reaction rates -what is meant by rate of reaction? -how do we calculate reaction rate?

Reaction Rates

-What is meant by rate of reaction?-How do we calculate reaction rate?

• 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

Collision theory

• 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

To increase the rate of a reaction we can:

Increasing Rate

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

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

Rate of Reaction

• The rate of reaction is the change in concentration of a reactant or product per unit time.

• rate=

• Units: mol dm-3 s-1

change in concentration of reactant or product

time for the change to take place

Concentration

• Concentration in mol dm-3 is represented by square brackets.

• [HCl] means the ‘concentration of HCl in mol dm-3’.

Average rate

• The equation gives an average rate of reaction over the entire time period.

• Rate at a given time is measured using a graph.

Rate at this point in time

X

Y

= Y / X

= Gradient of tangent

Time / s

Concn / mol dm-3

How to Draw a Tangent Line

• Choose a point on the curve.

• Adjust the angle of the ruler so that near the point it is equidistant from the curve on either side.

Initial Rate

• The initial rate of reaction is the change in concentration of reactant, or product, per unit time at the start of a reaction when t = 0.

• The rate of reaction changes as the reaction proceeds.

Measuring Rate from a Graph

Sulphur dichloride, SO2Cl2, decomposes according to the equation: SO2Cl2 (g) SO2(g) + Cl2(g)

The concentration of SO2Cl2 was measured over time. Draw a concentration – time graph and calculate the rate of reaction at t = 0 and t = 3000Time t/s 0 500 1000 2000 3000 4000

[SO2Cl2]/mol dm-3

0.50 0.43 0.37 0.27 0.20 0.15

0 500 1000 2000 3000 40000

0.1

0.2

0.3

0.4

0.5

0.6

[SO2Cl2]/mol dm-3

[SO2Cl2]/mol dm-3

Measuring Rates Again

2H2O2(aq) 2H2O(l) + O2(g)

Plot a concentration – time graph for this reaction.Calculate the initial rate, rate at t = 20 and t = 90

Time t/s 0 15 30 60 100 180

[H2O2]/mol dm-3

0.40 0.28 0.19 0.07 0.03 0.01

0 15 30 60 100 1800

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

[H2O2]/mol dm-3

[H2O2]/mol dm-3

Rate Orders and Rate Constant

-What is the order of reaction?-What is the rate constant?

Rate Orders

• In a chemical reaction the concentration of each reactant can affect the overall reaction rate.

• This effect is called the order of reaction.• For a reactant A:

– Rate [A]x

– Where x = order with respect to A

Rate Orders

• Zero order– The rate is unaffected by concentration. Rate [A]o

• Any number to the power of zero = 1

• First order– Order is 1 with respect to B. Rate [B]1

– If [B] is increased by 1 then rate increases by 1• Second order

– Order is 2 with respect to C. Rate [C]2

– If [C] increases by 3 then rate increases by 32=9

Rate Constant

• Combining the rates from the previous slide from the reaction A + B + C products gives us:

rate [A]0 [B]1 [C]2

The rate constant, k, links the rate with the concentration, giving the equation:

rate = k [B]1 [C]2

(remember [A]0= 1)

Overall Order

• The overall order of the reaction is the sum of the individual orders.

• For rate = k [B]1 [C]2

• Overall order = 1 + 2 = 3

• The rate order comes from experimental results, not balanced equations!

Units of Rate Constants

• Units for k depend on the overall order.

Rate Determining Steps

-What affects the order of reactions?

R.D.S

• Some reactions occur as a series of separate steps.

• Each step in the reaction has it’s own rate and rate constant.

• The overall rate is governed by the slowest step.

• This is the rate determining step (R.D.S.)

RATE DETERMINING STEP

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]

R.D.S

• If the reactant appears in the rate equation, that reactant is involved in the rate determining step.

rate = k [NO2]2

The R.D.S involves two molecules of NO2

NO2 + NO2 Slow, R.D.S

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 CONCENTRATION

RA

TE

OF

RE

AC

TIO

N /

mo

l dm

-3 s

-1

CONCENTRATION / mol dm-3




RA

TE

OF

RE

AC

TIO

N /

mo

l dm

-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 dm

-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 dm

-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.

RATE 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 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 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]


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 kSubstitute 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 kSubstitute 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 [C] 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



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


Rate equation is r = k[C]2[D]2

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


Rate equation is r = k[C]2[D]2

Value of kSubstitute 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]

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]

Value of kSubstitute 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

reaction rates -what is meant by rate of reaction? -how do we calculate reaction rate?

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