Chapter 6 – Chemical Kinetics Chemical kinetics – the study of controlling reaction rates Rate of reaction – the speed at which a chemical change occurs, generally expressed as change in concentration per unit time

Can measure the rate at which the product is formed, or the rate at which a reactant is consumed

Units are mol/L•s or molL-1s-1

Example: CH4(g)+ Cl2(g) CH3Cl(g) + HCl(g)

The concentration of the reactant Cl2(g) would fall during the progress of the reaction (see graph below)

Average rate of reaction – the change in concentration of a reactant or product over time; is the absolute value of the slope of the secant (a line between two points on a curve)

Instantaneous rate of reaction – the speed at which a reaction is proceeding at a particular point in time; is the slope of the curve at that point (the slope of the tangent to a curve at a particular point)

Textbook pg. 358 - 417

Evidence shows that for most reactions, the concentration changes are more rapid near the beginning of the reaction and the rate decreases with the time elapsed

o From the graph, at point B the slope is less steep – an indication that the reaction is slower

Plotting the data for the concentration of a product will result in a rising curve with a steadily decreasing positive slope

The rate of reaction for the methane-chlorine system above could be represented by any of the following expressions:

Note – the negative (-) sign indicated a rate of consumption of reactant, and a positive (+) sign indicated a rate of production of products

The rate of reaction in terms of the reactants and products is the same for this chemical reaction because the mole ratios are all 1 to 1

Comparing Rates of Consumption and Production

Must use a balanced chemical equation

Concentration of the reactant may not change in a 1:1 ratio with the concentration of the product

Ex. Given the reaction: 2A + 3B D + 4C

o If B is being consumed at a rate of 6.0 x 10-4 mol/L•s, then A is being consumed at a rate of:

6.0 x 10-4 mol B/L•s x

= 4.0 x 10-4 mol A/L•s

o Similarly, D is produced at a rate of:

6.0 X 10-4 mol B/L•s x

= 2.0 x 10-4 mol D/L•s

o And C is produced at a rate of:

6.0 x 10-4 mol B/L•s x

= 8.0 x 10-4 mol C/L•s

Sample Problem Consider the reaction of iodate, iodide, and hydrogen ions to yield iodine and water. IO3

-(aq) + 5 I-

(aq) + 6 H+(aq) 3 I2(aq) + 3 H2O(l)

What are the rates of reaction with respect to the various reactants and products? The rate of reaction with respect to iodate ions (rate of consumption of IO3

-(aq)) is determined experimentally to be

3.0 x 10-5 mol/(L•s). Solution Rate of iodate ions: [IO3

-(aq)] = 3.0 x 10-5 mol/(L•s)

Rate of iodide ions: [I-

(aq)] = Rate of hydrogen ions: [H+

(aq)] = Rate of iodide: [I2(aq)] = Rate of water: [H2O(l)] = Measuring Reaction Rates

Different methods can be used to measure reaction rates depending on the type of reaction

Reactions that produce a gas - collect the gas and measure its volume and/or pressure

Reactions that involve ions - the conductivity of the solution changes as the reaction proceeds and can be measured and plotted as a function of time

o Ex. (CH3)3CCl(aq) + H2O(l) (CH3)3COH(aq) + H+(aq) + Cl-(aq)

Reactions that change colour - the intensity (strength of colour) can be measured of a coloured reactant or product using a spectrophotometer

o Can measure wavelengths of light that are invisible to the human eye

o Ex. ClO-(aq) + I-

(aq) IO-(aq) + Cl-(aq)

colourless colourless yellow colourless

In this example, the yellow colour of the aqueous hypoiodite ions (IO-(aq)) appears initially, then

becomes more intense as the reaction progresses and more ions form

Factors Affecting Rates of Reaction There are 5 factors that affect the rate of a reaction:

1. Nature of the reactants

Different elements react at different rates (ex. Au and Ag react slowly in air while Na and K react so quickly in air that they are rarely found naturally in their elemental state)

Elements in the same group react similarly but at different rates (ex. Zn, Fe and Pb all react with HCl acid to produce H2 gas, but even when all the other conditions are the same, the rates of reaction are different)

This is demonstrated in the activity series

Most reactions with monatomic ions (Ex. Ag+ and Cl-) are extremely fast, while reactions of molecular substances are often slower (ex. glucose, C6H12O6) as geometry becomes a factor

2. Concentration

The more concentrated, the faster the reaction

Experiments suggest that if the initial concentration of a reactant is increased, then the reaction rate generally increases

Ex. when one adds a metal to acid, a higher initial concentration of acid increase the rate of gas production

Implies “concentrated is better and faster” 3. Temperature

When the temperature of a system increases, the molecules move faster and the reaction rate increases (the opposite is true when the temperature is cooler)

An increase of 10ºC causes the reaction rate to double or triple

Ex. this is seen in changes that occur when food is cooked 4. Catalysts

Catalysis is the effect of a catalyst

Catalyst – a substance that alters the rate of a chemical reaction without being permanently changed itself

Enzymes – a molecular substance (protein) found in living cells that controls the rate of a specific biochemical reaction

Ex. lactase is an enzyme that facilitates the breakdown of the sugar lactose in our bodies

Enzymes are often very sensitive to temperature and pH and therefore conditions are critical for their use

5. Surface Area

When a reaction system is heterogeneous (solid in liquid, etc), a reaction occurs when the two different phases make contact, so the amount of exposed surface area where contact can be made affects the reaction rate

Higher surface area provides a higher area of contact, so the faster the rate of reaction

Ex. powdered sugar dissolves more quickly in water than do sugar cubes

In general, the reaction rate increases proportionally with the increase in surface area

Rate Laws and Order of Reaction

Experimental evidence has shown that the rate of a reaction is exponentially proportional to the product of the initial concentration of the reactants (other factors held constant)

Consider the reaction: aX + bY products The Rate Law – the rate, r, will always be proportional to the product of the initial concentrations of the reactants, where these concentrations are raised to some exponential values r α [X]m [Y]n Where: r = rate of reaction

m and n are some exponents determined from experimental data (any real number) – NOT related to the coefficients (a and b) in the chemical equation

Using proportionality techniques, one can form the mathematical equation: Rate Law Equation: r = k [X]m [Y]n Where: k is the rate constant or proportionality constant that is determined

experimentally and valid only for the temperature specified

Ex. The empirically determined rate law equation for the reaction between nitrogen dioxide and fluorine gas:

2 NO2 + F2 2 NO2F is r = k[NO2]

1[F2]1

In this reaction, the rate depends equally on the initial concentrations of each reactant (m = 1 and n = 1), despite the fact that their reaction coefficients are different.

The exponents are also called the individual orders of reaction that show the dependence of the rate on the initial concentration of a particular reactant

o Ex. If m = 1, the order of reaction is 1 with respect to X If n = 2, the order of reaction is 2 with respect to Y

The rate constant will assume units based on the orders of reaction to yield units for the rate in mol/(L•s)

The overall order of the reaction is the sum of the individual orders of reaction for each reactant o Ex. For the above example (m = 1 and n = 2) the overall order of reaction is (1 + 2) = 3

The Effect of an Order of Reaction

To understand the effect of an order of reaction, consider the following chemical equation: 2X + 2Y + 3Z (products) Experimental evidence provides the following rate law equation: r = k [X]1[Y]2[Z]0 The orders of reaction for each reactant can be interpreted by looking at the proportionality relationship between rate and each reactant

r α [X]1 if [X] is doubled, then the rate will double (multiply by 2 or 21) if [X] is tripled, then the rate will triple (multiply by 3 or 31)

r α [Y]2 if [Y] is doubled, then the rate is multiplied by 4 (or 22) if [Y] is tripled, then the rate is multiplied by 9 (32)

r α [Z]0 if [Z] is doubled, then the rate is unchanged (multiplied by 1 or 20) if [Z] is tripled, then the rate is unchanged (multiplied by 1 or 30) The overall order of the reaction is (1 + 2 + 0) = 3 The rate law equation can be simplified to: r = k [X]1[Y]2 as [Z] has NO effect

Finding the Rate Equation from Experimental Data Ex. When aqueous bromate and bisulfide ions react to produce bromine, the overall equation is 2 BrO3

-(aq) + 5 HSO3

-(aq) Br2(g) + 5 SO4

2-(aq) + H2O(l) + 3 H+

(aq) Consider the series of experiments recorded in Table 2, in which initial reactant concentrations are varied and rates are compared. From the evidence provided, determine a rate equation.

Solution The reaction rate, r, is proportional to the initial concentrations of bromate ions and of bisulfate ions.

1. Write the general formula for the reaction rate: r = k [BrO3

-(aq)]

m[HSO3-(aq)]

n where k, m, n are to be determined.

The key to solving this problem is to look for pairs of data in which the initial concentration of only one reactant changes (and the other remains the same). Then, compare the change in concentration of that reactant, to the change in rate that occurred to find the relationship between concentration change and reaction rate.

2. To find m, compare Trial 1 vs. Trial 2 (initial concentration of BrO3-(aq) changed while the

concentration of HSO3-(aq) remained the same)

[BrO3

-(aq)] change =

Rate change = Order of reaction with respect to BrO3

-(aq) (m) =

3. To find n, compare Trial 2 vs. Trial 3 (initial concentration of HSO3-(aq) changed while the

concentration of BrO3-(aq) remained the same)

[HSO3

-(aq)] change =

Rate change = Order of reaction with respect to HSO3

-(aq) (n) =

4. To find the rate constant, k, enter the values from Trial 1 (or any trial) into the rate equation, with the concentrations expressed in mol/L.

r = k[BrO3

-(aq)]

1[HSO3-(aq)]

2 The reaction is third-order overall - first order with respect to bromate ion and second order with respect to bisulfite ion. The rate equation is:

r = k[BrO3-(aq)]

1[HSO3-(aq)]

2 where k = 1.1 x 104 L2/(mol2•s)

Relating Reaction Rate to Time

Studies have shown that the rate of reaction is inversely proportional to the elapsed time of a reaction:

rav α

Therefore, if the rate of a reaction in which some reactant A is consumed is rav α [A]n Then

α [A]n

How will the graph of

vs [A] appear based on the order of reaction?

Chemical Kinetics and Half-Life

The rate of reaction for many reactions is a first order reaction o Ex. decomposition reactions or nuclear decay

In a first order reaction, the quantity of reactant remaining in a sample follows a predictable pattern

Many times we refer to the half-life of the reaction, t1/2, which is the time required for half the sample to react

Ex. Consider the reaction: nA (products) The rate law equation for a first order reaction will be:

rate =

= k[A]1

We can rewrite the equation as:

= kΔt

Using integral calculus, we can derive the following equation:

ln

= -kt

Where: [A]t = concentration (or mass) of the reactant at any time t [A]0 = initial concentration (or mass) of the reactant [A]t/[A]0 = fraction of reactant remaining at time t ln = natural logarithm (ln e = 1) To calculate the half-life of the reactant A, we set the ratio at ½ ln ½ = -kt1/2 or ln 2 = kt1/2 Working with natural logarithms, one would find: kt1/2 = 0.693

Note – This equation only relates to first order reactions. Sample Problem A radioisotope has a half-life of 24 a (years) and an initial mass of 0.084g. a) What mass of radioisotope will remain after 72 a? b) Approximately how many years will have passed if only 10% of the radioisotope remains?

Collision Theory and Rates of Reaction

The Collision Theory was developed to understand how reactions occur and why they occur at different rates

Molecules are held together by chemical bonds and, according to collision theory, chemical reactions can occur only if energy is provided to break those bonds

o The source of that energy is the kinetic energy of the molecules Concepts of the Collision Theory

A chemical system consists of particles (atoms, ions, or molecules) that are in constant random motion at various speeds

The average kinetic energy of the particles is proportional to the temperature

A chemical reaction must involve collision of particles with each other or the walls of the container

An effective collision is one that has sufficient energy and correct orientation of the colliding particles so that bonds can be broken and new bonds formed

Ineffective collisions involve particles that rebound from the collision, essentially unchanged

The rate of a given reaction depends on the frequency of collisions and the fraction of those collisions that are effective Rate = frequency of collisions x fraction of collisions that are effective

Increasing either factor will yield an increase in the rate Transition State Theory

The transition state of a reaction is when molecules collide and break apart

At this point, an activated complex is formed consisting of an unstable grouping of reactant molecules with bonds in the process of being broken and bonds in the process of being formed

o It is unstable because it possesses the maximum potential energy possible

A potential energy diagram can be used to show energy change as a reaction proceeds

The reaction passes through a transition state where an activated complex is formed o This can become the products, or an intermediate product if the reaction occurs in

more than one step

The difference between the initial energy of the reactants and the transition state is the activation energy (Ea) – the minimum energy with which the molecules must collide before they can rearrange in structure, resulting in an effective collision

Exothermic Reaction forward reaction NO(g) + O2(g) → NO2(g) + O2(g) ∆Hº = -200kJ/mol Ea = +10kJ/mol

Endothermic Reaction reverse reaction NO2(g) + O2(g) → NO(g) + O2(g) ∆Hº = +200kJ/mol Ea = +210kJ/mol

Exothermic collision – the molecules of the product will have a lower potential energy than the reactants and therefore a higher kinetic energy

o When these products collide with other molecules in the system and surroundings they will increase the speed of the molecules they collide with, resulting in an increase in the temperature of the system

Endothermic collisions – the molecules of the product will have a higher potential energy than the reactants and therefore a lower kinetic energy

o When the products subsequently collide with other molecules in the system and in the surroundings, they will decrease the speed of the molecules they collide with, resulting in a drop in the temperature of the system

Reaction Mechanisms

Collision theory suggests that collisions of three particles at the same time must be less frequent than two particle collisions

o Four particle collisions would be even more rare

Reactions with a single reactant are a special case (ex. Cl2 + light 2Cl•) o A single molecule can hit a container wall or other molecules such that enough energy is

converted from kinetic to potential energy for the molecule to decompose

It is believed that most chemical reactions occur as a sequence of elementary steps

Elementary step – a step in a reaction mechanism that only involves one-, two-, or three-particle collisions

Reaction mechanism – a series of elementary steps that makes up an overall reaction

Ex. Consider the oxidation of hydrogen bromide (at 400ºC to 600ºC) 4 HBr(g) + O2(g) → 2H2O(g) + 2Br2(g)

The rate law equation has been proven experimentally as: r = k[HBr][O2]

Although the number of molecules of HBr involved in the reaction is four times that of O2, it has an equal influence on the rate of reaction. This has been explained by theorizing that the reaction occurs in the following elementary steps:

HBr(g) + O2 → HOOBr(g) (slow) HOOBr(g) + HBr(g) → 2HOBr(g) (fast) 2{HOBr(g) + HBr(g) →H2O(g) + Br2(g)} (fast) ____________________________

4 HBr(g) +O2(g) → 2H20(g) +2Br2

The potential energy diagram appears as:

The theoretical explanation is that the first elementary step is relatively slow because it has a the highest activation energy of all three steps

o This is shown in the energy level diagram (first peak) o The rate of the overall reaction is basically controlled by this step

The slowest reaction step in any reaction mechanism is called the rate-determining step

The chemicals HOOBr and HOBr are not present when the reaction is complete and so are called reaction intermediates (short-lived products in a reaction mechanism)

Unstable activated complexes occur at the peaks of the potential energy diagram, while slightly more stable reaction intermediates occur in the valleys

Experimental determination of the rate law equation is closely linked to the slowest step in the reaction mechanism The rate law equation was: r = k[HBr][O2]

Both reactants are found in the slowest step – they are not found together in any other reaction step

The exponents of the reactants in the rate law equation are also related to the coefficients of the reactants in the slowest step of the reaction mechanism

o Ie. The exponents are both one (1) because the coefficients are one (1) for each of the reactants in the slowest step

In general: If the rate law equation is found to be: r = k[molecule X]n[molecule Y]m then, the rate-determining step can be identified as: nX + mY products or reaction intermediates Sample Problem Consider the overall reaction involving three elements as reactants, and a compound as the product: X + 2Y + 2Z XY2Z2 When a series of reactions is performed with different initial concentrations of reactant, the results are as follows: Doubling the concentration of X has no effect on the overall rate Doubling the concentration of Y multiplies the overall rate by 4 Doubling the concentration of Z doubles the overall rate State a) The rate law equation for the system b) The rate-determining step c) A possible mechanism, indicating the slow step. d) A possible reaction intermediate in your mechanism. Solution a) From the empirical information provided, r α [X]0, r α [Y]2, r α [Z]1, giving the rate equation of

r = k[Y]2[Z]1 b) Therefore, the rate-determining step must be 2Y + Z some product(s) c) Any mechanism consistent with the above rules is acceptable. For example, one possibility is X + Z XZ (fast) 2Y + Z Y2Z (slow) XZ + Y2Z XY2Z2 (fast) X + 2Y + 2Z XY2Z2 d) Two possible reaction intermediates are XZ and Y2Z

Factors Affecting the Frequency of Collisions and Fraction of Effective Collisions 1. Chemical Nature of Reactant

Must consider the atomic structure of reactants and the nature of their bonds, as well as the type of reaction occurring

Molecules in a substance are not all moving at the same speed – the distribution of kinetic energies is called a Maxwell-Boltzmann distribution and has been experimentally found to fit the pattern below:

In a chemical reaction, an effective collision requires a minimum energy – energy of collision – which is converted into potential energy (the activation energy) as the activated complex is formed

o This minimum energy is called the threshold energy

The chemical nature of reactants affects the activation energy in two ways: o Some molecules have bonds that are relatively weak and have small activation energy

barriers, so the threshold energy (activation energy) is relatively low and a large fraction of molecules are capable of colliding effectively. Other molecules have strong bonds and high activation energy barriers resulting in ineffective collisions.

o Collision geometry – some reactions involve complicated molecular substances or complex ions that are often less reactive because more bonds have to be broken and the collision must occur in the correct orientation

2. Concentration

A higher concentration means a larger number of particles per unit volume, which are more likely to collide

Ex. If there are twice as many particles, then there is twice the probability of an effective collision

3. Surface Area

Only applies to heterogeneous reactions – gas with a solid or solid with a liquid

Surface area affects collision frequency because reactant can collide only at the surface where the reactants are in contact

Ex. The number of particles per square millimeter of surface of a solid is fixed, but if this solid is finely divided there will be more particles exposed to the surface

4. Temperature

Higher molecule speed increases the number of collisions – as the average kinetic energy of molecules increases, more collisions will have enough energy to overcome the activation energy

Consider the distribution of molecular energy at two different temperatures:

If temperature T2 is greater than temperature T1, then the distribution of molecules with kinetic energy greater than the activation energy, Ea, increases

o The number of collisions with sufficient energy will increase, resulting in an increase in the rate of reaction

An increase in temperature results in an increase in both the collision frequency and the fraction of collisions that are effective

Increasing temperature causes molecules to collide both more often and with more force

With respect to Ea, a much larger fraction of molecules have the required kinetic energy at a higher temperature than at a lower temperature

5. Catalyst

A catalyst is a substance that increases the rate of reaction without being consumed itself

A catalyst decreases the activation energy of a reaction and also provides additional (alternate) reaction steps

o A catalyst allows the reaction to occur by a different mechanism, inserting different intermediate steps, but resulting in the same products overall

If the new pathway has a lower activation energy, a greater fraction of molecules possesses the minimum required energy and the reaction rate increases

Types of Catalysts o Homogeneous catalyst – catalyst is in the same physical state as the reactants

NO2(g)

2 SO2(g) + O2(g) SO3(g)

o Heterogeneous catalyst – catalyst is in a different state than the reactants

Pt(s) H2(g) + ½ O2(g) H2O(l)

Also available are inhibitors to slow down the rate of reaction