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Chapter 26 Reaction Kinetics

Rate Equations

1. The rate equation shows exactly how the rate is dependent on the concentration of
reactants; it relates the rate of reaction to the concentration of reactants raised to the
appropriate power.

2. The general rate equation is rate = k [Reactant]n k = rate constant

n = order of reaction
c = concentration of reactant

3. Rate equation can only be obtained experimentally.

Order of Reaction

1. Definition: The power to which the concentration of that reactant is raised in the rate
equation.
Or the value of a in the expression rate = k [A]a

2. The order of reaction shows _________________________________________________

and can only be found experimentally.

3.

e.g. aA + bB → cC + dD

The rate equation is rate = k [A]m [B]n

Rate Constant

1. The rate constant, k, is a constant of proportionality in the rate equation.

It is constant for a given reaction at a particular temperature.

The units of k depend on the order of reaction.

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Half Life

1. The half-life of a reaction, t1/2, is the time taken for the initial concentration of a reactant
to fall to half its value.

Zero Order Reaction ( n = 0 )

1. In zero order reaction, Rate α [Reactant]0.

Graph

[Reactant] Rate

Time [Reactant]

2. The concentration-time graph is a straight-line graph showing that rate is constant.

e.g Decomposition of NH3 on heated tungsten at high pressure.

2NH3(g) → N2(g) + 3H2(g)

The rate equation is

Unit of k =

• The rate of reaction is independent of the concentration of NH3.

• Changing the concentration of NH3 will not affect the rate of reaction.
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First Order Reaction ( n = 1 )

1. In a first order reaction, Rate α [Reactant]1

Graph

[Reactant] Rate

Time [Reactant]

2. The concentration-time graph is a curve with a constant half-life.

t1/2 = constant or t1 = t2 = t3

3. Half life is related to the rate constant by the expression:

k = 0.693
t1/2 (in s)

e.g. Decomposition of hydrogen peroxide

2H2O2 → O2 + 2H2O

The rate equation is

Unit of k =

• The rate of reaction is directly proportional to [H2O2].

• Doubling [H2O2] will double the rate of reaction.

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Second Order Reaction ( n = 2 )

1. In a second order reaction, Rate α [Reactant]2

Graph

[Reactant] Rate

Time [Reactant]

2. The concentration-time graph is a curve with a non constant half-life

t1  t2  t3 or t1/2  constant

e.g. Decomposition of NO2

2NO2(g) → 2NO(g) + O2(g)

The rate equation is

Unit of k =

• The rate of reaction is proportional to [NO2]2.

• Doubling [NO2] will increase the rate of reaction four times.

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TRY!
MJ 2022 Paper 41 Q3

MJ 2021 Paper 42 Q7
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Experimental Technique For Studying Reaction Rates

Titrimetric Method

1. Titrimetric method is particularly useful for reaction in solutions.

2. The concentration of reactants/products is determined at different times by withdrawing

known volumes of the reaction mixture and titrating with a suitable standard solution.

3. The sample volume withdrawn must be quenched by either

(i) Adding a large volume of cold water to slow down the reaction, or
(ii) Adding a “quenching reagent”, which would react immediately with one of the
reactants and hence, stop the reaction.

Example:

Acid catalyzed reaction of propanone with iodine

H+
CH3COCH3(aq) + I2(aq) CH2ICOCH3(aq) + HI(aq)

To study the rate of reaction, the amount of unreacted iodine in the reaction mixture is
determined at different times.

Portions of the reaction mixture are pipetted (at regular time intervals) into aqueous NaHCO3
solution which “quenches” the reaction by neutralizing the acid catalyst. The quenched
mixture is then titrated against a standard solution of sodium thiosulphate (VI) to determine
the amount of unreacted iodine.

2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)

The graph of “volume of S2O32-(aq) against time” is a straight-line graph showing that rate is
constant as [I2] varies (since volume of S2O32- α [I2]).
Thus, the reaction is zero order with respect to I2.
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Measurement of Volume of Gaseous Product

1. For reaction that involve formation of gaseous product, the rate of reaction may be
studied by determining:

(a) Volume of gas produced per unit time (where the gas is collected in a syringe), or
(b) Loss in mass per unit time (where the gas is allowed to escape).

Reaction between dilute HCl and marble chips

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

Marble chip

Two possible methods to study the rate of reaction:

1. Determine the volume of CO2(g) product per unit time.

• CO2(g) is collected in a gas syringe and its volume measured at regular time
intervals during the reaction.
• The rate of reaction may be found by plotting “volume of CO2 against time”. At
any time, t, the rate of reaction is given by the gradient of the graph at time t.

2. Determine the loss in mass of the conical flask and its contents per unit time (where the
gas is allowed to escape).

The rate of reaction may be found by plotting “loss in mass (of conical flask and contents)
against time”.At any time, t, the rate of reaction is given by the gradient of the graph at time t.

[Cotton wool is placed at the mouth of the conical flask to prevent the splashing out of
reaction mixture from the conical flask.]
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Colorimetry Method

1. Since the colour intensity of a compound is proportional to its concentration, the rate of
reaction may be studied by following the change in colour intensity.

H+
e.g: CH3COCH3 + I2 → CH2ICOCH3 + HI

• The rate of reaction can be measured by recording the reduction in [I2] by the
decrease in the intensity of its red-brown colour as measured in colorimeter.
• The rate of reaction maybe found by plotting [I2] against time. Since colour intensity
is proportional to [I2]. The rate of reaction is given by the gradient of the graph.

• To study the order of reaction, the experiment is repeated by varying the initial
concentration of each of the reactants in turn, keeping the other two constant and the
corresponding graphs of [I2] against time are plotted.

Expt. Initial concentration of reactants / mol dm-3 Initial rate /

[CH3COCH3] [I2] [H+] mol dm-3 s-1
1 0.01 0.01 0.01 2.0 x 10-6
2 0.03 0.01 0.01 6.0 x 10-6
3 0.03 0.02 0.01 1.2 x 10-5
4 0.03 0.02 0.02 1.2 x 10-5

The order of a reaction can be deduced by the initial rates method – by comparing the initial
rates of two reactions at known initial concentrations.
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TRY!
1) Given the variation of initial reaction rate with [NO2] for the reaction:
NO2(g) + CO(g) → NO(g) + CO2(g)

Experiment Rate / mol dm-3 s-1 [NO2] / mol dm-3

1 0.01 0.15
2 0.04 0.30
3 0.16 0.60
4 0.36 0.90

2) The rate of decomposition of 3.0 mol dm-3 H2O2(aq) was measured by withdrawing 10 cm3
portions at various times and titrating with acidified 0.1 mol dm-3 KMnO4(aq).
The results obtained are given in the table.

Time / min Volume of

KMnO4 / cm3
0 30.0
5 23.4
10 18.3
15 14.2
20 11.1
25 8.7
30 6.8
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3) RBr + NaOH → ROH + NaBr

The order of reaction may be deduced by plotting graphs of [RBr] vs time.

From graph A,
Initial rate = gradient g1
= 5.5 X 10-5

From graph B,
Initial rate = gradient g2
= 8.3 X 10-5
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Reaction Mechanism

A reaction mechanism is a sequence of simple steps proposed in theory to account for the
overall chemical reaction that takes place and it must be consistent with the observed
kinetics; it must satisfy both the rate equation and the stoichiometric equation.
We can find order of reaction and balanced equation from mechanism.

Rate-Determining Step

1. The slowest step in the reaction mechanism is called the rate-determining step (r.d.s.) and
it determines the rate of the entire multi-step process.

2. The order of reaction could be predicted from the rate-determining step of a given
reaction mechanism.
The rate equation for the overall reaction is obtained from the rate-determining step.

e.g. 1 CH3CH2Br + OH- → CH3CH2OH + Br-

- Both OH- ion and CH3CH2Br molecule are involved in the slow step (rate determining
step).

Hence, the rate equation is:

rate = k [CH3CH2Br] [OH-]

- Reaction is first order with respect to CH3CH2Br, and first order with respect to OH-.
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e.g. 2 (CH3)3C-Br + OH- → (CH3)3C-OH + Br-

- This reaction takes place in two steps:

Step 1: (CH3)3C-Br → (CH3)3C+ + Br- (slow)

Step 2: (CH3)3C+ + OH- → (CH3)3C-OH (fast)

- Only the (CH3)3C-Br molecule is involved in the slow rate-determining step.

Hence, the rate equation is:

rate = k [(CH3)3C-Br]

- Reaction is first order with respect to (CH3)3C-Br, and zero order with respect to OH-

- The OH- does not appear in the rate equation and it does not take part in the rate-
determining step.

Example (The balanced equation)

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Question 1

H+
I2(aq) + CH3COCH3(aq) → CH2ICOCH3(aq) + I-(aq)

Question 2

5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l)

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The effect of temperature change on the rate constant and the rate of reaction

The Arrhenius Equation

k = rate constant
A = the frequency factor / the pre-exponential factor
e = mathematical quantity (2.71828)
R = gas constant
T = temperature in K
Ea = Activation energy

If the temperature increases by 10°C , for example, 20°C to 30°C (293 K to 303 K)
The frequency factor, A, in the equation is approximately constant for such a small
temperature change. Let’s look at how e-(EA / RT) changes

Assuming an activation energy of 50000 J mol-1

At 20°C (293 K) the value of the fraction is

At 30°C (303 K) the value of the fraction is

As temperature increases, rate constant increases and therefore rate of reaction

increases.

Nov 2017 Paper 42 Q1

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Homogenous and Heterogeneous Catalysis

Homogenous Catalysis
• Both catalyst and reactants are in the same physical state.

• Depends on its ability to exist in variable oxidation sates.

e.g. Catalytic oxidation of SO2 by NO2

SO2(g) + ½ O2(g) SO3(g)

Step 1: SO2 + NO2 → SO3 + NO

Step 2: NO + ½ O2 → NO2

e.g. Oxidation of I-(aq) by S2O82-(aq)

Fe3+(aq)
S2O82-(aq) + 2I-(aq) 2SO42-(aq) + I2(aq)

- The activation energy of this reaction is high without the presence of catalyst because

_____________________________________________________________________

- Fe3 +(aq) ion catalyses the reaction to increases the rate of reaction.

Step 1: 2Fe3+ + 2I- → 2Fe2+ + I2

Step 2: 2Fe2+ + S2O82- → 2Fe3+ + 2SO42-

Heterogeneous Catalysis
• Both catalyst and reactants are in different physical states.

• Depends on the availability of partially filled 3d orbitals, which allows the adsorption of
reactant onto the catalyst surface.
The adsorption weakens the bonds in the reactant molecules and lowers the activation energy.

e.g 1. Haber process

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e.g.2. Catalytic convertor

2NO(g) + 2CO(g) → 2CO2(g) + N2(g)

2CO(g) + O2(g) → 2CO2(g)

CxHy(g) + (x + y/4)O2(g) → xCO2(g) + y/2H2O(g)

Catalyst: Pt, Pd, Rh(s)

e.g. 3. Contact Process

SO2(g) + ½ O2(g) SO3(g)

Catalyst: V2O5(s)

e.g. 4 Hydrogenation of alkene

CH2=CH2 + H2 → CH3CH3

Catalyst: Pt or Ni(s)
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TRY!
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TRY!

The rate is first order with respect to Cr(CO)6 and zero order with respect to PR3.
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Nov 2014 Paper 41 Q1

MJ 2019 Paper 41 Q5
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May/June 2006
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May/June 2010 Paper 43

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MJ 2012 Paper 42 Q2
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May/June 2004