 mol dm-

3

The rate is negative becausethe reaction is slowing down

The variation in rate can be investigated by measuring the change in concentration of one reactants or product, plotting a graph and then finding the gradients of tangents to the curve at different concentrations.


RATE CALCULATION



y

x

gradient = y / x

The gradients of tangents at several other concentrations are calculated.

Notice how the gradient gets less as the reaction proceeds, showing that the reaction is slowing down.

The tangent at the start of the reaction is used to calculate the initial rate of the reaction.

The variation in rate can be investigated by measuring the change in concentration of one of the reactants or products, plotting a graph and then finding the gradients of the curve at different concentrations.

FIRST ORDER REACTIONS AND HALF LIFEFIRST ORDER REACTIONS AND HALF LIFE

One characteristic of a FIRST ORDER REACTION is that it is similar to radioactive decay. It has a half-life that is independent of the concentration.

It should take the same time to drop to one half of the original concentration as it does to drop from one half to one quarter of the original.

The concentration of a reactant falls as the reaction proceeds

The concentration of reactant A falls as the reaction proceeds

The concentration drops from

4 to 2 in 17 minutes



4 to 2 in 17 minutes2 to 1 in a further 17 minutes




4 to 2 in 17 minutes2 to 1 in a further 17 minutes1 to 0.5 in a further 17 minutes



A useful relationship

k t½ = loge 2= 0.693

where t½ = the half life


Half life = 17 minutes

k t½ = 0.693

k = 0.693 t½

k = 0.693 = 0.041 min-1 17

ORDER OF REACTION – GRAPHICAL DETERMINATION ORDER OF REACTION – GRAPHICAL DETERMINATION

The order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order.



PLOTTING RATE AGAINST CONCENTRATIONPLOTTING RATE AGAINST CONCENTRATION

RA

TE

OF

RE

AC

TIO

N /

mo

l d

m-3

s-1

CONCENTRATION / mol dm-3




RA

TE

OF

RE

AC

TIO

N /

mo

l d

m-3

s-1


ZERO ORDER – the rate does not depend on the concentration. The line is parallel to the x axis.




RA

TE

OF

RE

AC

TIO

N /

mo

l d

m-3

s-1



FIRST ORDER – the rate is proportional to the concentration so you get a straight line of fixed gradient. The gradient of the line equals the rate constant for the reaction.




RA

TE

OF

RE

AC

TIO

N /

mo

l d

m-3

s-1



FIRST ORDER – the rate is proportional to the concentration so you get a straight line of fixed gradient. The gradient of the line equals the rate constant for the reaction.

SECOND ORDER – the rate is proportional to the square of the concentration. You get an upwardly sloping curve.



PLOTTING RATE AGAINST TIMEPLOTTING RATE AGAINST TIME

RA

TE

OF

RE

AC

TIO

N /

mo

l d

m-3

s-1

TIME / s

ZERO ORDERA straight line showing a constant decline in concentration.

FIRST ORDERA slightly sloping curve which drops with a constant half-life.

SECOND ORDER The curve declines steeply at first then levels out.

ORDER OF REACTIONORDER OF REACTION

GRAPHICAL GRAPHICAL DETERMINATION DETERMINATION

Calculate the rate of reaction at1.0, 0.75, 0.5 and 0.25 mol dm-3

Plot a graph of rate v [A]

Calculate the time it takesfor [A] to go from...1.00 to 0.50 mol dm-3 0.50 to 0.25 mol dm-3

Deduce from the graphthat the order wrt A is 1

Calculate the value andunits of the rate constant, k

RATE DETERMINING STEPRATE DETERMINING STEP

Many reactions consist of a series of separate stages.

Each step has its own rate and rate constant.

The overall rate of a multi-step process is governed by the slowest step (like a production line where overall output can be held up by a slow worker).

This step is known as the RATE DETERMINING STEP.

If there is more than one step, the rate equation may not contain all the reactants in its format.


THE REACTION BETWEEN PROPANONE AND IODINE

Iodine and propanone CH3COCH3 + I2 CH3COCH2I + HIreact in the presence of acid

The rate equation is... r = k [CH3COCH3] [H+]

Why do H+ ions appear inthe rate equation?

Why does I2 not appear in the rate equation?





Why do H+ ions appear in The reaction is catalysed by acidthe rate equation? [H+] affects the rate but is unchanged overall

Why does I2 not appear The rate determining step doesn’t involve I2

in the rate equation?





Why do H+ ions appear in The reaction is catalysed by acidthe rate equation? [H+] affects the rate but is unchanged overall

Why does I2 not appear The rate determining step doesn’t involve I2

in the rate equation?

The slowest step of any multi-step reaction is known as the rate determining step and it is the species involved in this step that are found in the overall rate equation.

Catalysts appear in the rate equation because they affect the rate but they do not appear in the stoichiometric equation because they remain chemically unchanged


HYDROLYSIS OF HALOALKANES

Haloalkanes (general formula RX) are RX + OH- ROH + X-

hydrolysed by hydroxide ion to give alcohols.

With many haloalkanes the rate equation is... r = k [RX][OH-] SECOND ORDERThis is because both the RX and OH- mustcollide for a reaction to take place in ONE STEP


HYDROLYSIS OF HALOALKANES

Haloalkanes (general formula RX) are RX + OH- ROH + X-

hydrolysed by hydroxide ion to give alcohols.

With many haloalkanes the rate equation is... r = k [RX][OH-] SECOND ORDERThis is because both the RX and OH- mustcollide for a reaction to take place in ONE STEP

but with others it only depends on [RX]... r = k [RX] FIRST ORDER

The reaction has taken place in TWO STEPS...- the first involves breaking an R-X bond i) RX R+ + X- Slow- the second step involves the two ions joining ii) R+ + OH- ROH Fast

The first step is slower as it involves bond breaking and energy has to be put in.

The first order mechanism is favoured by tertiary haloalkanes because the hydroxide ion is hindered in its approach by alkyl groups if the mechanism involves the hydroxide ion and haloalkane colliding.


The reaction H2O2 + 2H3O+ + 2I¯ I2 + 4H2O takes place in 3 steps

Step 1 H2O2 + I¯ IO¯ + H2O SLOW

Step 2 IO¯ + H3O+ HIO + H2O FAST

Step 3 HIO + H3O+ + I¯ I2 + 2H2O FAST

The rate determining step is STEP 1 as it is the slowest


The reaction H2O2 + 2H3O+ + 2I¯ I2 + 4H2O takes place in 3 steps

Step 1 H2O2 + I¯ IO¯ + H2O SLOW

Step 2 IO¯ + H3O+ HIO + H2O FAST

Step 3 HIO + H3O+ + I¯ I2 + 2H2O FAST

The rate determining step is STEP 1 as it is the slowest

The reaction 2N2O5 4NO2 + O2 takes place in 3 steps

Step 1 N2O5 NO2 + NO3 SLOW

Step 2 NO2 + NO3 NO + NO2 + O2 FAST

Step 3 NO + NO3 2NO2 from another Step 1 FAST

The rate determining step is STEP 1 rate = k [N2O5]

OTHER TOPICSOTHER TOPICS

AutocatalysisA small number of reactions appear to speed up, rather than slow down, for a time. This is because one of the products is acting as a catalyst and as more product is formed the reaction gets faster. One of the best known examples is the catalytic properties of Mn2+(aq) on the decomposition of MnO4¯(aq). You will notice it in a titration of KMnO4 with either hydrogen peroxide or ethanedioic (oxalic) acid.

Molecularity The number of individual particles of the reacting species takingpart in the rate determining step of a reaction

e.g. A + 2B C + D molecularity is 3 - one A and two B’s need to collide

A 2B however has a molecularity of 1 - only one A is involved

REVISION CHECKREVISION CHECK

What should you be able to do?

Explain how the rate changes during a chemical reaction

Recall and understand the terms... rate equation, individual order and overall order

Interpret and calculate data to derive a rate equation

Construct graphs to find the order of reaction

Understand the importance of the rate determining step

CAN YOU DO ALL OF THESE? CAN YOU DO ALL OF THESE? YES YES NONO

You need to go over the You need to go over the relevant topic(s) againrelevant topic(s) again

Click on the button toClick on the button toreturn to the menureturn to the menu

WELL DONE!WELL DONE!Try some past paper questionsTry some past paper questions

THE RATE EQUATION A guide for A level students KNOCKHARDY PUBLISHING.

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