Chemistry

Key unit competency:

Measure the rates of reaction and formulate simple rate equations using the experimental results.

Learning objectives:
By the end of this unit the learners should be able to:

• State and explain kinetic conditions for a chemical reaction to take place.
• Explain the effect of the temperature and catalysts on the rate of the reaction using Boltzmann distribution of energies (and of collision frequency).
• Differentiate between SN1 and SN2 mechanisms.
• State and explain the rate determining steps for multi-step reactions.
• Deduce the order of reaction from appropriate experimental data.
• Calculate the initial rates and the rate constants of reactions from the experimental data.
• Perform practical activities to show how different reactions have different rates.
• Interpret the graphs which show the change in activation energy with the catalyst.
• Calculate the half-life of chemical reaction.
• Perform practical activities to measure the rates of reaction by observing the changes in physical quantities (e.g. volume, mass and colour change).
• Use collision theory to predict if the reaction will go faster or slower.
• Construct rate equations of the form (Rate = k [A]n[B]m) limited to simple cases involving zero,

  first and second order reactions.
• Interpret graphs of concentration against time and those of concentration against rate for zero

  and first order reactions.
• Develop a spirit of team work, analysis, and self-confidence while discussing exercises

  and performing the experiments.
• Appreciate the contributions of Arrhenius and Boltzmann on the effect temperature and activation

   energy of different substances and number of molecules

Introductory activity

One of the mission of chemistry consists in making substances from other substances. Referring to the following six figures; some processes are observed during a chemical reaction from each Figure.

Observe these figures carefully and answer the related questions.

 

(1)                                                                                              (2)

 

                 (3)                                                                            (4)

 

                                      (6)

                 (5)

i. Suggest what is happening in each image.

ii. Classify the processes into slow or fast processes, and explain why?

14.1 Theories of reaction rates

Activity 14.1

1. Draw the energy profile diagram for

a. An exothermic reaction

b. An endothermic reaction

2. Using textbook or internet read and analyze the content about collision theory of reaction rates and make a summary to be presented to the class.

In this work you should be able to explain contribution of Arrhenius and Boltzmann on the effect of temperature and activation energy on rates of reaction.

In the previous unit, the factors that influence the rate of a reaction have been discussed.

It is known that during a chemical reaction, the reactant species are converted into the new substances called products. This means that during a chemical reaction, the concentration of reactants decreases as they are consumed and that of products increases as they are in formation.

In a chemical reaction, how quickly or slowly reactants turn into products is called the rate of reaction.

Two theories known as “collision theory” and transition state theory” explain the rate of reaction.

14.1.1. Collision theory

According to the collision theory, a chemical reaction takes place due to the effective collisions of reacting molecules. For an effective collision to happen, the reactant species must be oriented in space correctly to facilitate the breaking of old bonds and forming of new bonds. Molecules which did not participate in effective collisions do not favor the reaction. During a chemical reaction species that collide must possess also a minimum amount of kinetic energy called “activation energy (Ea)”. The activation energy varies depending on the reaction. Therefore, the rate of reaction depends on the activation energy; a higher activation energy means that fewer molecules will have sufficient energy to undergo an effective collision.

Consider the reaction: A-A + B-B → 2(A-B)

For the reaction to take place, A-A and B-B must possess the activation energy, Ea that is required for breaking their bonds in order to form the new compound 2(A-B). The energy changes that occur during a chemical reaction can be shown in a Figure 14.1.

Figure 14.1: The energy profile diagram

The activation energy, (Ea), is the minimum energy required to initiate a reaction. Only colliding molecules with energy greater than or equal to Ea can lead to the reaction process. When two reactant molecules approach one another, their velocity will be slowed down due to the repulsion between their electrons clouds. However, if the initial kinetic energy of the colliding molecules is high, they can overcome the repulsion force and approach one another leading to the formation of products by merging the electron clouds.

For a chemical reaction to occur, there are three main conditions that must be fulfilled.

i. The reactants must collide: no reaction is observed if there are no collisions between reactant compounds.
ii. The molecules must have sufficient energy (activation energy) to initiate the reaction.
iii. The molecules must have proper orientation. Unless the reactant particles possess this orientation when they collide, the collision will not be an effective one.

Application examples:
1. Consider the reaction: A + B→ AB
As the two reactant molecules approach one another, their velocity will be slowed down due to the repulsion between their electron clouds.

Repulsion between electron clouds

If the initial kinetic energy of the colliding molecules is low, the molecules will repel one another and no product will be formed.

If the initial kinetic energy of the colliding molecules is high, they can overcome the repulsion force and approach one another so that electron clouds merge to form the products.

2. Consider the reaction (CO(g) + O₂(g) → CO₂(g)) represented as follows:

These examples show that when an effective collision requires a proper orientation of the reacting

  particles to form the products.

3. The following examples (a) and (b) show also that the probability of a reaction to occur depends

    not only on the collision energy but also on the spatial orientation of the molecules when they collide.

The collision theory is based on the kinetic theory and assumes a collision between reactants before a reaction can take place. It explains how temperature and activation energy affect the rate of reaction.

a. The speed of reaction and theory of collision
The activation energy depends on the nature of chemical bonds, which are broken during the reaction.

  The stronger the bonds, the greater is the activation energy.

For an elementary reaction, the collision theory shows that the speed (rate, R) of a reaction

   is given by R = f x p x z

Where: R = the rate or speed of the reaction
             f = the fraction of molecules having kinetic energy that can cause a reaction
             p = the probable fraction of collisions with proper (effective) orientation
             z = the frequency of collision

b. Effect of increase in temperature on the rate of reaction
An increase in temperature typically increases the rate of reaction.

An increase in temperature will raise the average kinetic energy of the reactant molecules.

Therefore, a greater proportion of molecules will have the minimum energy necessary for an effective collision. For example, an increase in temperature by 10 °C (or 10 K), doubles the speed of reaction.

From the kinetic theory, it is known that the kinetic energy (K) of a gas is directly proportional to its temperature.

Where: m = mass of the gas
v = speed of the gas molecule

N_A = Avogadro’s number = 6.6023 x10²³ particles for one mole of a gas at standard conditions

             = The Boltzmann constant

T = Absolute temperature in Kelvin (K)

R = Universal (molar) gas constant = 0.08206 L atm/mol K or 8.314 m³Pa/mol K
The numerical values of the gas constant, R, can be expressed using different units as

       indicated in Table 14.1.
Table14.1. Gas constant R converted in various units

When the temperature is increased, there is an increase in the total number of particles with energy

 equal to or greater than the activation energy (Ea).

The combined effects cause the rate of reaction to increase as illustrated by the Maxwell-Boltzmam distribution curve (Figure 14.2).

Figure 14.2: The Maxwell-Boltzmann distribution curve

Note:
• The area under the curve is the same in both cases because the number of molecules remains the same.
• All the fractions add up to unity
• The fraction of molecules with energy E ≥ Ea, is greatest at higher temperature, T₂.
That is when temperature increases,

i. the average kinetic energy of the molecules increases resulting in more collisions per unit time,
ii. The fraction of molecules with E ≥ Ea increases
At temperature T₁, the number of particles having energy equal to or greater than the activation energy (Ea) is given by the area ABC (Figure 14.2).
At T₂ the number of particles having energy equal or greater to Ea increases about two times by the area ADF (Figure 14.2).

c. Temperature, rate of reaction and Arrhenius equation
The relationship between the rate constant (k) of a reaction and temperature (T) of the system was first proposed by Arrhenius in 1889 and it is commonly known as the Arrhenius equation. It is written as:

                                  (1)
Where: k = rate constant
Ea = activation energy
R is the gas constant = 8.314 (J mol ^-1 K ^-1)
T = absolute temperature in Kelvin degrees (0°C = 273 K)

A = Arrhenius constant or frequency factor
A = p · Z, where Z is the collision frequency and p is called the steric factor (always less than 1)

and reflects the fraction of collisions with effective orientations.
e^-Ea/RT represents the fraction of collisions with sufficient energy to produce a reaction.
The Arrhenius equation (1) is often written using the natural logarithm of both sides in order to determine

the activation energy (Ea) for a chemical process.

From the equation (2), it is possible to determine the activation energy using the ratio of two rate constants (k1, k2) and the two corresponding reaction temperatures (T₁, T₂).

The following equations may be obtained:

Subtracting the equation (4) from the equation (3), we have:

Simplifying the equation (5) and rearranging it gives:

The activation energy can be deduced from the equation (8) if k₁, k₂, T₁ and T₂ are known.

Example 1:
For a reaction with activation energy of 55 kJ/mol, by what factor will the rate constant go up with a rise

  in temperature from 300 K to 310 K?

The above example showed that an increase of the temperature leads to the increase of the rate of the reaction. The rate of the reaction doubles for any 10K rise in temperature.

Example 2:
What is the activation energy of a reaction whose rate quadruples when the temperature is raised from 293 K to 313 K?

Solution:

The Arrhenius relation can be plotted to obtain a straight line.

The graph of ln k versus 1/T (Figure 14.3) is a straight line with a negative slope.

It shows the dependence of the rate of reaction with temperature.

Figure 14.3: The Arrhenius plot of ln k against 1/2

A part from the collision theory that explain the effect of temperature on the rate of reaction, the transition state theory also is based on the temperature effect on the rate of reaction.

14.1.2 The transition state theory

The transition state theory has been developed by Henry Erying in 1935. The theory supposes that a collision between reactants does not lead immediately to products. This theory supposes that the molecules collide and remain held for a certain interval of time, forming intermediate species or activated complexes (transition state) whose energy is higher than the average energy of the reactants. These activated complexes get then dissociated by generating either the reaction products or the starting reactants.

The activated complex is an unstable grouping of atoms, formed during a fruitful collision that breaks apart to form reaction product(s).
The energy needed to form an activated complex is equal to or greater than the activation energy Ea of the reaction.

Example:

(* ) The exponent is used to designate the activated complex.
The transition state theory provides a way to calculate the rate constant for the reaction.

According to the transition state theory:

(1) During collision, the attraction between reacting molecules decreases and the kinetic energy of the molecules is converted into potential energy.

(2) When the molecules approach, the interaction of the electronic configuration allows the rearrangement

    of valence electrons.

(3) A temporary bond between atoms A and B is formed where as the bond for a bond B-C is weakered, which leads to the formation of activated complex. The activated complex is momentary as it decomposes

  to give the products (A-B + C)

Figure 14.4: The diagram of energy profile for an exothermic reaction

The difference in potential energy between the reactants and the activated complex corresponds to the activation energy of the reactants, Ea
Eb is the energy required for the products to give reactants (for reverse reaction)
The change in enthalpy of the reaction ΔH can be found as follows:

Example: Consider the diagram profile for the following reversible reaction:

ΔHr_xn = Ea (rev) – Ea (fwd) > 0, the reaction is exothermic

If the potential energy of the products is less to the energy of reactants, the obtained energy (for the transition state to change to products will be greater than the activation energy.. Hence the reaction will be exothermic. If the potential energy of products is greater than that of reactants, the energy released for the activated complex to give products will be less than the activation energy. Hence the reaction will be endothermic.

For any reaction, there exist a barrier of energy (activation energy) which must be overcome for a reaction to take place.

The use of a catalyst can reduce the activation energy as it provides an alternative reaction pathway with the minimum activation energy as a result, more reactant molecules possess the energy required for a successful collision. As the speed of the reaction is proportional to the effective collisions, the presence of a catalyst increases the speed of both forward and reverse reaction but it is regenerated after reaction.

It should be noted that even if a catalyst lowers the activation energy, the difference in energy, ΔH, between products and reactants remains constant. For an endothermic reaction, the activation energy of the reactants is greater than that of products and ΔH of the reaction is negative (Figure 14.5).

Figure 14.5: Potential energy for an endothermic reaction.
Example:

Consider the following diagram profile of the reaction:

ΔH_rxn = Ea (rev) – Ea (fwd) < 0, the reaction is endothermic

Checking up 14.1

1. Given the following equation for the decomposition of nitrous oxide:

2N₂O(g) → 2N₂(g) + O₂(g) and given that the activation energy for the forward reaction is + 251 kJ and

  that the change in enthalpy for the reaction is +167 kJ.

a. Sketch and label a potential energy diagram for the decomposition of nitrous oxide:

b. What is the activation energy for a reverse reaction?

2. For the reaction between A-A and B-B to give AB +AB               i.e A-A + B-B → AB +AB,

Draw a situation showing effective collision and another situation which is not effective.

14. 2. Measuring the rates of reaction by observing the mass changes, colour changes and

          volume changes

Activity 14.2

Experiment 1: Measuring the rate of reaction by observing the volume change of the gas produced. Apparatuses and chemicals: 3.0M HCl, calcium carbonate in form of marble chips, graduated measuring cylinder or flexible graduated syringe, balance, funnel, con-ical flask, stop clock, stopper for closing the conical flask, retort stand .

Procedure:

1. Weigh 40g of marble chips (calcium carbonate) and put them in the conical flask

2. Measure 100 cm³ of 3.0M HCl and transfer all the acid into the conical flask using a funnel. Close the mouth of the conical flask with a stopper

3. Connect the conical flask to a flexible graduated syringe. Clamp the flexible graduated syringe.

Start the stop clock and record the total volume of carbon dioxide collected in the flexible graduated syringe. Record your results in the table below:

Measuring the rate of reaction by observing the volume change

Questions:

a. Plot a graph of gas volume of carbon dioxide evolved on y-axis against time on the x-axis.

b. Determine the rate of evolution of gas at 10 seconds, 20 seconds, and 100 seconds. What can you conclude about the reaction rate as time progresses?

Experiment 2: Measuring the rate of reaction by obsering color change

You are provided with 0.2M HCl(aq) and 0.05M Na2S2O3(aq).By changing the concentration of Na2S2O3 and keeping constant the concen-tration of HCl.

a. On a piece of paper, draw a cross (X) using a marker pen and place this paper under a conical flask.

b. Complete the following table of results:

i. Write an equation between Na2S2O3(aq) and HCl(aq).

ii. Plot a graph of time for the cross to disappear against volume of Na2S2O3(aq)

iii. Write down the color changes in the reaction above.

iv. Describe the chemical test of a gas produced in the reaction

Measuring the rates of reaction uses different techniques depend on the reaction.

The rate of reaction can be measured by using the change in mass, color change, volume and other properties. By analyzing the reaction mixture at suitable time intervals, one can determine the concentration of both reactants and products at particular time, hence determining the reaction rate.

Sometimes it is easier to measure the change in the amount of a reactant that has been used up; sometimes it is easier to measure the change

in the amount of a product that has been produced.

The method used to analyze the reaction mixture depends on the reaction under consideration.

The concentration is not measured directly in many cases, it is determined based on the signal related

 to the change in concentration. For example in the case of the reaction producing the colour, the intensity

 of the colour can be measured and related to the amount of products formed. If a reaction gives off gas,

 the change in volume or in mass can be measured. In this case, we will discuss about the change in volume or mass and change in color to measure the rate s of the reaction.

14.2.1. Measuring the rates of reaction by observing the volume changes
To measure the reaction rate by change in volume of gas produced is a convenient method if one of the products is a gas. A gas syringe can be used for this purpose (Figure 14.7).

Figure 14.7: The gas syringe method.

This method consists of a ground glass and a plunger, which moves outwards as the gas collects and is calibrated to record the volume directly. As more gas is produced, the plunger is pushed out and the volume of the gas in the syringe can be recorded. By measuring the volume at different time intervals, we can plot the data by following the change in volume against time and hence determine the rate of the reaction (Figure 14.8).

Figure 14.8: Graph showing the volume of gas collected against time.

Examples of reactions that produce gas

i. Reaction that produce hydrogen gas
When a metal reacts with an acid, hydrogen gas is produced. For example, magnesium reacts with

 sulphuric acid to produce magnesium suphate and hydrogen as follows.

The hydrogen can be collected in a test tube. A glowing splint can be used to test the presence of hydrogen. The ‘pop’ sound shows that hydrogen is present.

ii. Reaction that produce carbon dioxide
When a carbonate reacts with an acid, carbon dioxide gas is produced. When carbon dioxide is passed through limewater, it turns the limewater milky. A burning splint will also stop burning (be extinguished)

 in the presence of CO₂ gas. These are simple tests for the presence of carbon dioxide.

iii. Reaction that produce oxygen
Hydrogen peroxide decomposes in the presence of a manganese (IV) oxide as catalyst to produce oxygen and water.

14.2.2. Measuring the rate of reaction by observing change in mass

Many reactions involve a change in mass and it may be measured directly.

For a reaction that produces gas, the decrease in mass can be measured by standing the reaction mixture directly on a balance (Figure 14.9).The mass loss indicates the amount of gas that has been produced and escaped from the reaction vessel. The method allows continuous reading and a graph can be plotted directly of sample mass and mass lost against time (Figures 14.10 and 14.11), respectively. The release of carbon dioxide from the reaction between a carbonate and a diluted acid can be measured by this method and the rate is determined

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

To start the reaction, the flask is gently lent to one side, causing the card to fall and the marble chips and acid to mix. A piece of cotton wool placed in the neck of the flask allows carbon dioxide gas to escape. As the gas escapes the mass of the flask reduces within an interval of time. The rate of the reaction can be deduced as the change in mass over the change in time.

Figure 14.9. Measurement of the reaction rate following change in mass of reactants against time

Since the reaction rate is the change in the amount of a product or a reactant per unit time, any property

that is related to amount of product or reactant present can be used to measure the rate of reaction.

Figure 14.10. A graph of the sample mass versus time Figure 14.11. A graph of the mass loss versus time.

property that is related to amount of product or reactant present can be used to measure the rate of reaction

14.2.3. Measuring the rate of reaction by observing color change

Measuring the rate of reaction by observing color change can be applied if one of the reactants or products is colored and gives characteristic absorption in the visible region (about 320-380 nm of wavelength). Sometimes, an indicator can be added to generate a coloured substance that can be followed in the reaction. A colorimeter or spectrophotometer can be used to measure the intensity of the light transmitted by the reaction compounds. As the concentration of the coloured substances increases, it means that it absorbs more light and is less transmitted.

Amount of light absorbed depends on the amount of absorbing species present. If a reactant or product absorbs light, then the color intensity will vary as the amount of this species changes. An example is the reaction between sodium thiosulphate and hydrochloric acid.

Na₂S₂O₃ (aq) + HCl(g) →SO₂(g) + 2NaCl(aq) +H₂O(l) + S(s)

The solution changes its color from colorless to yellow due to sulphur precipitate which is formed, and the concentration of yellow precipitate increases with time as more sulphur is being formed indicating that the reaction has occurred.

14.3. Experimental determination of orders of reaction and rate laws

Activity 14.3

Using this textbook or internet or any other relevant book, read and analyze the content about experimental determination of orders of reactions and rate laws and make summary to be presented in the class. In this work you should also do the exercises of calculation.

• Rate law or rate equation
The rate of the reaction changes with time. The rate of reaction does not depend on the choice of substance, but it does depend on the way in which the chemical reaction is written. Therefore, a rate must be specified with a specific time unit (Mole/L. sec or M.s^-1.

The rate law or rate equation of a chemical reaction is a mathematical equation that links the reaction rate with concentration or pressure of reactants and a constant rate.

In a chemical reaction, the stoichiometric coefficients indicate the reacting mole ratio of the reactant and the number of mole of product that can be obtained.

Let us consider the following equation:

aA+ bB→ cC+ dD

In the reaction, a mole of A will combine with b moles of B to form c moles of C and d moles of D.

The rate law of this reaction can be expressed as:

Where:
k: rate constant

x and y are integers: small whole numbers (Usually 0,1 or 2) called orders of reaction

[A] and [B] are molar concentrations of A and B respectively.

The proportionality constant, k, is known as the rate constant and is specific for the reaction shown at a particular temperature. The rate constant changes with temperature and its units depend on the sum of the concentration term exponents in the rate law. The exponents (x and y) must be experimentally determined and do not necessarily correspond to the coefficients in the balanced chemical equation.

Note: When the concentration of A is changed by a factor of x, the rate will increase by a factor of x. A reaction’s rate law may be determined by the initial rates method.

• Reaction order
The sum of the concentration term exponents in a rate law equation is known as its reaction order.  We can also refer to the relationship for each reactant in terms of its exponent as an order. All the concentrations of all reactants taking part in a reaction cannot determine the order of reaction. The order of reaction is given by the number of atoms or molecules whose concentrations vary during the chemical change. In this part, we will be dealing with zero order, first order, and second order reaction.

   • For a zero order reaction any change in concentration of reactant does not affect the rate of reaction.
   • For a first order reaction, increasing the concentration of reactant two or three times also increase the

       rate two or three times (i.e. if the concentration of a reactant changes by a given factor,

       the rate must also change by the same factor) .

    • For a second order reaction, doubling or tripling the concentration of a reactant increases the rate four

      or nine times respectively (i.e. any change in concentration of a reactant by a factor x changes the

      rate by x² times.

Experimental determination of order of reaction
The order of the reaction can be determined using the initial rate method.
Let us consider the following reaction:

aA + bB → cC + dD

The rate law = k [A]ⁿ [B]^m

Where n and m are the order of reactants A and B respectively. The order n can be equal to a or not, and the order m can also be equal to b or not.

As the order of the reaction is determined by experiment, the initial rate method can be used to determine the order of the reaction which is the sum (n + m), n is the partial order with respect to the reactant A and m is the partial order with respect to the reactant B

In order to determine the order with respect to A, two experiments are carried out separately keeping the concentration of B constant in both cases while changing the concentration of A. The initial rate for both the experiments are then determined.

Any change in the initial rate can only be due to the change in the concentration of A and not B.

To determine the order with respect to B, the concentration of A is taken as constant while the concentration of B is changing.

Application examples:

Example 1: The following results were obtained for Q run between A and B

a. What is the order of reaction with respect to A and with respect to B.
b. What is the rate equation for the reaction?
c. Calculate the rate constant.

Solution
A+B →products

The rate of reaction =k[A]ⁿ[B]^m

Let us compare run (a) and run (b) where [A]is constant

The order of reaction with respect to B=2

The order of reaction with respect to A=1
Rate equation =k [A]ⁿ [B]^m Then
Rate equation =k [A]¹[B^]2
Rate =[A][B]²

Example 2: P, Q, and R reacted together to form products.

P +Q +R → Products

The table shows the results of the experiments carried out to investigate the kinetics of the reaction.

a. Deduce the order of reaction with respect to P, Q and R.
b. Write the rate equation
c. Calculate the rate constant k for the reaction and state the units.

Solution:
a. Let us write the rate as follows:

For finding the order of reaction with respect to P:

Comparing experiments (1) or (2) where [Q] and [R] are constant, we have,

Simplifying the equation we get

2 = 2x
x = 1

Simplifying the equation we get
4 = 2^y
2^y = 2²
y= 2

For finding the order of reaction with respect to R

Comparing experiments (3) or (4) where [P] and [R] are constant,, we have

Simplifying the equation we get

b) The rate equation,

c. The rate constant can be found as follows:

From the rate law

You can use experiment (1)

Checking up 14.3The table below shows the experimental data for the following reaction :

A+2B → C

a. Determine the order of reaction with respect to A and B.

b. Write the rate equation for the reaction

c. Calculate the rate constant for the reaction and give its units.

d. Calculate the rate of reaction when the concentrations of A and B are

       8.50 x10-³ and 3.83 x10-³ mol dm-³, respectively.

14.4. Relationship between reactant concentrations and time for zero, first and second order reaction

Activity 14.4

1. For hypothetical reaction: n A → B

Write the equation of the:

a. Average rate of reaction in terms of A

b. Rate law for the reaction

c. Compare the equation found in (a) and the equation found in (b)

d. Replace the exponent n by the following figure and in each case integrate from 0 to the figure.

i. 0

ii. 1

iii. 2

2. Using this textbook or internet or any other relevant book, read and analyze the content about relation between reactant concentrations and time for zero order reaction, first order reaction and second order reaction and make summary to be presented to the class. In this work you should do the exercises of calculation and sketch different graphs for order 0, 1, and 2.

Rate laws express the rate as a function of the reactant concentration. Rate laws

can also be converted into the equations that express the concentrations of the reactants or products at any given time during the course of a reaction.

14.4.1. Zero order reaction
In a zero –order reaction, the rate is independent of the concentration of the reactant.
Consider the reaction: A→ Products

The rate law is R = k [A]⁰

Let us consider [A]₀ as the initial concentration of A , [A]t as the concentration of A at time t and [A] as the concentration of A that has reacted. Therefore [A] = [A]₀ - [A]_t

The rate of the reaction is expressed as the change in concentration over time.

Integrating from the time t = 0 to time t = t, we have:

For a zero order reaction, the concentration-time graph is a straight line, showing a constant rate

(Figure 14. 12). The gradient of the line = k. The rate-concentration graph is a horizontal line (Figure 14.13).

Figure 14.12: Concentration-time graph for a zero order reaction

Figure 14.13: Rate-concentration graph for a zero order reaction

Examples of zero order reactions are:
The reaction of the iodination of propanone

CH₃COCH₃(aq) + I2 (aq) → CH₃COCH₂I(aq) +HI(aq)

The reaction is said to be zero order with respect to iodine.

The rate law equation can be given as follows: R = k [CH₃COCH₃]

1. The adsorption of gaseous reactants in gaseous reactions:

Sometimes, reactions between gases are zero order with respect to one of the reactants. This often indicates that this reactant has been adsorbed on the surface of the vessel. The rate of reaction then depends on the frequency with which molecules of the adsorbed gas collide with the inside of the vessel. This frequency is proportional to the concentration of the non-adsorbed reactant.

2. Reaction of sodium thiosulphate and HCl is zero order with respect to HCl

14.4.2. First order reaction
A first order reaction is a reaction whose rate depends on the concentration of a single reactant raised

 to the first power.
Consider the first order reaction A →products
The rate equation can be written as: R = k [A]

Integrating for time t=o to t=t, we have

Therefore for all first order reactions, half-life (t_1/2) is independent of initial concentration [A]₀ of the reactant.

For a first order reaction, the graph concentration-time (Figure 14.14) is a curve showing the decreasing

rate with concentration while the graph rate-concentration (Figure 14.15) is a straight line passing

 through the origin.

                                

Figure 14.15: Concentration-time graph for            Figure 14.16: Rate-concentration graph for a first order reaction a first order reaction.

14.4.3. Second order reaction
A second order reaction is a reaction involving two reacting species.
For the following reaction: A+B → Products,

Integrating from time, t = 0 to time, t = t:

For a second order reaction, the graph concentration-time (Figure 14.17) is curve while the graph rate-concentration is a parabola. Characteristic of the square function (Figure 14.18).

                                                  

Figure 14.17: Concentration-time graph for           Figure 14.18: Rate-concentration for a second order graph  for a second order reaction.                        reaction

14.4.4. Pseudo-first order reaction

It is known that the order of reaction depends on the dependency of the rate of reaction on the concentration of reactants. If the rate is independent of the concentrations of reactants, the order of reaction is zero.

A reaction which is not first-order reaction naturally but made first order by increasing or decreasing the concentration of one or the other reactant is known as Pseudo first order reaction. Pseudo-first order reaction is a reaction dependent upon the concentrations of both the reactants, but one of the components is present in large excess and thus its concentration hardly changes as the reaction proceeds.
Consider the reaction: A + B → Products

This reaction is dependent upon the concentrations of both A and B but if the component B is present in large excess and the concentration of B is very high compared to that of A, the reaction is considered to be pseudo-first order reaction with respect to A. similarly, if the component A is in large excess and the concentration of A is very high as compared to that of B, the reaction is considered to be pseudo first order with respect to B.

I. If [B] is in excess, the rate equation can be written as follows:   Rate   =k[A] [B]The concentration of B practically remains constant during the reaction, and therefore the rate law can

 be written:

Rate =k’[A]

Where the rate constant, k’= k[B]

The real order of that reaction is 2 but in practice it will be order 1.

Example:
Let us consider the elementary reaction where water is in excess:

CH₃CH₂COOCH₃ (l) + H₂O (l) + H+ (aq) → CH₃CH₂COOH(aq) +CH₃OH(aq)

The rate law for the reaction is: Rate = k’ [CH₃CH₂COOCH₃], where k’ = k [H₂O]

At low concentration of H₂O, water is first order. But when water is in excess, water does not affect the

 rate of reaction.

• Rate constant and its units

The rate of the reaction, as seen above, is the change in concentration over the change in time.

Therefore, the reaction rate is expressed in M^s-1 or mol dm^-3 s^-1. The unit of concentration is mol/dm³ and

that of time is seconds (s).

The units of the rate constant depend on the form of the rate law in which it appears.
Units of k for a zero order reaction

Note: Time can be expressed in seconds, minutes, days, hours, months, years, etc.

Units of k for a first order reaction

Units of k for second order reaction

Checking-up 14.4

1. The half–life of a first-order reaction is 2.5 min, calculate the time taken for the amount of reactant to 

    decrease to10% its original value.

2. Consider the following reaction: A + B  products

This is first order with respect to A and second order with respect to B.

a. Write an expression of the reaction.

b. Draw and label graphs that would allow the rate constant to be determined from a series

of experiments in which

i. [A] is kept constant, but [B] is varied.

ii. [B] is kept constant, but [A] is varied.

14.5. Difference between order of reaction and molecularity

Activity 14.5

Using this textbook, internet or any other relevant book, read and analyze the content about order of reaction and molecularity and make summary to be pre-sented to the class.

 In this work you should also do give examples of reactions for each.

Molecularity is the number of reacting species (e.g: molecules, ions) that participate (take part) simultaneously in the formation of the transition state. That is, the number of species involved in the rate determining step for the reactions occurring in stages.

It can be the number of species that participate as reactants simultaneously in an elementary reaction or elementary processes (reaction occurring in single event or step).

If a single molecule is involved simultaneously in an elementary reaction, the reaction is unimolecular.

For example: N₂O₅→NO₃ + NO₂, One reactant molecule is decomposed into two product molecules.

If two molecules are involved simultaneously in an elementary reaction, the reaction is bimolecular.

Examples:
i. NO (g) + O₃ (g) → NO₂ (g) + O₂ (g)
ii. Mechanisms for the reactions of primary alkylhalides (1⁰ R-X ) with nucleophiles, S_N2, involve two molecules in the rate determining step.
If three molecules are involved simultaneously in an elementary reaction, the reaction is termolecular.

Termolecular reactions are less probable than unimolecular or bimolecular reactions and are rarely encountered because the chance that three particles collide at the same time, with proper orientation

  and sufficient energy, are considered extremely small.

An example of termolecular reaction is thought to occur during the formation of ozone from oxygen in

 the outer atmosphere:

                        2 O₂ + N₂ → O₃ + O. + N₂*

The rate laws for most reactions have the general form:

Rate =k[reactant A]^x[reactant B]^y…..
The exponents x and y in a rate law equation are called reaction orders.
For the reaction aA + bB+ cC → Product…….

The order of a reaction is the sum of powers to which the concentration terms raised in the rate equation

Rate =k[A]^x [B]y[C]^z

For the reaction rate; x, y, z are the powers of the considered concentration of A, B and C respectively with respect to each reactant.
x is the order of the reaction with respect to A
y is the order of reaction with respect to B
z is the order of reaction with respect to C

The sum of x, y and z gives the overall order of reaction.
x, y, z are obtained through experiments only and not from the number of moles as written in the stoichiometric equation.
The difference between the order and molecularity of reaction are shown in the Table 14.1.
Table 14.1. Differences between order of reaction and molecularity

Checking-up 14.5

It has been proposed that the conversion of ozone into O₂ proceeds by a two –step mechanism:

O₃(g) → O₂ (g)+ O. (g)

O₃(g) + O.(g) → 2O2(g)

a. Describe the molecularity of each elementary reaction in this mechanism.

b. Write the equation for the overall reaction

c. Identify the intermediates

14.6. Reaction mechanisms and kinetics

Activity 14.6

1. a. Five people have to run 100 m. Two of them do not perform well in running (they are slow in running). You are asked to organize them in order to perform that exercise so that the people leave the same place and reach the same destination at the same time.

b. Differentiate between SN₁ and SN₂ reactions. Show also the mechanism of reaction.

2. Using this textbook or internet or any other relevant book, read and analyse the content about the mechanism of reaction and make summary to be presented to the class

Reaction rates provide information regarding how a chemical process occurs as well as the mechanism by which a reaction happens at molecular level. Most chemical reactions occur via a series of steps called the reaction mechanism. The individual steps, called elementary steps, cannot be observed directly. A reaction mechanism consists of a set of proposed elementary steps involving molecular species – reactants as well as reaction intermediates. A reaction mechanism explains how a given reaction might take place and from which a rate law can be derived, which must agree with the one determined experimentally.

If the mechanism consists of more than one elementary step, the sum of these steps must be equal to the overall balanced equation for the reaction.

A reaction mechanism describes in great detail the order in which bonds are broken and formed and the changes in relative positions of the atoms in the course of the reaction.

The Table 14.2 summarizes the types of elementary steps and the rate laws that they follow. A and B represent the reactants or reaction intermediates. Typically these steps are usually either unimolecular or bimolecular.

Table 14.2. Elementary steps and rate laws

Example: N₂ (g) +3H₂ (g) →2NH₃ (g)
Molecularity of reaction is equal to 1+3 = 4 but the order can be derived from the rate determining step.

The reaction mechanism is the step-by-step process by which reactants actually become products.

In the reaction mechanisms, some steps are fast while others are slow.

A reaction cannot proceed faster than the rate of the slowest elementary step, the slowest step in a mechanism establishes the rate of the overall reaction which is known as the rate determining step.

When a proposed mechanism consists of more than one elementary step, the one with the slowest rate

 will determine the overall rate of reaction.

Examples:
1. Consider the reaction: NO_2(g) + CO_(g) → NO_(g) + CO_2(g)
    This reaction is believed to take place in two steps.

Step-1: NO₂ + NO2 → NO₃ + NO; (slow rate-determining)
Step-2: NO₃ + CO → CO₂ + NO; (fast)
The overall chemical equation is obtained by adding the two steps and canceling any common species

 to both sides.
The overall reaction is given by:

If the reaction follows a one elementary step mechanism, the rate law would be:
Rate = k[NO₂][CO]However, the experimentally determined rate law is:
Rate = k[NO₂]²

The rate law for the rate-determining step: Rate = k₁[NO₂]², which is identical in form to the rate law obtained experimentally. The second step, which occurs very fast, does not influence the overall rate.

An intermediate substance is neither a reactant nor a product in the overall reaction. It is formed in one step and consumed in the next step. In the above example, the intermediate is NO₃ because it is produced in the first step and consumed in the second one.

2. Consider the reaction: 2NO₂(g) + F₂(g) → 2NO₂F(g)
The experimental rate law is: Rate = k [NO₂][F₂]

The reaction is first order with respect to each reactant.

This reaction occurs follow two steps (slow and fast):

F is an intermediate species, it is produced in the first step and consumed in the next step.
Since the overall rate of the reaction is determined by the slow step, it seems logical that the observed rate law is Rate = k₁ [NO₂][F₂].

SN₁ and SN₂ mechanisms and kinetics

Chlorine, bromine, and iodine (X) possess a higher electronegativity than carbon.

As a result, the bonding electrons in C-X bond are unevenly distributed. The carbon atom is partially positively charged (δ+) while the halogen atom is partially negatively charged (δ-).

The polarity of carbon-halogen bonds forms the basis of two frequently found reaction types of this compound family, namely nucleophilic substitutions (S_N) reactions and elimination reactions.

In a nucleophilic substance, or electron rich species, nucleophile, selectively attack a positive or partially positive charge of an atom or group of an atom or group of atoms to replace the leaving group.

Nucleophilic substitutions (S_N) reactions can be of:

Second-order reactions: Occur with primary alkyl halides.
 
First-order reactions: Occur with tertiary alkyl halides.

Secondary alkyl halides can undergo second-order reactions or first-order reactions.

In both kinetic cases of substitutions, a leaving group (halide ion) is substituted by a nucleophile.

1. Second-order nucleophilic substitution (SN² reactions)
A nucleophile attacks a positively polarized carbon atom. The attack of the nucleophile results in the heterolytic cleavage of the carbon-ligand bond, where the bonding electron pair is completely passed

onto the ligand (X). The substrate, along with the nucleophile, participates in the rate-determining step. Thus, the reaction rate depends on both the substrate and the nucleophile’s concentration. Therefore,

 this reaction type is called bimolecular nucleophilic substitution reaction (SN²).

Example:
CH₃CH₂Br + HO- → CH₃CH₂OH +Br –

The rate of hydrolysis of a primary alkyl halide is proportional to the concentration of both

  halogenoalkane and hydroxide ions.

This reaction is second order and its rate expression is, Rate = k[RX] [OH-]

2. First-order nucleophilic substitution (SN₁ reactions)

The carbon-halogen bond is cleaved and involves the halogenoalkane alone, forming a halide ion

and a carbocation. The attack of the nucleophile on the carbocation yields the substitution product.

Thus, the reaction rate depends only on the halogenoalkane concentration.

Therefore, this reaction type is called unimolecular nucleophilic substitution reaction (SN¹ reaction).

Example:
Let us consider a hydrolysis reaction of a tertiary halogenoalkane:

CH₃)₃CBr + NaOH → CH₃)₃COH + NaBr. The experimental results show that the reaction is first order

  with respect to the halogenoalkane:

Rate=k[CH₃)₃CBr]
[1]

The rate of hydrolysis of a tertiary alkyl halide is proportional to the concentration of halogenoalkane

 but does not depend on hydroxide ions. The reaction is 1^st order with respect to the halogenoalkane.

Rate =k[CH₃)₃CBr]

Checking-up 14.6

1. Differentiate between SN¹ and SN² reactions.

2. The decomposition of nitrous oxide, N₂O ,is believed to occur by two –step mechanism:

N₂O (g) +N₂(g) + O. (g) (slow)N₂O(g) +O (g) → N₂(g)+O₂(g) (fast)

a. Write the equation for the overall reaction.

b. Write the rate law for the overall reaction.

END UNIT ASSESSMENT

1. Collision theory states that particles must collide with each other in correct orientation and

    sufficient energy to

       A. attract

       B. repel

       C. react

       D. respond

2. According to Collision theory, particles must

A. collide every where

B. collide with correct orientation

C. must be kept under immense pressures

D. shall not be below their melting points

3. Some particles collide but bounce back afterwards it is called

A. successful collision

B. unsuccessful collision

C. successful reaction

D. unsuccessful reaction

4. The rate law for a reaction is k [A][B]2Which one of the following statements is false?

A. The reaction is first order in A.

B. The reaction is second order in B.

C. The reaction is second order overall.

D. k is the reaction rate constant

5. A reaction was found to be zero order in A.

Increasing the concentration of A by a factor of 3 will cause the reaction rate to

A. remain constant

B. increase by a factor of 27

C. increase by a factor of 9

D. triple

6. The rate law of the overall reaction is k[A][B]0.

Which of the following will not increase the rate of the reaction?

A. increasing the concentration of reactant A

B. increasing the concentration of reactant B

C. increasing the temperature of the reaction

D. adding a catalyst for the reaction

7. When a lit match is touched to the wick of a candle, the candle begins to burn.

When the match is removed, the candle continues to burn. The match

A. behaves as a catalyst

B. supplies the activation energy

C. is part of the rate determining step

D. lowers the activation energy barrier

8. Which step of a reaction is the rate-determining step?

A. the fastest step

B. the last step of the reaction mechanism

C. the first step

D. the slowest step

9. Consider the following mechanism:

Step 1: Cl. + O₃ → ClO. + O₂

Step 2: O. + ClO→ Cl + O₂

The reaction intermediate is

A. Cl

B. O₂

C. O₃

D. ClO.

10. The table below shows the rates of reaction between substance A and B at different concentrations

a. Determine the order with respect to A and B.

b. Write the rate expression or rate law.

c. Calculate the rate constant indicating clearly its units

11. a. A reaction between two reactants, P and Q, gave the rate-concentration graphs shown below:

b. The rate equation for a reaction between R and S is Rate = K[R]2[S]

i. Determine the order with respect to R and S.

ii. Sketch the rate-concentration graphs for reactants R and S.

12. Consider the following reaction: A + B P + R

i. Determine the rate equation and the rate constant.

ii. Find the value of Z.

13. In the reaction: P + Q + R → Product (s), deduce the order of reaction with respect to P, Q and R if:

a. The rate does not change when [R] is doubled.

b. The rate increases by 9 when [Q] is tripled

c. The rate doubles when [P] is doubled

14. In the reaction: A + B → C, Rate = [A][B]2. State how the rate will change if:

a. The concentration of A is halved.also halved

b. The concentration of A is tripled and that of B is doubled.

15. In the reaction: 2NO + Br₂ →2NOBr, when the concentration of NO is reduced from 4x10^-2M to 2x10^-2M the rate falls by a factor of 4. What is the order of reaction with respect to NO? Give a reason to your answer.

16. In the reaction: 2NO₂→ 2N₂ + O₂ , when the concentration of NO₂ decreases from 5x10^-3M to 2x10^-3M, the rate of decomposition falls by a factor of 2.5. What is the order of reaction? Give a reason to your answer.

17. The reaction between 2-bromo–2-methylpropane (CH3)3C–Br and aqueous hydroxide ions OH– takes place by the following outline mechanism:

Step 1 (CH₃)₃C–Br (CH₃)₃C+ + Br– (slow)

Step 2 (CH₃)₃C+ + OH– (CH₃)₃C–OH (fast)

a. What is the order of reaction with respect to (CH₃)₃C–Br?

b. What is the total order of reaction?

c. Write the rate equation for this reaction.

18. a. Paracetamol has a biological half-life of 380 seconds.

How long will it take for the level of paracetamol in the body to fall to one-sixteenth of its origi-nal value?

b. 50% of a first order reaction is complete in in 23 minutes.

Calculate the time required to complete 90% of the reaction.