Chapter 25 The Rates of chemical reactions

Contents Empirical chemical kinetics 25.1 Experimental techniques 25.2 The rates of reactions 25.3 Integrated rate laws 25.4 Reaction approaching equilibrium 25.5 The temperature dependence of reaction rates Accounting for the rate laws 25.6 Elementary reactions 25.7 Consecutive elementary reactions 25.8 Unimolecular reactions

Assignment for chapter (b),25.6(a),25.10(b),25.15(a) 25.2,25.5,25.11,25.23

St. John’s wort is an herb that is thought to create a sense of tranquility. Herbs and other medicines have been used through the ages to cure disease and to relieve pain. In many cases, the medicine is effective because it controls the rates of reactions within the body. In this chapter, we examine the rates of chemical reactions and the mechanisms by which they take place. Empirical chemical kinetics

Monitoring the progress of a reaction 2N 2 O 5 (g)  4NO 2 (g)+O 2 (g) Initial pressure of n moles of N 2 O 5 is p 0. Progress of the reaction: N 2 O 5 NO 2 O 2 Total Initial n 0 0 n At time t n(1-a) 2an 0.5an n(1+1.5a) Progress of the reaction: N 2 O 5 NO 2 O 2 Total Initial p p 0 At time t n(1-a) p 0 2an p 0 0.5an p 0 n(1+1.5a) p 0

Classroom exercise 2NOBr(g)  2NO(g)+Br 2 (g) Initial pressure of n moles of N 2 O 5 is p 0. Progress of the reaction: NOBr NO Br 2 Total Initial n 0 0 n At time t n(1-a) an 0.5an n(1+0.5a) Progress of the reaction: N 2 O 5 NO 2 O 2 Total Initial p p 0 At time t n(1-a) p 0 an p 0 0.5an p 0 n(1+0.5a) p 0

Other properties to be monitored in the progress of chemical reactions Absorption of radiation (spectrophotometry) Ions in solution (Electrical conductivity) Emission of radiation (emission spectroscopy) Mass of ions (mass spectrometry) Adsorption of molecules (gas chromatography) Absorption of radiofrequency radiation (NMR spectroscopy) Absorption of microwave radiation (ESR spectroscopy) Etc.

Experimental techniques: flow technique

Experimental techniques: stopped-flow technique

Experimental techniques: flash photolysis Cl

Experimental techniques: flash photolysis Cl

Experimental techniques: quenching methods Chemical quench flow method: quench by chemical reactants such as acids Freeze quench method: quench by rapid cooling

Definition kA  mC+nD Rate=change in concentration of reactant / time interval 2HI(g)  H 2 (g)+I 2 (g)

The activity of penicillin declines over several weeks when it is stored at room temperature in the absence of stabilizers. The shape of this graph of concentration of penicillin as a function of time is typical of the behavior of chemical reactions, although the time span may vary from fractions of a second to years.

This graph shows two examples of how the rate of consumption of penicillin can be monitored while it is being stored. The red line shows the average rate calculated from measurements at 0 and 10 weeks, and the blue line shows the average rate calculated from measurements at 2.5 and 7.5 weeks. The instantaneous rate at 5 weeks is the tangent to the curve at that time (not shown).

To calculate the instantaneous reaction rate, we draw the tangent to the curve at the time of interest and then calculate the slope of this tangent. To calculate the slope, we identify any two points, A and B, on the straight line and identify the molar concentrations and times to which they correspond. The slope is then worked out by dividing the difference in concentrations by the difference in times. Notice that this graph shows the concentration of a product.

Exercise

Instantaneous Rate of Reaction A+2B  3C+D v is the same for all species in a reaction.

The initial rate of reaction for the decomposition of N 2 O 5 in five experiments. The initial rate of disappearance of a reactant is determined by drawing a tangent to the curve at the start of the reaction. 2N 2 O 5 (g)  4NO 2 (g)+O 2 (g)

Instantaneous Rate of Reaction For a homogeneous reaction, For a heterogeneous reaction, The surface density of species J

The plot of the initial rate of decomposition of N 2 O 5 as a function of initial concentration for the five samples in previous figure is a straight line. The linear plot shows that the rate is proportional to the concentration. The graph also illustrates how we calculate the rate constant, k, from the slope of the straight line. Initial rate=k x initial concentration

(a) The instantaneous reaction rates for the decomposition of N 2 O 5 at five different times during a single experiment are obtained from the slopes of the tangents to the line at each of the five points. (b) When these rates (the slopes) are plotted as a function of the concentration of N 2 O 5 remaining, the result is a straight line with a slope equal to the rate constant. In (b), we have indicated the rates by redrawing the tangents. rate=k x concentration

Illustration In the reaction 2NOBr(g)  2NO(g)+Br 2 (g), the rate of formation of NO is 0.16 mmol/L/s. Calculate the rate of consumption of NOBr.

Classroom exercise In the reaction 2CH 3 (g)  CH 3 CH 3 (g), the rate of consumption of CH 3 is -1.2 mol/L/s. Calculate the rate of the reaction and the rate of formation of CH 3 CH 3.

a: order of reaction, determined by experiment. k: rate constant, determined by experiment. Order of a Reaction

(a) The concentration of the reactant in a zero-order reaction falls at a constant rate until the reactant is exhausted. (b) The rate of a zero-order reaction is independent of the concentration of the reactant and remains constant until all the reactant has been consumed, when it falls abruptly to 0. v=constant

There are no images in this section of the chapter.

More Complicated Rate Laws Order=a+b+…-p-q… n A A+n B B+n C C+ …  n P P+n Q Q+n R R+ …

More Complicated Rate Laws Order=a for A b for B,-p for P, -q for Q… n A A+n B B+n C C+ …  n P P+n Q Q+n R R+ …

(a) When sulfur dioxide and oxygen are passed over hot platinum foil, they combine to form sulfur trioxide. (b) The sulfur trioxide forms dense white fumes of sulfuric acid when it comes into contact with moisture in the atmosphere. The rate law for the formation of sulfur trioxide shows that its rate of formation decreases as its concentration increases, so the sulfur trioxide must be removed as it is formed if the reaction is to proceed rapidly. 2SO 2 (g)+O 2 (g)  2SO 3 (g) SO 3 (g)+H 2 O(g)  H 2 SO 4 (l)

The determination of rate law Method of initial rates: Only one reactant is left limited, all the rest are in large excess. True rate law: By making B in large excess, [B]=[B] 0 The rate law with respect to A can be determined.

The determination of rate law Isolation method: Only one reactant is left limited, all the rest are in large excess. True rate law: By making B in large excess, [B]=[B] 0 The rate law with respect to A can be determined.

Using the method of initial rates The initial rates of the reaction of 2I(g)+Ar(g)  I 2 (g)+Ar(g) were measured as follows: [I] 0 /10 -5 molL v 0 /molL-1s - 1 a) [Ar]=1 mmolL x x x x b) [Ar]=5 mmolL x x x x c) [Ar]=10 mmolL x x x x Determine the orders of reaction with respect to I and Ar and the rate constant.

Log[I] 0 =0 Logv 0 =8.94,8.88,8.75

Integrated Rate Law Find the concentration of a reactant at time t from reaction order and initial concentration. First order reaction:

The characteristic shape of the graph showing the time dependence of the concentration of a reactant in a first-order reaction is an exponential decay, as shown here. The larger the rate constant, the faster the decay from the same initial concentration.

How to find k (first order reaction):

Half-Lives of First Order Reactions Time constant:

The half-life of a reactant is short if the first-order rate constant is large, because the exponential decay of the concentration of the reactant is then faster.

Second Order Integrated Rate Laws Prove above equation (classroom exercise).

The characteristic shapes (orange and green lines) of the time dependence of the concentration of a reactant during two second- order reactions. The gray lines are the curves for first-order reactions with the same initial rates. Note how the concentrations for second- order reactions fall away much less rapidly than those for first-order reactions do.

Second Order Integrated Rate Laws For A+B  P

First order reaction close to equilibrium

First order reaction close to equilibrium: general cases

Relaxation methods At T=T 1 At T=T 2

Exercise The H 2 O(l)  H + (aq)+OH - (aq) reaction relaxes to equilibrium with a relaxation time of 37 μ s at T=298 K and pH=7, pK w = Given that the forward reaction is first- order and the reverse is second-order overall, calculate the rate constants for the forward and reverse reactions.

An Arrhenius plot is a graph of ln k against 1/T. If, as here, the line is straight, then the reaction is said to show Arrhenius behavior in the temperature range studied. The activation energy for the reaction is obtained by setting the slope of the line equal to  E a /R.

Example T/K k/L mol -1 s The rate of the second-order decomposition reaction of CH 3 CHO: Find its activation energy and pre-exponential factor K /T ln(k/L mol -1 s -1)

More values are given in the Data Section

The variation of the rate constant with temperature for different values of the activation energy. Note that the higher the activation energy, the more strongly the rate constant varies with temperature.

This sequence of images illustrates the motion of two reactant molecules (red and green) in solution. The blue spheres represent solvent molecules. We see the reactants drifting together, lingering near each other for some time, and then drifting apart again. Reaction may occur during the relatively long period of encounter. To highlight the positions of the reactant molecules, the insets show the solvent molecules as a solid blue background.

We join this sequence of images at the moment when the reactant molecules are in the middle of their encounter. They may acquire enough energy by impacts from the solvent molecules to form an activated complex, which may go on to form products. Once again, the insets highlight the reactants by showing the solvent molecules as a solid blue background.

The reaction profile for the activated complex theory of reactions in solution, in which the reactants form an activated complex, provided they encounter each other with at least the activation energy.

A catalyst provides a new reaction pathway with a lower activation energy, thereby allowing more reactant molecules to cross the barrier and form products. Notice that E a for the reverse reaction is also lowered on the catalyzed path.

A small amount of catalyst—in this case, potassium iodide in aqueous solution—can accelerate the decomposition of hydrogen peroxide to water and oxygen. This effect is shown (a) by the slow inflation of the balloon when no catalyst is present and (b) by its rapid inflation when a catalyst is present.

The reaction between ethene, CH 2  CH 2, and hydrogen on a catalytic metal surface. In this sequence of images, we see the ethene molecule approaching the metal surface to which hydrogen molecules have already adsorbed: when they adsorb, they dissociate and stick to the surface as hydrogen atoms. After sticking to the surface, the ethene molecule meets a hydrogen atom and forms a bond. At this stage (center), a ·CH 2 CH 3 radical is attached to the surface by one of its carbon atoms. Finally, the radical and another hydrogen atom meet, ethane is formed, and the product escapes from the surface.

The internal structure of a zeolite is like a honeycomb of passages and cavities. As a result, a zeolite presents a huge surface area. It can also permit the entry and exit of molecules of a certain size into the active regions within the holes. This zeolite is ZSM-5.

The structure of a typical catalytic converter for an automobile exhaust. The gases flow through a honeycomblike porous ceramic support covered with catalyst.

The lysozyme molecule shown here is a typical enzyme molecule. Lysozyme occurs in a number of places, including tears and the mucus in the nose. One of its functions is to attack the cell walls of bacteria and destroy them. In this illustration the winding ribbon represents the long chain that makes up the molecule.

In the lock-and-key model of enzyme action, the correct substrate is recognized by its ability to fit into the active site like a key into a lock. In a refinement of this model, the enzyme changes its shape slightly as the key enters.

(a) An enzyme poison (represented by the mottled green rectangle) can act by attaching so strongly to the active site that it blocks the site, thereby taking the enzyme out of action. (b) Alternatively, the poison molecule may attach elsewhere, so distorting the enzyme molecule and its active site that the substrate no longer fits.

Case Study (a) A plot of Michaelis-Menten enzyme kinetics. At low substrate concentrations, the rate of reaction is directly proportional to substrate concentration. However, at high substrate concentrations, the rate is constant, as the enzyme molecules are “saturated” with substrate.

Case Study (b) A diagram of a synapse. The triangles represent neurotransmitters that travel from the neuron on the left to the receptors in the neuron on the right. The concentration of neurotransmitters in the synapse is controlled by enzymes.

Dopamine NH 2 OH

Case Study (c) The neurons in the human brain affect how we think and feel and how we perceive reality, including chemistry books.

Accounting for the rate laws 25.6 Elementary reactions 25.7 Consecutive elementary reactions 25.8 Unimolecular reactions

Elementary reaction: unimolecular reaction

Elementary reaction: bimolecular reaction CH 3 I(alc)+CH 3 CH 2 O - (alc)  CH 3 OCH 2 CH 3 (alc)+I - (alc) CH 3 I+CH 3 CH 2 O -  CH 3 OCH 2 CH 3 +I -

Consecutive elementary reactions

Rate Determining Step(RDS) The elementary reaction that is much slower than the rest and governs the overall reaction

The rate of a reaction is controlled by the rate-determining step (RDS). (a) If the rate-determining step is the second step, then the rate law for that step determines the rate law for the overall reaction. The orange curve shows the reaction profile for such a mechanism, with a high activation energy for the slow step. The concentrations of intermediates can usually be expressed in terms of reactants and products by taking into account the steps preceding the RDS. (b) If the rate-determining step is the first step, then the rate law for that step must match the rate law for the overall reaction. Later steps do not affect the rate or the rate law. (c) If two parallel paths lead to products, the faster one (in this case, the lower one) determines the rate of the reaction.

Example

The steady state approximation

Using steady state approximation

Pre-equilibrium

The kinetic isotope effect

The chemical equations for elementary reactions show the individual events that take place when atoms and molecules encounter one another. This illustration shows two of the steps believed to occur during the formation of hydrogen iodide from hydrogen and iodine vapor. In one, I 2  I 2  I 2  I  I, a collision between two iodine molecules results in the dissociation of one of them. In the second step, I  H 2  H  HI, one of the I atoms produced in the first step attacks a hydrogen molecule and forms a hydrogen iodide molecule and a hydrogen atom.

Many reactions occur by a series of elementary reactions. The molecularity of an elementary reaction indicates how many species are involved in that step.

In a chain reaction, the product of one reaction step is a reactant in a subsequent step, which in turn produces species that can take part in subsequent reaction steps. Initiation Propagation

Example (I)

This flame front was caught during the rapid combustion that occurs inside an internal combustion engine every time a spark plug ignites gasoline vapor. This radical chain reaction occurs in automobile engines. The products, which are hot gases, push a piston out, initiating a chain of events that ultimately moves the vehicle.

Example (II): Branching Chain Reaction

Unimolecular reactions

The activation energy of a composite reaction