Calculating Rate Expressions

7/31/2019 Calculating Rate Expressions

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16.1 Rate Expression

Logic is an important source of reliable knowledge for scientists. The rate of a chemical reaction is

determined empirically by experiment. It cant be determined theoretically using stoichiometry.

Using inductive reasoning specific data is collected from experiments and then combined into a

relationship called a rate expression or rate law. When sufficient reliable data has been collected a

series of inferences or general statements are made relating the concentration of each reactant to

the initial rate of the reaction. These inferences are combined and expressed as a rate expression.

Consider the reaction

A + B C

reactants product

Rate of reaction, R = k [ A]

m

x [ B]

n

Rate expression / rate law

/ rate equation

Where:

R, the rate of the reaction is measured in mol dm-3 s-1 [ ]is the concentration of the reactant measured in mol dm-3 m and n are the order of A and B in the reaction and the powers to which the concentration

needs to be raised: they usually have the values of 0,1 and 2.

k is the rate constant. It relates the rate of the reaction to the concentration of thereactants. The rate constant of a reaction increases with increasing temperature. Its units

vary depending on the order of the reaction.

The overall order of the reaction = m + n

[concentration of reactant]order

= Rate

A little light relief: The Constant Chemist

Chemists have to put up with a certain amount of mockery from other scientists, particularlyphysicists, over the word constant. When a physicist speaks of a constant, the word often means

something which has a universal, unchanging value such that you can look it up in a reference book.

Chemists often define a constant and then immediately go into detail about how it varies! The gas

constant, R, is a physicists constant but the rate constant, k, is only constant for a particular

reaction and then only at a particular temperature.


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Finding the Order of a Reaction

The order of a reaction indicates how the rate of a reaction varies with concentration. Catalysts are

ignored, because their concentrations do not change in a reaction. The normal method is to carry

out a number of separate experiments with different initial concentrations of each reactant and the

initial rate of each reaction determined. To find the rate expression the order of the reaction is

determined from the experimental data. Rate = [concentration of the reactant]order

1. First order reaction [ A ]1

Examples:

a) Doubling the concentration of reactant, A doubles the rate.


= rate

2order

= 2

21

= 2

b) Quadrupling the concentration of reactant, A quadruples the rate

[concentration of reactant ]order = rate

4order

= 4

41

= 4

The reaction is first order [ A ]1. The rate is proportional to the concentration raised to the power

of one. A first order reaction is often written as either [ A ]1

or [ A ]

Rate = k [ A ]1

2. Second Order [ A ]2

Examples:

a) Doubling the concentration of reactant, A quadruples the rate.

[concentration of reactant ]order

= Rate

2order

= 4

22

= 4

b) Quadrupling the concentration of reactant A increase the rate by a factor of 16 (16x).


= rate

4order

= 164

2= 16

The reaction is second order [ A ]2. The rate is proportional to the concentration raised to the

power of one. Rate = k [ A ]2


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3. Zero Order [ A ]0

Example:

a) Doubling the concentration of reactant, A does not change the rate.


= rate

2order = 1

20

= 1

The reaction is zero order [ A ]0. The rate is proportional to the concentration raised to the power

of zero. Sometimes a zero order reaction is written as [ A ]0

and sometimes it is left out of the rate

expression entirely. Note: any number raised to the power of zero is equal to one. Rate = k

Overall Order

The overall order of a reaction is the sum of the orders in the rate expression. For example if the

rate expression is:

R = k [ A]1

x [ B]2

The overall order of the reaction = 1 + 2 = 3 or third order.


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Experimental determination of the rate equation using the Initial Rates method

Example 1


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Example 2

For the reaction

A + B + C D + E

reactants products

A series of different experiments were carried out using different initial concentrations reactants A,

B and C. For each experiment a graph of concentration vs. time was plotted

From the graph the initial rate was determined by drawing a tangent to the curve when the time is

zero and determining the slope of this tangent (c t in mol dm-3

s-1

).

And the experimental data

Experiment [ A ]

(mol dm-3

)

[ B ]

(mol dm-3

)

[ C ]

(mol dm-3

)

Initial rate

(mol dm-3

s-1

)

1 0.4 1.6 0.06 4.86

2 0.8 1.6 0.06 9.72

3 0.4 0.8 0.06 4.86

4 0.8 1.6 0.18 87.5

Determine: (a) Rate expression for the reaction

(b) Value for the rate constant, k in experiment 1 and its units

(c) Overall order of the reaction

Steps:

1. Start with reactant A. Find the order of A by comparing two experiments where the

concentration of B and C are constant (do not change between the two experiments).

NOTE: There could be more than one option of comparison in the experimental data. Anyone will do.

2. Find the order of B by comparing two experiments where the concentration of Aand C are constant. There could be more than one option of comparison in the

experimental data.

3. Find the order of C by comparing two experiments where the concentration of Aand B are constant. There could be more than one option of comparison in the

experimental data.

4. Write the rate expression and determine in the units for k.

5. Find the overall order of the reaction.


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Determining the Order of a Reaction Qualitatively

1. Using Rate vs. Concentration of reactant Graphs

There are three types of rate vs. concentration graphs used to graphically determine the order of a

reaction.

The rate is directlyproportional to the

concentration


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2. Using Concentration of reactant vs. time graphs


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Half Life

In the graph above, the first order reaction shows an exponential decrease in the concentration of

reactants with time. The time for the concentration to fall from its initial value to half its initial

value is equal to the time required for it to fall from half to one quarter of its initial value. This time

is called the half life, t1/2 and is defined as the time for the concentration of a reactant to reach half

its initial value. The half life is constant because a first order reaction is independent of the original

concentration.

The rate expression for the reaction would be, R = k [ A ]1

For a first order reaction, to find the rate constant, k when t1/2 is knownuse the formula:

t = 0.693

k

This equation is in the IB Chemistry data booklet. The unit for k is s-1

.

For example, the graph below show the amount (moles) of H2O2 remaining at different times in

during the decomposition of H2O2.

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

The graph below shows

how the amount of H2O2

decreases with time.


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The graph can be used to determine the half life of the reaction i.e the time taken for half the H2O2

to get used up.

The graph shows that

whatever the starting amount

(moles/concentration), t1/2 is always 27 s. Therefore the decomposition of H2O2 is first order with

respect to the H2O2. The same rule applies for the half lives of all first order reactions the half life

is constant no matter what the starting amount because a first order reaction is independent of the

original concentration. When determining the half life of first order reactions graphically

extrapolate the line back to the y axis. Look for consistency in the first two half lives. Zero order

and second order reactions do not have constant half lives.

Questions

Questions 1 to 4 refer to the rate expression given below

Rate = k x [ A ] x [ B ]2x [ H

+]

1. Which one of the following statements is not true about this reaction?A It is first order in A

B It is second order in B

C It is first order in H+

D It is third order overall

2. The units of the rate constant (k) will beA mol dm-3 s-1B mol s

-1

C dm3

mol-1

s-1

D dm9mol

-3s

-1


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3. If the concentration of A and B are both doubled, but the concentration of H+ remainsConstant, the rate of the reaction would increase by a factor of:

A 2

B 4

C 8

D 16

4. Which one of the following would lead to the greatest increase in reactions rate ?A doubling the concentration of A only

B doubling the concentration of B only

C doubling the concentration of A and H+

only

D doubling the concentration of B and H+

only

5. Draw a labeled sketch graph to show a reaction is zero order with respect to the reactantwhose concentration was being varied. [3]

6. The data given below refers to the hydrolysis of a 0.002 mol dm-3 solution of an ester by 0.2mol dm-3 aqueous sodium hydroxide.

Time (s) 60 120 180 240 300 360 420 480

[ ester x 10-3

]

(mol dm-3

)1.48 1.10 0.81 0.60 0.45 0.33 0.24 0.18

a) Plot a suitable graph to determine the order of the reaction with respect to theester. Explain your answer. [6]

b) Use the graph to determine the half life of the reaction and use it to determine thevalue of the rate constant. [5]

c) Other than using half life, use the graph to find k.7. The following data refers to the acid catalysed iodination (adding iodine) of propanone.

CH3COCH3 + I2 + H+

I-CH2COCH3 + HI

Solution[ CH3COCH3]

(mol dm-3

)

[ I2]

(mol dm-3

)

[ H+]

(mol dm-3

)

Rate

(mol dm-3

s-1

)

1 0.2 0.008 1 4 x 10-6

2 0.4 0.008 1 8 x 10-6

3 0.6 0.008 1 1.2 x 10-5

4 0.4 0.004 1 8 x 10-6

5 0.4 0.002 1 8 x 10-6

6 0.2 0.008 2 8 x 10-6

7 0.2 0.008 4 1.6 x 10-5

a) Give the order with respect to CH3COCH3 , I2 and H+ and the overall order. [3]


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b) From the data in the table derive the rate expression for the reaction explaining theevidence for the dependency of each of the species. [2]

c) Use the data from solution 1 to calculate the value of the rate constant. [3]

8. Consider some of the possible uncertainties or limitations in relying on inductive reasoningto determine a rate expression.

9. Since a rate expression can only be determined experimentally is it still reliable knowledge?Explain.


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Answer for Question 2

Order for A

Comparing experiments 1 and 2

[ A ] doubles and the rate doubles A is first order [ A ]1

21

= 2

[ B ] and [ C ] are constant

Order for B


[ B ] halves and the rate stays the same B is zero order [ B ]0

0

= 1

[ A ] and [ C ] are constant

Order for C


[ C ] increases by 3 x and the rate increases 9 x C is second order [ C ]2

32

= 9

[ A ] and [ B ] are constant

Therefore:

(a) rate expression = k x [ A]1

x [ B]0

x [ C]2

or rate = k x [ A]1

x [ C]2

(b) overall order = 1 + 2 = 3 third order

(c) rate = k x [ A]1

x [ C]2

k = rate

[ A]1

x [ C]2

k = 4.86

(0.4)1

x (0.06)2

k = 4.86 units mol dm-3

s-1

0.4 x 0.0036 mol dm-3

x (mol dm-3)

2

k = 3375 mol-2

dm6

s-1

k = 3 x 103 mol

-2dm

6s

-1(1SF)


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16.1 Rate Expression ANSWERS

1. D (1)

2. D (1)

3. C (2)1

x (2)2

= 2 x 4 = 8

A is first order, If you double A its rate will double, so (2)1

= 2

B is 2nd order, If you double B its rate will quadruple, so (2)2

= 4 (1)

2 x 4 = 8

4. D (1)

5. In a first order reaction the rate is unaffected by

changes in concentration.

The rate = constant or rate = [conc]0

Line (1)

x and y axis labeled correctly (x = conc and y = rate) (1)

x (mol dm-3

) and y (moldm-3

min-1

) axis correct units (1)

6. a) Graph

i) x and y axis labeled correctly (x = time and y =

concentration) (1)

ii) x (s) and y (moldm-3

) axis correct units (1)

iii) Title relates x and y axis (1)iv) Line graph with points joined with smooth curve thru points (1)v) Appropriate scale on axis (1)

vi) Suitable size, fills graph paper/plot area (1)

Reaction is First order (1)

The concentration of the ester halves in equal time intervals (1)

(as seen by the first two half lives which are equal at 142s)


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b) t1/2 = 142 st1/2 = 0.693 k = 0.693 = 0.693 = 4.95 x 10

-3s

-1

k t1/2 142

t1/2 = 142 s (1)

equation for half life (1)

answer (1)

correct sig figs (1)

correct unit , s-1

= (1)

c)Intial rate can be found by finding the slope of a tangent at zero seconds

Tangent at t = 0 = Initial rate = [ester] = 0.013 moldm-3

s-1

time

The initial concentration of the ester can be found by extrapolating the line back to

the x axis = 1.83 moldm-3

dm-3

Since reaction is 1st

order

Rate = k [ester]1

k = Rate = 0.013 moldm-3

s-1

= 0.0071 s-1

(2SF)

[ester]1

1.83 mol dm-3

7) a) CH3COCH3 is first order (1)

When the concentration CH3COCH3 doubles its rate doubles so 2x= 2 - first order

I2 is zero order (1)

When the concentration I2 halves its rate remains the same so x= 1 - zero order

H+

is first order (1)

When the conc H+ doubles its rate doubles so 2x= 1 - first order

b) Rate = k [CH3COCH3]1

[I2]0

[H+]

1(1)

overall order = 1 + 0 + 1 = 2nd order (1)

d) Rate constant, K for solution 1 is:


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K = rate = 4 10-6

= 2 10-5

mol-1

dm3s

-1

[CH3COCH3]1

[H+]

10.2 x 1

Equation (1), correct answer (1), correct unit (1)

Intial rate can be found by finding the slope of a tangent at zero seconds

Tangent at t = 0 = Initial rate = [ester] = 0.013 moldm-3

s-1

time

The initial concentration of the ester can be found by extrapolating the line back to

the x axis = 1.83 moldm-3

dm-3

Since reaction is 1st

order

Rate = k [ester]1

k = Rate = 0.013 moldm-3

s-1

= 0.0071 s-1

(2SF)

[ester]1

1.83 mol dm-3

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Calculating Rate Expressions