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Medicinal Chemistry - INTRODUCTION TO KINETICS lecture notes supplement (2010) Lecturer: Dr. Cheang Khoo A. Introduction Chemical kinetics is the study of the rate (or speed) of chemical reactions and the factors influencing the rate. The rate of a chemical reaction is important for several reasons. For example, if an explosive reaction is required then a reaction that produces a great deal of energy in a very short time (µs→ms) is required. Sometimes a slightly slower reaction rate is required – for example when rocket fuel is burned to produce gases that propel the rocket, the reaction needs to be fast but an explosive reaction is not desired. A slow reaction is desired if the aim is to prevent reactions like rusting (an oxidation reaction) or food spoilage. The rate of a chemical reaction may be described by the rate of the reactant(s) disappearing or the rate of the product(s) (of the reaction) appearing. The rate of reactant disappearance or product appearance will depend on the reaction stiochiometry (or reaction ratios) as shown in the given example: Example 1: For the general reaction: 2A + B → 3C + D where A and B are the reactants and C and D are the products of the reaction and the numbers indicate the reaction stiochiometry. The rate expression (do not confuse with rate law) is given by the expression: Rate = - 1 d[A] = - d[B] = + 1 d[C] = + d[D] -------------- (i) 2 dt dt 3 dt dt where d[A] = change in concentration of A ; with similar definitions for B, C, D. time interval for change A and B are reactants and are therefore decreasing in concentration with time, hence the (-)ve sign while C and D are products and therefore increasing in concentration, hence the (+)ve sign. Since 2 A’s react with 1 B, for every decrease of 1 unit of B, 2 units of A will decrease. Thus if we want to write a mathematical expression equating the rate of change of A and B, we have to multiply the rate of change of A by ½ , as shown in equation (i), or multiply the rate of change of B by 2. Comparing the change in concentration of products and reactants over time, the following plot may be observed:

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The curves are symmetrical about the horizontal intersection point if their reaction stoichiometries (and therefore rate coefficients) are equal (as is the case for B and D). Exercise questions: Exercise A1: For a general reaction 2A + B → 3C, express the rate of the reaction in the differential form with respect to each of the species involved. (Answer: -1/2 d[A]/dt = -d[B]/dt = 1/3 d[C]/dt ) Exercise A2: For the reaction A + 3B → 2C + 2D, write the rate expression. (Answer: -d[A]/dt = -1/3 d[B]/dt = ½ d[C]/dt = ½ d[D]/dt) Exercise A3: For the reaction A + ½B → 3C + ¼D, write the rate expression. (Answer: -d[A]/dt = -2 d[B]/dt = 1/3 d[C]/dt = 4 d[D]/dt) Exercise A4: Write a stoichiometric equation for the general reaction whose rate is expressed as: Rate = -d[A] = -½ d[B] = -2 d[C] = d[D] = 2 d[E] dt dt dt dt dt (Answer: A + 2B + ½ C → D + ½ E) Exercise A5: Write a stoichiometric equation for the general reaction whose rate is expressed as: Rate = -d[A] = -¼ d[B] = ½ d[C] = d[D] = 3 d[E] dt dt dt dt dt (Answer: A + 4B → 2C + D + 1/3 E) Exercise A6: The following reaction occurs consecutively:

Concentration of reactant B

Concentration of product D

Time

Concentration Axis of symmetry when stoichiometries of B & D are the same.

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k1 k2 A → B → C where the reaction rate constants k1 = k2 (that is, the reaction rates are the same) Draw a graph for the time variation of the concentrations of A, B and C (concentration y-axis, time x-axis). B. Rate Law and Reaction Order 1. Definition and introduction For the general reaction: aA + bB + cC + ……… → Product(s), where A, B, C…. are the reactants and a, b, c are the reaction stoichiometries, the rate law expression is: Rate = k[A]x[B]y [C]z

where k = rate constant, x,y,z… are the orders of the reaction with respect to A, B, C…. respectively. The overall order of the reaction is the sum of the individual orders, namely x + y + z +…… The values of x, y, z are usually (though not always) small integers (whole numbers) and can also include zero. If x = 1, then the order of the reaction with respect to A is said to be first order. If x = 2, then it is second order, if x = 3 it is third order and so on. Note that the orders of the reaction (x, y, z…) do not necessarily follow the reaction stiochiometries (a, b, c…). The orders of the reaction can only be determined experimentally and is not obtained by theoretical calculation. If the order of the reaction is the same as their respective reaction stiochiometries (that is Rate = k[A]a [B]b [C]c) then the reaction may occur via a single step – that is the a-number molecules of A, collide with b-number of molecules of B and c-number of molecules of C to give the products. Single step reactions are uncommon as they are statistically unfavourable. 2. Rate law determination The rate law can only be determined experimentally. It cannot be calculated from theory. That is, the values of k and the orders of the reactant(s) can only be found by conducting an experiment. Example 1: Determine the rate law for the reaction:

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A + 2B → C + D from the experimental data obtained: ___________________________________________________ Concentration (M) Rate of formation of C (M s-1)___ A B 0.10 0.10 2.0 x 10-4

0.10 0.20 4.0 x 10-4

0.20 0.20 1.6 x 10-3

0.30 0.20 3.6 x 10-3______________ That is, for the expression: Rate = k[A]x [B]y, determine the values of k, x, and y. Answer: From the first and second experimental data, it is observed that keeping [A] constant (0.10 M) while [B] is doubled (from 0.10 to 0.20 M), the rate also doubles (2.0 x 10-4 to 4.0 x 10-4). That is, ‘ [B](concentration) x 2’ leads to ‘rate x 2’; or 2j

(concentration) = 2(rate), therefore j = 1 since 21 = 2. That is, rate α [B]1. (α is the symbol for ‘proportional to’) The order for [A] may be obtained by using any combination of the last three data sets. Using the second and third data sets, if [B] is kept constant at 0.20 M and [A] is doubled from 0.10 to 0.20 M, the rate changes by [(1.6 x 10-3) ÷ (4.0 x 10-4)] or 4 times. Thus 2j

(concentration) = 4(rate), therefore j = 2 (since 22 = 4) That is rate α [A]2. The same relationship for [A] may be obtained by using the last two data sets where, if the concentration of [A] is increase 1.5 times (from 0.30 ÷ 0.20), the rate increases 2.25 times (from (3.6 x 10-3) ÷ (1.6 x 10-3)). Thus 1.5j

(concentration) = 2.25(rate), therefore j = 2 since 1.52 = 2.25. That is rate α [A]2, which is the same result as obtained for [A]x earlier. The rate expression obtained so far is: Rate = k[A]2 [B] The value of the rate constant ‘k’ may be obtained by substituting the [A], [B] and rate values from any one of the data sets. The same k value should be obtained for any data set used. Using the second data set as an example, k = __Rate_ = __4.0 x 10-4_(M s-1)____ = 0.20 M-2 s-1

[A]2 [B] [0.10]2 [0.20] ([M]2 [M]) Note therefore that the units for k will depend on the order of the rate expression.

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Exercise questions: Exercise B1: For the decomposition of hydrogen peroxide in solution at 25 oC: 2 H2O2 → 2 H2O + O2 (g), the following data was obtained: Experiment [H2O2] -d[H2O2]/dt number mmol L-1 mmol L-1 s-1 __________________________________ 1 14.45 6.70 x 10-3 2 28.90 13.3 x 10-3 3 57.80 26.7 x 10-3_ a. What is the order of the reaction? b. What is the rate equation for the reaction? c. Determine the rate constant for the reaction (at 25 oC) d. Determine the decomposition rate (at 25 oC) when [H2O2] = 10.0 mmol L-1. e. Does the rate increase, decrease or remains unchanged as the reaction proceeds? (Answer: c. 4.6 x 10-4 s-1; d. 4.6 x 10-6 mol L-1 s-1) Exercise B2: At 27 oC, the reaction: 2 NOCl → 2 NO + Cl2 is observed to exhibit the following dependence on rate of reaction: Initial NOCl concentration Initial rate of NO formation [mmol dm-3] [mmol dm-3 s-1] 0.30 3.60 x 10-9 0.60 1.44 x 10-8 0.90 3.24 x 10-8

a. What is the rate law for the reaction? b. What is the rate constant? c. By what factor would the rate increase if the initial concentration of NOCl was increased from 0.30 to 0.45 M? (Answer: b. 4.0 x 10-8 L mmol-1 s-1; c. 2.25 times faster) C. Measurement of reaction rates Three methods will be discussed. These are the ‘initial rate method’, the ‘integrated rate law method’ and the ‘half-life’ method. 1. Initial rate method For this method the initial rate of reaction is determined for various combinations of reactant concentration. The rate of a reaction is not constant over time since initial reactant(s) concentration(s) change as the reaction proceeds. If only the initial rate is measured (that is, the rate at the start of the reaction), the known initial reactant(s)

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concentration(s) can be assumed to not have changed significantly and the initial concentration(s) can be used in the rate calculations. The rate of reaction may be monitored either by measuring the rate of appearance of a product or the rate of disappearance of a reactant. Example 1: Determine the rate for the reaction: 2I- + S2O8

2- → I2 + 2S2O82-

by measuring the initial rate of formation of the product, I2. We therefore need a method to determine how long it takes for a certain amount of I2 to form, or the amount of I2 formed in a given time. In the experiment that will be performed in the laboratory class, the method that will be used for rate determination for this reaction is the time it takes for a given amount of I2 to form. This is achieved by placing some starch solution in the reaction mix (that is, starch + I- + S2O8

2-) so that the formation of I2 can be easily detected as it reacts with starch to form an intense blue colour. However this arrangement does not enable us to know when a certain concentration of I2 has been reached. This problem is solved by placing in the reaction mix, a small; accurately known amount of a substance (which is S2O3

2-) which quickly reacts with I2 as it is formed (that is the S2O3

2- uses up the I2 as soon as it is formed). When all the S2O3

2- has all been used up, free I2 enters the solution and it reacts with the starch to give a dark blue colour. Thus the time it takes for the initially clear solution to go dark, is the time taken for a certain amount of I2 to form. That certain amount of I2 formed is directly related to the amount of S2O3

2- added. This amount of S2O3

2- takes part in what is referred to as the ‘indicator reaction’. These reactions are summarised by the following equations: 1. 2I- + S2O8

2- → I2 + 2SO42- } Reaction being measured

2. I2 + 2S2O32- → 2I- + S4O6

2- } Indicator reaction Reaction 2 is much faster than reaction 1. As mentioned previously, the rate of the reaction is measured by the time it takes for a predetermined quantity of I2 to form. This quantity of I2 is equal to half the quantity of S2O3

2- added as shown in equation 2. That is: Rate (of I2 formation) = ½ d[S2O3

2-] dt The rate is determined for a given set of [I-] and [S2O8

2-] concentrations. During the course of the reaction, [I-] remains absolutely constant while [S2O8

2-] remains approximately constant. The concentration [S2O8

2-] is, of course, known. Refer to your practical notes for a full discussion of this experiment.

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2. Integrated rate law method This method of rate law determination shows how the concentration(s) of reactant(s) vary with time. For a general reaction: A → products The half-life (t1/2) of a chemical reaction is the time it takes for the concentration of a reactant to reach half of its initial concentration. (a) Zero order rate law: This is when the order for the reaction is zero. For a zero order rate law reaction (see also later discussion on page 9), the reaction rate is independent of the concentration of the reactants as described by the equation: Rate = k (reaction depends on the rate constant k only) The integrated rate equation is: [A]t = -kt + [A]o Compare with y = -mx + c which is the equation for a straight line and k is given by the slope as shown in the figure: (b) First order rate law: For a first order reaction where [A] is the concentration of reactant A and k is the rate constant, then Rate = d[A] = k[A] dt dt From calculus, ln [A]t = -k1t + ln [A]o Compare with y = -mx + c which is the equation for a straight line.

time (t)

[A]

Slope = -k (mol L-1 s-1)

t1/2 = [A]o/2k

t = time; [A]t = conc. A after time t; [A]0 = conc. A at time 0 (initial conc.)

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Where [A]t = concentration of A after time t [A]o = concentration of A at time 0 (that is initial concentration of A) Thus for a first order reaction, a plot of ln [A] versus t will give a straight line with the rate constant obtained from the slope of the line (without the (-) sign) as shown in the figure: (c) Second order rate law: For a second order reaction the rate law expression is: Rate = k2[A]2. Or _1_ = k2t + _1_ [A]t [A]o which is comparable to the equation for a straight line: y = mx + c Thus a plot of 1/[A]t versus t yields a straight line with the rate constant, k2, obtained from the slope of the line as shown in the figure: 3. Half life method: The half life, t1/2, of a chemical reaction is the time taken for a given reactant to be reduced to half its initial concentration. (a) For a zero order reaction where [A] is the concentration of reactant A and k is the rate constant, the rate expression is:

time (t) ln [A]

Slope = -k (s-1)

t1/2 = 0.693/k

_1_ [A]

time (t)

_1__ k[A]o Slope = k (L mol-1 s-1) t1/2 =

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Rate = k (mol L-1 s-1) and t1/2 = [A]o/2k (b) For a first order reaction where [A] is the concentration of reactant A and k is the rate constant, the rate expression is: Rate = k1[A] From calculus, ln [A]t = -k1t + ln [A]o Or ln [[A]o/[A]t] = k1t At t1/2 [A] = ½[A]o Therefore, ln [[A]o/½ [A]t] = k1t1/2 That is, ln 2 = k1t1/2

Or 0.693 = k1t1/2 Or, t1/2 = 0.693 k1 Therefore it will be observed that for a first order reaction, t1/2 is independent of the concentration of A as the term “[A]” is absent from the expression. (c) For a second order reaction: t1/2 = __1__ k2[A]o so that for a second order reaction, t1/2 is dependent on the initial concentration of A (that is, the t1/2 is dependent on [A]o) . If [A]o (the initial concentration of A) is doubled, t1/2 is halved. General comment on zero order reactions: It is generally observed that the concentrations of reactants affect the rate of a chemical reaction but there a few reactions whose rates are independent of concentration. In such cases, the reaction proceeds at a constant rate. That is, rate of reaction = k = constant. This can arise if all the orders of the reactants are zero. The reaction is said to follow zero order kinetics. A plot of concentration of reactant versus time gives a straight line with a negative slope:

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[A] decreases from a maximum value of [A]o at time t = 0 till [A] = 0. The rate constant, k = - slope. That is, the rate of reaction which remains constant throughout and the reaction rate is the negative of the slope of the straight line graph. The units of k are the same as the units of the rate of a reaction, namely mol L-1 (time)-1, for example, M s-1. Summary of graphical method to determine order of a reaction: 1. If a plot of [A] versus time gives a straight line, the reaction is zero order. 2. If a plot of ln[A] versus time gives a straight line, the reaction is first order. 3. If a plot of 1/[A] versus time gives a straight line, the reaction is second order Exercise questions: Exercise C1: The half life for the 1st order reaction: SO2Cl2 → SO2 + Cl2 is 8.0 min. How long will it take for the concentration of SO2Cl2 to be reduced to 1% of the original concentration? (hint: the relevant formula is: ln [[A]o/[A]t] = -k1t) (Answer: 53 min) Exercise C2: The 1st order rate constant for the conversion of cyclobutane to ethylene at 100 oC is 87 s-1:

a. What is the half life of this reaction at 1000 oC?

C C

C C H

H

H

H

H

H

H

H H

H

C C H

H 2

cyclobutane ethylene

[A]

Time

[A]o

[A] = 0

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b. If the initial quantity of cyclobutane is 1.00 g, how long will it take to consume 0.70 g of it? c. How much of an initial quantity of 1.00 g cyclobutane remain after 1.00 x 10-2 s? (Answer: a. 8.0 x 10-3 s; b. 0.0138 s; c. 0.419 g) Exercise C3: The molecule (CH2CO2H2)2CO undergoes 1st order decomposition in aqueous solution according to the equation: (CH2CO2H2)2CO → (CH3)2CO + 2 CO2 a. Give the rate law expression b. If the rate constant, k1 is 5.48 x 10-2 s-1 at 60 oC, what is the t1/2 at 60 oC? c. If the rate constant, k2 is 2.46 x 10-5 s-1 at 0 oC, what is the t1/2 at 0 oC? d. Are the calculated t1/2’s in accordance with the statement: “In general, the speed of a chemical change approximately doubles for each 10 oC rise in temperature” (Answer: a. 12.65 s; b. 2.8 x 104 s) D. Factors influencing the rate of a chemical reaction 1. Introduction: There are four factors that influence the rate of a chemical reaction. These are: (i) The chemical nature of the reactants and products (ii) The concentration of the reactants (iii) The temperature of the reaction environment (iv) The presence of a catalyst 2. Collision rates: Not all collisions between reactants lead to product formation. For a successful collision (that is, one that leads to product formation) the molecules must collide with: (i) Correct orientation and (ii) Sufficient energy (i) Correct orientation: Consider the reaction of nitrogen oxide with ozone: NO(g) + O3(g) → NO2(g) + O2(g) Orientation leading to product formation:

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Orientation not leading to product formation:

(ii) Collision with sufficient energy: Even with the correct orientation, the reactants must collide with sufficient energy for product to form. Otherwise the colliding molecules will simply bounce apart without reacting. Collision with sufficient energy is necessary because, for new bonds to form to give the product, old bonds must be broken and bond breaking requires energy. If the collision process does not provide sufficient energy for this bond breaking to occur, the product does not form and the colliding molecules move apart chemically unchanged. This process is illustrated in the given diagram:

N

O O O

O O

O

N O

O O

N O

Before collision Collision After collision

O

O No reaction

O N

O O

O O

O O

N O

O O

N O

O

Before collision Collision After collision

Reaction

Nitrogen oxide

Ozone

Figures showing requirement for correction orientation of colliding reactant molecules

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Whether the reaction is exo- or endothermic, the activated complex, which is a transition state product, will have P.E. greater than either the P.E. of the reactant(s) or product(s). In the course of a reaction, bond breaking occurs as new ones are formed. The process of bond breaking requires energy input whereas the process of bond formation releases energy. The difference in energy absorbed and energy released is the enthalpy change (∆H) for the reaction. If more energy is released during bond formation that was put in for the process of bond breaking, then a (-)ve ∆H is observed and vice versa. The activated complex is the transitory molecular arrangement at the point of bond breaking, just before new bonds are formed. It is therefore the point where it has the highest P.E. The activation energy (EA) comes from the kinetic energy (K.E.) of the reacting molecule. Note that K.E. = ½ mv2 (m = mass, v = velocity), thus the kinetic energy is strongly dependent on the velocity of the molecule. Since the K.E. is the energy the molecule has due to its motion, the K.E. is therefore temperature dependent (higher temperature, higher velocity, higher K.E.). At the point of collision, the K.E. is converted to P.E., thereby providing the energy necessary to climb the EA ‘hill’. The P.E. plot for an endothermic ((+)ve ∆H) has the P.E. of the products higher than the P.E. of the reactants as shown in the given figure:

∆H Reactants

Products

:O:: :O:

O:

:O:

O:

::O: N:. O::

:O:

N.

.:N=O:: +

Activated complex

:O: +

::O: :O:

EA(F)

EA(R)

Reaction coordinate

Potential Energy (P.E.)

New bond forms

Old bond breaks

EA(F) = Energy of activation for the forward reaction EA(R) = Energy of activation for the reverse reaction ∆H = Enthalpy of reaction which is (-)ve in this example (exothermic reaction, P.E. of products lower than P.E. of reactants) Potential energy plot for an exothermic ((-)ve ∆H) reaction

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Spontaneous endothermic reactions are driven by entropy (∆S) since for a spontaneous reaction to occur, the Gibbs free energy (∆G) must be negative (note: ∆G = ∆H - T∆S), thus the ∆S must be large enough (or positive enough) to overcome the positive ∆H. (iii) Presence of a catalyst A catalyst is a substance that increases the rate of a chemical reaction without being used up in the reaction. The catalyst takes part in the chemical reaction but it is regenerated at the end, resulting in no net change in the amount of catalyst present. A catalyst functions by lowering the activation energy, EA, by providing an alternative (lower EA) pathway or mechanism for the reaction to proceed. A catalyst does not alter the equilibrium constant (K) for the reaction because the forward and reverse reaction rates are increased equally. The enthalpy of reaction (∆H) also remains unchanged. The effect of a catalyst on EA on the P.E. versus reaction coordinate plot is shown in the given figure:

Reactants

Products

Reaction coordinate

Potential Energy (P.E.)

∆H

EA(F)

EA(R)

Potential energy plot for an endothermic ((+)ve ∆H) reaction

Activated complex

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Homogeneous and heterogeneous catalysts: Catalysts may be classified as homogeneous (where the catalyst is in the same phase as the reactant(s)) or heterogeneous (where the catalyst is in a different phase to the reactants). An example of a homogeneous catalyst is the catalytic decomposition of vitamin C by Cu2+ ions in solution. An example of a heterogeneous catalyst is the use of finely divided Ni metal for the hydrogenation of oils in the production of margarine. Examples of mechanisms of how catalysts work are given in the following examples. Example of how a homogeneous catalyst works: Formic acid can decompose to water and carbon monoxide according to the equation:

In the un-catalysed reaction the process occurs as shown in the figure:

H-C-O-H → H2O + CO

O

∆H Reactants

Products

Activated complex

EA(without catalyst)

EA(with catalyst)

Reaction coordinate

Potential Energy (P.E.)

Catalyst lowers EA

Potential energy plot for an exothermic ((-)ve ∆H) reaction in presence of a catalyst

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The decomposition of formic acid to water and carbon monoxide is catalysed by acid. In the catalysed reaction the process occurs as shown in the figure:

Example of how a heterogeneous catalyst works: The exact mechanism of how a heterogeneous catalysts is not completely clear but it appears that the availability of d-electrons and d-orbitals on atoms at the surface of the catalyst plays an important role. A large number to transition elements and their compounds show catalytic activity. For a heterogeneous catalyst to work, the reactants (which may be gas or liquid) must be adsorbed on the surface of the catalyst. Not all atoms at the surface of the catalyst is effective – those that are called active sites. The sequence of events that take place in the action of a catalyst is:

OH

O

+ HOH

O O

Activated complex

O

C C

C C

HOH H+

H +

H H

H+

Potential Energy (P.E.)

HOH

Figure showing catalysed decomposition of formic acid

&

Reaction coordinate

EA is lowered in the presence of the catalyst (H+)

EA

Reaction coordinate

Potential Energy (P.E.)

C H

O C = O

OH

OH

C

O

& HOH

H

Activated complex

Figure showing un-catalysed decomposition of formic acid

EA

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Step 1. Adsorption of the reactants Step 2. Diffusion of the reactants along the catalyst surface Step 3. Reaction at an active site of the catalyst to form the adsorbed product Step 4. Desorption of the product This sequence of events is illustrated in the example involving the rhodium (Rh) catalysed reaction between carbon monoxide and nitric oxide to produce carbon dioxide and water according to the reaction: 2CO + 2NO → 2CO2 + N2 Step 1: Two molecules each of CO and NO are adsorbed on the Rh surface:

Step 2: The adsorbed NO molecules dissociate into adsorbed N and O atoms and diffuse along the surface of the catalyst:

Step 3: The adsorbed CO molecules and O atoms combine to form CO2 molecules which desorb from the catalyst surface as the gas. The two adsorbed N atoms combine and desorb as the N2 molecule:

Enzymes as catalysts:

O O O O O O O O

C C C N N N N C Rh Rh

Rh Rh O

O

O

O O O

C C N N N N

Rh Rh

O O

O O

CO2

C C N N

N2

CO2

Rh

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Catalysts associated with chemical reactions in living organisms are called enzymes which are high molecular weight proteins. Enzymes are very specific in the reactions they catalyse. For example, in alcoholic fermentation, the six-carbon compound glucose is broken down to give two molecules of ethanol. This is a 12 enzymic step process, the last of which catalysed by the enzyme alcohol dehydrogenase. In the simplest mechanism of enzyme action, known as the Michaelis-Menten mechanism, a reactant species called the substrate (S) attaches itself to an active site on the enzyme (E). The result is an enzyme-substrate complex (ES) which dissociates to produce a product species (P) and the originals enzyme (E). This two step mechanism may be summarised in the given equations: S + E ↔ ES ES ↔ E + P A plot of reaction rate as a function of time for an enzyme catalysed reaction which follows the Michaelis-Menten mechanism of enzyme action is shown:

Along the rising portion of the curve the reaction is 1st order with respect to substrate concentration S because the rate at which the complex [ES] forms is proportional to [S]. That is, Rate = k[S] At high substrate S concentration the reaction is zero order. The enzyme E is saturated and adding more substrate does not increase reaction rate, therefore: Rate = k1[S]0 = k1 Toxicokinetics: This is the study of changes of toxin or drug concentration in the body as a function of time.

Concentration of substrate [S]

Rate of reaction

First order

Zero order

Figure showing Michaelis-Menten mechanism of enzyme action

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Most drugs used in modern medicine bind very loosely with their receptors. For example cimetidine binds to the histamine receptor. The structure of cimitidine is:

The substances can usually be easily washed off their receptors which the stops to trigger the effects of the substance. However some xenobiotics (for example, organophosphate pesticides) are able to form covalent bonds with the receptors and the effects of these are more difficult to reverse. Therefore the physiological effect of a toxin may result from either 1. a reversible reaction between an active agent and a receptor (the target molecule) or 2. an irreversible interaction between the two. In the first example the intensity of the toxic effect depends on the number of target molecules (T) that have been bound to the toxin or active agent (A): A + T A-T → toxic effect The position of the equilibrium depends on the concentration of A and the strength of the A-T bond. The blood concentration of A at the site of the action changes with time, therefore the intensity of the effect as a function of time can also be described by the change in blood concentration of A as a function of time. If a toxic effect is a result of covalent A-T bond, then the reaction is essentially irreversible and the number of adduct molecules (molecules formed by combination of precursor molecules, for example, A-T) and hence by the amount of active agent (A). The mechanism by which a chemical species reaches a particular blood concentration in the body is difficult to describe as it depends on a variety of factors, for example nature of the substances involved, its absorption, distribution, biotransformation and excretion. Despite the complexity described the body can often be presented schematically by a so-called ‘compartment model’ for the purposes of fitting to a mathematical model. This toxicokinetic (or pharmacokinetic) concept of compartment applies to all tissues, organs and fluids in the body that do not differ from each other in terms of kinetics. Toxicokinetics mainly deals with reversible reactions though toxicology often involves irreversible reactions (which involve covalent bonding). If a substance, after entering the circulation, is rapidly distributed throughout the organism, and if the substance is readily exchanged between the circulating blood and the rest of the body (for example the fatty or adipose tissue), then the organism can be described as a two-compartment model. In this case the two compartments are the adipose tissue (often referred to as the peripheral or tissue compartment, T) and the circulation

HN N

H3C CH2-S-CH2-CH2-NH-C-NH-CH3

N-CN

20

plus other organs and tissues (often referred to as the central compartment, C). Whether a number of organs, tissues and body fluids constitute a compartment depends of the nature and extent of the organs and tissues and the physiochemical properties (for example, hydrophobic/hydrophilic nature, binding affinity) of the substance concerned. (iv) Effect of temperature on reaction rates and the measurement of EA An increase in the temperature of the reaction environment leads to increased K.E. of reactant molecules, thus increasing the fraction of molecules with K.E. > EA, thereby increasing the number of molecules that can have a collision that leads to product formation. This increase in the fraction of molecules with P.E. > EA, with increased temperature is illustrated in the given figure:

The Arrhenius equation gives the relationship between the rate constant, k, energy of activation, EA, and temperature (in K):

where R = gas constant e = base of natural log A = proportionality constant whose value is determined by the collision frequency and molecular orientation during collision Taking logs on both sides, the equation becomes:

k = Ae -EA/RT

EA

Number of molecules

Kinetic energy

Kinetic energy distribution of molecules at two different temperatures

K.E. distribution at lower temperature

K.E. distribution at higher temperature

Number of molecules with P.E. > EA

21

ln k = ln A - EA . 1 R T y = c - m . x which is the equation for a straight line. Thus by plotting ln k versus 1/T, a straight line should be obtained, with a slope equal to -EA/R and ln A is obtained where the line cuts the y-axis. This is illustrated in the given figure:

The slope is measured, the gas constant, R, is known and therefore can be determined. Other than the graphical method just described, the value of EA can also be determined by mathematical calculation if the rate constant, k, is determined at two different temperatures. The calculation for this is shown: If k1 is the rate constant at temperature T1 (K) and k2 is the rate constant at temperature T2, then the following expressions may be written:

1/T (K-1)

ln k Slope = ∆ln k ∆(1/T) = - EA

R

Plot of ln k versus 1/T to determine EA

22

If EA is in joules (J), then R = 8.314 J K-1 mol-1. Examples of calculations using the Arrhenius equation: Example 1: The rate constant for a reaction was determined to be 1.3 x 10-5 L mol-1 s-1 at 15 oC and 8.0 x 10-3 L mol-1 s-1 at 50 oC. Determine the EA (in kJ mol-1) for the reaction. Answer: Using the equation Solving, EA = 142 kJ mol-1 (always check for consistency of units) Example 1: The EA for a particular reaction is 105 kJ mol-1. At 25 oC the rate constant is 2.0 x 10-3 s-1. Determine the rate constant at 50 oC. Answer: k1 = 2.0 x 10-3 s-1 at 25 oC (T1) k2 = ? at 50 oC (T2) EA = 105 kJ mol-1

k1 = Ae -EA/RT1

k2 = Ae -EA/RT2

k1 k2 = e

EA 1 - 1 R T1 T2

ln k1 k2 = EA 1 - 1

R T2 T1

OR

log k1 k2 = EA 1 - 1

2.303R T2 T1

ln k1 k2 = EA 1 - 1

R T2 T1

1.3 x 10-5 (L mol-1 s-1) 8.0 x 10-3 (L mol-1 s-1)

ln = EA __ ___ __ 1 - 1_____ 8.314 (J K-1 mol-1) 50 + 273 (K) 15 + 273 (K)

23

ln 2.0 x 10-3 (s-1) = ___105 (kJ mol-1)_____ ___1 _____ - ____1_____ k2 (s-1) 8.31 x 10-3 (kJ K-1 mol-1) 50 + 273 (K) 25 + 273 (K) = ___ 105 (kJ mol-1)_____ [-2.597 x 10-4 (K)] 8.31 x 10-3 (kJ K-1 mol-1) ln 2.0 x 10-3 = -3.282 k2 2.0 x 10-3 = 0.0376 k2 (s-1) k2 = 5.3 x 10-2 s-1 Note that R = 8.314 kJ K-1 mol-1 has been converted to kJ to be consistent with the units of EA which is in kJ. Exercise questions: Exercise D1: For the reaction: OCl- (aq) → Cl- (aq) + ½ O2 (g), the ∆Hreaction = -51 kJ mol-1 and the EA = 47 kJ mol-1. Draw a diagram plotting potential energy (P.E.) versus reaction coordinate, clearly labelling the diagram to show the ∆Hreaction, EA, the activated complex and the reactant and products. Exercise D2: The activation energy, EA for the decomposition of hydrogen iodide, HI: 2 HI (g) → H2 (g) + I2 (g) is 182 kJ mol-1. The rate constant, k, for the reaction at 700 oC is 1.57 x 103 dm3 mol-1 s-1. What is the rate constant at 600 oC? (Answer: 1.19 x 10-4 dm3 mol-1 s-1) Exercise D3: What activation energy, EA, should a reaction have so that raising the temperature from 0 oC to 10 oC would triple the reaction rate? (Answer: 7.06 x 104 J mol-1) Exercise D4: The following reaction follows 1st order kinetics: FClO2 (g) → FClO (g) + O (g) The EA for the reaction is 186 kJ mol-1 and the k at 322 oC is 6.76 x 10-4 s-1. (i) What is the k at 25 oC? (ii) What temperature of reaction is required to achieve a k = 4.00 x 10-2 s-1?

24

(Answer: (i) 3.38 x 10-20 s-1; (ii) 394 oC) E. Reaction mechanisms A balanced chemical equation does not provide information on how a reaction occurs. It is merely a summary of the reaction. Reaction mechanisms can be deduced by studying their rates and rate law. A balanced chemical equation can provides information on how a reaction occurs if it is a single step reaction – that is the product(s) are formed by direct collision of the molecules as written in the equation. However, most reactions follow a multi step mechanism. 1. Single step reactions: In a single step reaction the reactant molecules collide as written in the equation. For single step reactions, the rate law has orders of reaction which correspond to the relevant reaction stoichiometries. For the general reaction: aA + bB → cC Rate = k [A]a [B]b for a single step reaction. 2. Multi step reactions: For multi step reactions, the rate law does not correspond to the balanced equation for the reaction. For the reaction: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) The rate law was found to be Rate = k [N2O5] (rather than rate = k [N2O5]2 as would be observed if it were a single step reaction) The given reaction there does not occur via a single step but follows a multi step mechanism such as the one given: 2 [N2O5 → NO2 + NO3] (slow reaction) NO2 + NO3 → NO + NO2 + O2 (fast reaction) NO + NO3 → 2 NO2 (fast reaction) 2 N2O5 → 2 NO2 + O2 (overall reaction)

25

The different steps of a multi step may have very different rates. The overall rate cannot exceed the rate of the slowest step in the reaction mechanism. The slowest step is called the rate determining step. The rate determining step of a reaction mechanism gives the rate law for the reaction. Thus for the given example: Rate = k [N2O5] since it is the rate expression for the rate determining step. Where the rate expression for the slowest step does not involve a reactant that appears in the equation for the overall reaction, substitutions using the equilibrium constant (K) expressions may be used to place the reactant into the equation for the rate determining step. Each of the three individual steps shown in the example is called an elementary process, step or reaction. All the three steps taken together is called the reaction mechanism. Elementary steps are almost always unimolecular (that is, involving only one species) and bimolecular (that is, involving two species) and rarely termolecular (that is, involving three species). These are described in greater detail in the next section. In multi step reactions, species are often formed in the elementary steps which do not appear in the final products (as shown in the overall reaction). Such species (for example, NO3) are called reaction intermediates. Some reaction intermediates are odd-electron species called free radicals. (i) Unimolecular reactions These are reactions that result from the decomposition or internal rearrangement of a single molecule, for example:

The rate law was observed to be: Rate = k[cis-2-butene] A unimolecular reaction has first order rate law. (ii) Bimolecular reactions These reactions are a result of a successful collision between two molecules, for example: NO (g) + O3 (g) → NO2 (g) + O2 (g) The rate law was observed to be:

H3C C=C

CH3

H H CH3

H3C C=C

Cis-2-butene

H

H

Trans-2-butene

26

Rate = k [NO][O3] A bimolecular reaction has second order rate law. (iii) Termolecular reactions These occur by successful simultaneous collision between three molecules and would have a third order rate law. These reactions are uncommon because of probability considerations. 3. Chain reactions: A chain reaction is a multi step reaction in which a free radical is formed and this free radical initiates a series of two or more reactions in which the products are formed and the free radical is regenerated as shown in the given example: H2 + Br2 → 2 HBr (the overall reaction) which occurs via the following mechanism: Br2 → 2 Br. } chain initiation Br. + H2 → HBr + H. } chain propagation H. + Br2 → HBr + Br. } chain propagation Br. + Br. → Br2 } chain termination H2 + Br2 → 2 HBr______ } overall reaction Exercise questions: Exercise E1: The proposed mechanism for a gaseous reaction is: O3 → O2 + O O + O3 → 2 O2 a. What is the observed overall reaction being studied? b. What is the role of O in the reaction? Exercise E2: What is the overall reaction for the following postulated mechanism? Cl2 + AlCl3 → AlCl4

- + Cl+ Cl+ + C6H6 → C6H5Cl + H+ H+ + AlCl4

- → AlCl3 + HCl What is the role of AlCl3 in this reaction? Exercise E2: Ozone, O3 in the stratosphere can be decomposed by reaction with nitrogen oxide, NO (commonly called nitric oxide), from high flying jet aircraft, following the equation:

27

O3(g) + NO(g) → NO2(g) + O2(g) The rate expression is: Rate = k [O3][NO]. Which of the following mechanisms are consistent with the observed rate expression? A: NO + O3 → NO3 + O slow NO3 + O → NO2 + O2 fast___ _O3 + NO → NO2 + O2 overall B: NO + O3 → NO2 + O2 slow (one step) C: O3 → O2 + O slow O + NO → NO2 fast___ O3 + NO → NO2 + O2 overall_ D: NO → N + O slow O + O3 → 2 O2 fast O2 + N → NO2 fast__ O3 + NO → NO2 + O2 overall E: NO → N + O fast O + O3 → 2 O2 slow O2 + N → NO2 fast__ O3 + NO → NO2 + O2 overall Exercise E3: Do you expect the equation for the combustion of ethene, C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g) to represent the mechanism by which ethene burns (that is, would you expect the reaction to be a single step process)? Explain your answer. Exercise E4: The proposed mechanism for a gaseous reaction is given as: Step 1: NO + Cl2 → NOCl2 Fast Step 2: NOCl2 + NO → 2 NOCl Slow a. Write the equation for the overall reaction. b. What is the rate determining step in the mechanism c. What is the molecualrity of Step 1 and of Step 2? d. If the proposed mechanism is correct, would the rate equation be in terms of the overall reactants? (Hint: Use equilibrium constant (K) expression for Step 1 to get the rate expression to be in terms of the reactants in the overall equation) e. If the concentrations of both reactants NO and Cl2 were doubled, by what factor would the rate increase?

28

f. The rate constant, k, varies with temperature in a manner described by the Arrhenius equation:

A plot of ln k versus 1/T for the reaction gives a straight line with slope = -9.99 x 103 K. Calculate the EA for the reaction at 25 oC. (Answer: EA = 83 kJ mol-1) F. Nucleophilic substitution at a saturated carbon 1. Introduction: The term ‘nucleophilic’ means nucleus (that is, (+)ve charge) seeking. The nucleophilic species is therefore a (-)ve charged species and is called a nucleophile or nucleophilic reagent. The opposite of a nucleophile is an electrophile which is species which seeks after a (-)ve charge and it is therefore a (+)ve charged species. A saturated carbon is a carbon atom with four bonds attached to it. A substitution reaction is the substitution or replacement of one atom or a group of atoms in a molecule with another atom or group of atoms. Nucleophilic substitution reactions may be classified as: 1. Unimolecular (or first order) or 2. Bimolecular (or second order) If unimolecular, the reaction is referred to as a SN1 (substitution, nucleophilic, 1st order). If bimolecular, the reaction is referred to as SN2 (substitution, nucleophilic, 2nd order). The designations SN1 or SN2 do not reveal the reaction mechanism – it only states the type of nucleophilic substitution occurring at the saturated carbon. The favoured pathway (whether SN1 or SN2) is determined by the nature of the substituents on the saturated carbon and on the experimental conditions. 2. SN1 reactions: Consider the reaction whereby Br is substituted for OH in t-butyl bromide:

k = Ae -EA/RT

CH3 CH3

CH3

H3C H3C

CH3

C C Br Br - OH OH - + +

t-butyl bromide t-butyl alcohol

nucleophile

29

The rate law was found to have 1st order kinetics, that is: Rate = k [(CH3)3CBr] and not Rate = k [(CH3)3CBr][OH-]. Therefore the reaction rate is dependent on the concentration of t-butyl bromide only and independent of [OH-] concentration. The observed rate law is consistent with the following reaction mechanism:

Since the slow step (the rate determining step) determines the reaction rate, then the rate law expression is: Rate = k[t-butyl bromide] A substitution reaction will tend top proceed via the SN1 pathway if the molecular structure favours the formation of the carbocation species. Alkyl groups (for example, -CH3) are electron releasing and can thus help stabilise the (+)ve carbocation. 3. SN2 reactions: Consider the reaction of methyl bromide with the hydroxide ion (OH-):

+ + Br - Br

H3C C

CH3

Indicates movement of electron pair Carbocation species

[Slow step]

+ C

CH3

CH3

H3C + OH- nucleophile

CH3

C OH

t-butyl alcohol

H3C

CH3

[Fast step]

t-butyl bromide

CH3

C H3C

CH3 CH3

Figure showing mechanism for SN1 reaction

30

The rate law was found to have 2nd order kinetics, that is: Rate = k [CH3Br][OH-] Since the rate is influenced by both the [CH3Br] and [OH-] concentrations, the reaction probably proceeds in a single step involving a collision between the two reactant molecules. The observed law is consistent with the following mechanism:

The 3-H’s attached to the C-atom flip over to the other side of C and the product is thus inverted because the nucleophile (for example OH -) attacks from the back side of the

Br H H

H

H

+ Br -

Methyl bromide Methyl alcohol

nucleophile C

H

OH-

OH

OH-

H

H

H

H

H

C +

H

H

Br Br

Br-

-

Back of molecule Front of molecule

C

Activated complex

(the 3-H’s and C are on one plane)

H

H

+

Reactants

Products

C

(tetrahedral shaped molecule)

Figure showing mechanism for SN2 reaction

C

H

+ OH- OH

31

molecule. This aspect of inversion is not important in the given example as the product is the same whether or not there is inversion. The SN2 mechanism is favoured when the groups attached to the C-atom are not bulky (like, for example, H atom instead of CH3 or other bulkier alkyl groups) so that they can be easily flipped over to the other side of the C-atom. The rate of an SN2 reaction is largely influenced by stearic (or spatial) factors whereas the rate of an SN1 reaction is largely affected by electronic (or electrical charge distribution) factors. 4. SN1 versus SN2 pathways: Some reactions proceed via both the SN1 versus SN2 pathways. The reactivity towards each mechanism is given by the summary:

Most typically for halides, going along the series CH3, 1o, 2o, 3o, the following mechanisms apply:

(Note: 1o = CH3XH3, 2o = (CH3)2XH2, 3o = (CH3)3XH representing primary, secondary and tertiary methyl halides respectively) Exercise question: Exercise F1: Briefly describe what is meant by a SN1 and SN2 reaction. Give an example of each, showing clearly the mechanism of reaction in each case. Examples of multiple choice questions: 1. For the reaction between gaseous chlorine and nitrogen monoxide: 2NO + Cl2 → 2NOCl It was found that doubling the concentration of both reactants increased the rate by a factor of eight, but doubling the chlorine concentration alone only doubles the rate.

(i) What is the order of the reaction with respect to NO and Cl2? (ii) State the overall order and molecularity of the reaction.

(a) (i) Rate = k[NO]2[Cl], (ii) termolecular

CH3X > 1o > 2o < 3o SN2 SN2 Mixed SN1

RX = CH3X 1o 2o 3o SN1 increases

X = halide SN2 increases

32

(b) (i) Rate = k[NO] [Cl]2, (ii) termolecular (c) (i) Rate = k[NO] [Cl], (ii) bimolecular (d) (i) Rate = k[NO] [Cl]2, (ii) bimolecular 2. For the decomposition of hydrogen peroxide in solution at 25oC: 2H2O2 → 2H2O(g) , the following data was recorded: Experiment number: [H2O2] mmol L-1 -d[H2O2]/dt mmol L-1s-1 1 14.45 6.70 x 10-3 2 28.90 13.3 x 10-3 3 57.80 26.7 x 10-3 (i) What is the rate equation for the reaction? (a) Rate = k[H2O2]0 (c) Rate = k[H2O2]2 (b) Rate = k[H2O2] (d) Rate = k[H2O2]1/2 (ii) What is the order of the reaction? (a) 0 order (c) 2nd order (b) 1st order (d) 3rd order (iii) Calculate the rate constant (k) for the reaction at 25oC. (a) 1.3 x 10-2 mmol L-1 s-1 (c) 4.6 x 10-4 mmol L-1

(b) 1.3 x 10-2 mmol (d) 4.6 x 10-4 s-1 (iv) Calculate the decomposition rate (in mol L-1 s-1) at 25oC when [H2O2] = 10 mmol L-1. (a) 4.6 x 10-4 (c) 4.6 x 10-5 (b) 4.6 x 10-3 (d) 4.6 x 10-6 (v) How does the rate change as the reaction proceeds? (a) increases (c) increases then decreases (b) decreases (d) decreases then increases 3. At 27oC, the reaction 2NOCl → 2NO + Cl2, is observed to exhibit the following dependence of rate on concentration: Initial NOCl concentration [mol dm-3] Initial rate of NO formation [mol dm-3 s-1]

0.30 3.60 x 10-9 0.60 1.44 x 10-8

0.90 3.24 x 10-8 (i) What is the rate law for the reaction? (a) R = k[NOCl] (c) R = k[NOCl]1/2

(b) R = k[NOCl]2 (d) R = k[NOCl]3 (ii) What is the rate constant, k?

33

(a) 4.0 x 10-8 L mol-1 s-1 (c) 4.0 x 10-6 L mol-1 s-1 (b) 3.5 x 10-8 L mol-1 s-1 (d) 3.5 x 10-8 L mol-1 s-1 (iii) By what factor would the rate increase if the initial concentration were increased from 0.30 to 0.45 M (a) 1.5 times faster (c) 5.0 times faster (b) 0.675 times faster (d) 2.25 times faster (Answers: 1a; 2 (i)b, (ii)b, (iii)d, (iv)d, (v)b; 3 (i)b, (ii)a, (iii)d)