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### Chemical KineticsKhadijahHanim Abdul RahmanUniMAP

ERT 108 – Physical Chemistry Semester II- 2010/2011

Subtopics

- Experimental Chemical and Kinetics Reactions
- First Order Reactions
- Second Order Reactions
- Reaction Rates and Reaction Mechanisms
- Light Spectroscopy and Adsorption Chemistry (Experimental methods for fast reactions).

ERT 108 – Physical Chemistry Semester II- 2010/2011

Experimental Chemical and Kinetics Reactions

- Rates of chemical Reactions:
- the rate of speed with which a reactantdisappears or a productappears.
- the rate at which the concentration of one of the reactants decreases or of one of the products increases with time.
- typically, mol L-1 s-1.

ERT 108 – Physical Chemistry Semester II- 2010/2011

The rates of reactions

- Depends on composition and temperature of reaction mixture.
- Definition of rate
- as the slope of the tangent drawn to the curve showing the variation of conc with time

ERT 108 – Physical Chemistry Semester II- 2010/2011

Rate of Reaction: A variable quantity

- Rate of reaction is expressed as either:
or

[ Negative value ]

[ Positive value ]

ERT 108 – Physical Chemistry Semester II- 2010/2011

A + 2B 3C + D

- Rate of consumption – (one of the reactants, A or B) at a given time is d[R]/dt, R is A/B.
- Rate of formation – (products, C/D denotes as P) at a given time is d[P]/dt.
- These rates are positive values. The +/- sign- indicate that conc is increasing/decreasing.

ERT 108 – Physical Chemistry Semester II- 2010/2011

- From the stoichiometry:
- Rate of reaction is related to rates of change of concentration of products and reactants
- Undesirability of having different rates to describe the same reaction- using the extent of reaction, ξ.
- vJ is the stoichiometric coefficient for species J.
- Rate of reaction, r

ERT 108 – Physical Chemistry Semester II- 2010/2011

- For homogenous reaction the V can be taken inside the differential, [J]=nJ/V
- For heterogeneous reaction, use the surface area (constant), A as substitution to V, σJ=nJ/A
- Common units for r = mol dm-3 s-1 or related units for homogenous reaction
- For heterogeneous reaction= mol m-2 s-1

ERT 108 – Physical Chemistry Semester II- 2010/2011

- Example 1 differential, [J]=
- The rate of change of molar conc of CH3 radicals in the reaction 2CH3(g) CH3CH3(g) was reported as d[CH3]/dt= -1.2 mol dm-3 s-1 under particular conditions. What is
(a) the rate of reaction

(b) the rate of formation of CH3CH3?

ERT 108 – Physical Chemistry Semester II- 2010/2011

Rate laws and rate constants differential, [J]=

- Rate of reaction is often proportional to conc of reactants raised to a power.
r = k[A] [B]

Each conc raised to first power.

- k is the rate constant for the reaction
- k- independent of conc but depends on temp.
- Experimentally determined equation of this kind- rate law of the reaction

ERT 108 – Physical Chemistry Semester II- 2010/2011

- Rate law differential, [J]=- an equation that expresses the rate of reaction as a function of conc of all species present in overall chemical equation
r = f([A], [B],….)

- Rate law is determined experimentally- cannot be inferred from the stoichiometry of balanced chemical equation.
- Application of rate law:
- To predict the rate of reaction from the composition of mixture
- Guide to the mechanism of the reaction- for any proposed mechanism- must be consistent with the observed rate law.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Reaction order differential, [J]=

- Many reactions are found to have rate laws of the form
r = k [A]a[B]b…….

- The power to which the conc of a species is raised in a rate law- the order of the reaction with respect to that species.
- A reaction with rate law r = k [A] [B]
is first-order in A and first-order in B.

- The overall order of a reaction, second-order overall-the sum of the individual orders.
- Some reactions obey zero-order rate law- rate that is independent of conc of the reactant. Thus,
r = k

ERT 108 – Physical Chemistry Semester II- 2010/2011

Integrated rate laws differential, [J]=

- First-order rate law
is or [A] = [A]0e-kt

Where [A]0 is the initial conc of A at t = 0

- If ln ([A]/[A]0) is plotted against t- first-order reaction will give a straight line of slope= -k.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Half lives and time constants differential, [J]=

- Useful indication of the rate of a first-order chemical reaction- half-life,t1/2of a substance.
- Half-life: time taken for the conc of reactant to fall to half its initial value.
- Time for [A] to decrease from [A]0 to ½[A]0 in first-order reaction:
kt1/2 = -ln = -ln ½ = ln 2

∴ t1/2 =

- Another indication of the rate of a first-order reaction- time constant, τ
- time required for the conc of reactant to fall to 1/e of its initial value.
kτ = -ln = -ln 1/e = 1 ∴time constant, τ = 1/k

ERT 108 – Physical Chemistry Semester II- 2010/2011

Second-order reactions differential, [J]=

- Second-order rate law:
- is or
- Where [A]0 is the initial conc of A (at t = 0)
- To plot a straight line for second order reaction- plot 1/[A] against t. the slope= k.
- The half life for second order reaction is

ERT 108 – Physical Chemistry Semester II- 2010/2011

Half-life of differential, [J]=nth-order reaction

- In general for nth-order reaction (with n > 1) of the form A products, the half-life is related to the rate constant and the initial conc of A by

ERT 108 – Physical Chemistry Semester II- 2010/2011

Zero-order, First-order, Second-order Reactions differential, [J]=

ERT 108 – Physical Chemistry Semester II- 2010/2011

Zero-order, First-order, Second-order Reactions differential, [J]=

ERT 108 – Physical Chemistry Semester II- 2010/2011

Zero-order, First-order, Second-order Reactions differential, [J]=

Zero order

First order

Second order

ERT 108 – Physical Chemistry Semester II- 2010/2011

Example 2 differential, [J]=

(a) When [N2O5] =0.44M, the rate of decomposition of N2O5 is 2.6 x 10-4 mol L-1 s-1.

- what is the value of k for this first-order reaction?
(b) N2O5 initially at a concentration of 1.0 mol/L in CCl4, is allowed to decompose at 450C. At what time will [N2O5] be reduced to 0.50M?

ERT 108 – Physical Chemistry Semester II- 2010/2011

Example 3 differential, [J]=

- The data of the above table were obtained for the decomposition reaction: A → 2B + C.
- Establish the order of the reaction.
- What is the rate constant, k?

ERT 108 – Physical Chemistry Semester II- 2010/2011

Answer (Example 3) differential, [J]=

(a) Plot graph based on the data given in the Table.

(b) The slope of the 3rd graph:

ERT 108 – Physical Chemistry Semester II- 2010/2011

Determination of the rate law differential, [J]=

- Experimental data gives species conc at various times during the reaction.
- a few methods to determine the rate law from experimental conc vs. time data.
- Consider the following:
r = k [A]a[B]b…….

it is usually the best to find the order of a, b, … first and then the rate constant, k.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Determination of rate law- half-life method differential, [J]=

- Applies when the rate law has the form r = k[A]n. then this equation and apply.
- If n = 1, then t1/2 is independent of [A]0. If n ≠1, then
gives:

- A plot of log10 t1/2 vs. log10 [A] – gives straight line of slope = n-1.

ERT 108 – Physical Chemistry Semester II- 2010/2011

- To use the half-life method for determination of rate law: differential, [J]=
- Plot [A] vs. t
- Pick any [A] value, eg. [A]’ and finds the point where [A] has fallen to ½ [A]’. The time interval between this 2 points is t1/2 for the initial conc [A]’.
- Pick another point [A]” and determines the t1/2 for this A conc.
- Repeat this process several times
- Plot log10 t1/2 vs. the log of the corresponding initial A conc and measures the slope.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Example 4 differential, [J]=

- Data for the dimerization 2A A2 of certain nitrile oxide (compound A) is ethanol solution at 40oC follow:
Find the reaction order using the half-life method

ERT 108 – Physical Chemistry Semester II- 2010/2011

Method of Initial Rates differential, [J]=

- This simple method of establishing the exponents in a
rate equation involves measuring the initial rate of reaction, ro for different sets of initial concentration.

- Suppose we measure r0 for the 2 different initial A conc [A]0,1 and [A]0,2 while keeping [B]0, [C]o,… fixed.
- With only [A]0 changed and with the rate law assumed to have form r = k[A]a[B]b…[L]l, the ratio of initial rates for run 1 and 2 is
ro,2/ro,1 = ([A]o,2/[A]0,1)n

Example 5 differential, [J]=

- The data of three reactions involving S2O82- and I- were given in the below table.
(i) Use the data to establish the order of reaction:

with respect to S2O82-, the order with respect to I- & the overall order.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Example 5 differential, [J]=

(ii) Determine the value of k for the above reaction.

(iii) What is the initial rate of disappearance of in a S2O82- reaction in which the initial concentrations are [S2O82- ] =0.050M & [I-]=0.025M?

(iv) What is the rate of formation of SO42- in Experiment 1?

ERT 108 – Physical Chemistry Semester II- 2010/2011

The temperature dependence of reaction rates differential, [J]=

- Chemical reaction begins with a collision between molecules of A and molecules of B.
- Chemical reactions tend to go faster at higher temperature.
- slow down some reactions by lowering the temperature.
- Increasing the temperatureincreases the fraction of the molecules that have energies in excess of the activation energy.
- this factor is so important that for many chemical reactions it can lead to a doubling or tripling of the reaction rate for a temperature increase of only 100C.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Reaction rates: Effect of temperature differential, [J]=

- In 1889, Arrhenius noted that the k data for many reactions fit the equation:
where A & Ea are constants characteristics of the reaction & R = the gas constant.

- Ea – the Arrhenius activation energy (kJ/mol or kcal/mol)
- A – the pre-exponential factor (Arrhenius factor).
- the unit of A is the same as those of k.
- Taking log of the above equation:

ERT 108 – Physical Chemistry Semester II- 2010/2011

Reaction rates: Effect of temperature differential, [J]=

- If the Arrhenius equation is obeyed:
- a plot of ln k versus 1/Tis a straight line with slope = (-Ea/R) and A is the intercept of the line at 1/T = 0.
- This enables Ea and A to be found.
- Another useful equation:
(eliminate the constant A).

- T2 and T1 - two kelvin temperatures.
- k2 and k1 - the rate constants at these temperatures.
- Ea – the activation energy (J/mol)
- R – the gas constant (8.314 Jmol-1 K-1).

ERT 108 – Physical Chemistry Semester II- 2010/2011

- The activation energy, E differential, [J]=a: the minimum kinetic energy that reactants must have in order to form products.
- The pre-exponential factor, A: a measure of the rate at which collisions occur irrespective of their energy.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Reaction Mechanisms differential, [J]=

- The mechanism of reaction is the sequence of elementary steps involved in a reaction
- Most reactions occur in a sequence of steps called elementary reactions.
- A mechanism is a hypothesis about the elementary steps through which chemical change occurs.
- A typical elementary reaction is
H + Br2 HBr + Br

- Chemical equation for elementary reaction: equation only represents the specific process occurring to individual molecules
- Molecularity of an elementary reaction is the no of molecules coming together to react in an elementary reaction.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Reaction Mechanisms differential, [J]=

- Elementary processes in which a single molecule dissociates (unimolecular) or two molecules collide (bimolecular) much more probable than a process requiring the simultaneous collision of three bodies (termolecular).
- All elementary processes are reversible and may reach a steady-state condition. In the steady state the rates of the forward & reverse processes become equal. The concentration of some intermediate becomes constant with time.
- One elementary process may occur much more slower than all the others. In this case, it determines the rate at which the overall reaction proceeds & is called the rate-determining step.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Rate laws and equilibrium constants for elementary reactions differential, [J]=

- An overall reaction occurs as a series of elementary steps
- These steps constituting the mechanism of reaction
- This section consider the rate law for elementary reaction.
- Consider a bimolecular elementary reaction
A + B products, the rate of reaction, will be proportional to ZAB, the rate of A-B collisions per unit of time.

- ∴r for an elementary bimolecular ideal-gas reaction will be r = k[A][B]

ERT 108 – Physical Chemistry Semester II- 2010/2011

- For differential, [J]=unimolecular ideal-gas reaction B products, fixed probability that any particular B molecule will decompose/isomerize to products per unit time.
- The rate of reaction, r = k[B]
- Similar considerations apply to reactions in ideally/ideally dilute solution.
- In summary, in an ideal system, the rate law for the elementary reaction aA + bB products is
r = k[A]a[B]b, where a + b is 1,2 or 3.

- For an elementary reaction, the orders in the rate law equal the coefficients of the reactants.

ERT 108 – Physical Chemistry Semester II- 2010/2011

- Relation differential, [J]= between the equilibrium constant for a reversible elementary reaction and rate constants for forward and reverse reactions.
- Consider the reversible elementary reaction
aA + bB⇌ cC + dD

- rate laws for forward (f) and back (b) elementary reactions are rf = kf [A]a[B]b and rb = kb [C]c[D]d.
- At equilibrium, these opposing rates are equal:
rf,eq = rb,eq or

kf([A]eq)a([B]eq)b = kb([C]eq)c([D]eq)d and

- = Kc= kf/kb

kf

kb

ERT 108 – Physical Chemistry Semester II- 2010/2011

(b) The rate-determining-step approximation differential, [J]=

- In rate-determining step approximation- reaction mechanism assumed to consist 1 or more reversible reactions that stay close to equilibrium during most of the reaction.
- followed by relatively slow rate-determining step then in turn followed by 1 or more rapid reactions.
- As an example:
A⇌B⇌C⇌D

where step 2 (B⇌C)- assumed to be rate-determining step.

- For this assumption to be valid- k-1 >> k2

k1

k3

k2

k-1

k-2

k-3

ERT 108 – Physical Chemistry Semester II- 2010/2011

- The differential, [J]=slow rate of B C compared with B A – ensures that most B molecules go back to A rather than to C- ensuring that step 1 (A ⇌ B) remain close to equilibrium.
- k3 >> k2 and k3 >> k-2, to ensure that step 2 acts as ‘bottleneck’ and product D is rapidly formed from C.
- The overall rate is controlled by the rate-limiting step B C.
- Since we are examining rate of the forward reaction
A D, we further assume that k2[B] >> k-2[C].

- During early stage- the conc of C will be lower than B- this condition will hold. Thus, we neglect reverse reaction for step 2.

ERT 108 – Physical Chemistry Semester II- 2010/2011

- The relative magnitude of k1 compared with differential, [J]=k2 is irrelevant to the validity of the rate-determining-step approximation.
- ∴ the rate constant k2 of the rate-determining step might be larger than k1.
- However, the rate r2 = k2[B] of the rate-determing step must be smaller than r1=k1[A] of the first step.
- This follows from k2<<k-1 and k1/k-1 [B]/[A] (the conditions for step 1 –near equilibrium).
- For reverse overall reaction, the rate-determining step is the reverse of that for forward reaction.
- For example: the rate-determining step is C B. So,
k-2 << k3 (ensures that step D⇌C is in equilibrium) and k-1>> k2 (ensures that B A is rapid).

ERT 108 – Physical Chemistry Semester II- 2010/2011

The rate determining-step approximation differential, [J]=

- The rate law for the Br- - catalyzed aqueous reaction
H+ +HNO2 + C6H5NH2 C6H5N2+ + 2H2O

is observed to be

r = k[H+][HNO2][Br-] …..(1)

A proposed mechanism is

H+ + HNO2⇌ H2NO2+ rapid equilib

H2NO2+ + Br- ONBr + H2O slow ……(2)

ONBr + C6H5NH2 C6H5N2+ +H2O + Br- fast

Deduced the rate law for this mechanism and relate the observed rate constant, k in (1)to the rate constants in assumed mechanism (2).

k1

k-1

k2

k3

ERT 108 – Physical Chemistry Semester II- 2010/2011

- The second step in (2) is rate limiting. Since step 3 is much faster than 2, we can take d[C6H5N2+]/dt as = rate of formation of ONBr in step 2. therefore, the reaction rate is
r = k2[H2NO2+][Br-] …..(3)

(since step 2 is an elementary reaction, its rate law is determined by its stoichiometry. The species H2NO2+ in (3) is a reaction intermediate and we want to express r in terms of reactants and products. Since step 1 is in near equilib, gives:

Kc,1 = k1/k-1 =[H2NO2+]/[H+][HNO2]

And [H2NO2+] = (k1/k-1) [H+][HNO2]

Substitute in (3) gives

r = (k1k2/k-1)[H+][HNO2][Br-]

k = k1k2/k-1 = Kc,1k2

ERT 108 – Physical Chemistry Semester II- 2010/2011

The steady-state approximation much faster than 2, we can take d[C

ka

kb

- Multistep reaction mechanisms usually involve 1 or more species that do not appear in overall equation.
A I P

- After an initial induction period,[I] will start at 0, rise to max, [I]max, and then fall back to 0.
- During the major part of the reaction, the rates of change of conc of all reaction intermediates are negligibly small, therefore d[I]/dt = 0 for each reaction intermediate.
- The steady-state approximation assumes that rate of formation of reaction intermediate = rate of destruction.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Steady-state approximation much faster than 2, we can take d[C

- Apply steady-state approximation to the mechanism for
H+ + HNO2 + C6H5NH2 C6H5N2+ + 2H2O

Given in the preceding exercise to find the predicted law.

Br-

ERT 108 – Physical Chemistry Semester II- 2010/2011

- To apply for rate-determining step approximation much faster than 2, we can take d[C:
(a) Take the reaction rate, r = rate of determining step (divided by the stoichiometric no srds of rate-determining step, if srds≠ 1.

(b) Eliminates the conc of any reaction intermediates that occur in the rate expression obtained in (a) by using equilibrium-constant expressions.

- To apply steady-state approximation:
(a) Take the reaction rate, r = rate of formation of product

(b) Eliminate the conc of any reaction intermediates that occur in (a) by using d[I]/dt = 0 to find the conc of each I

(c) If step (b) introduces conc of other I, apply d[I]/dt = 0 to eliminate their conc.

ERT 108 – Physical Chemistry Semester II- 2010/2011

The Hydrogen-Iodine Reaction much faster than 2, we can take d[C

H2 (g) + I2 (g) → 2HI (g)

- Rate of formation of HI = k [H2][I2]
- The hydrogen-iodine reaction is proposed to be a two-step mechanism [Sullivan J. (1967). J.Chem.Phys.46:73].
- 1st step: iodine molecules are believed to dissociate
into iodine atoms.

- 2nd step: simultaneous collision of two iodine atoms
and a hydrogen molecule.

(this termolecular step is expected to occur

much more slowly – the rate-determining step).

ERT 108 – Physical Chemistry Semester II- 2010/2011

The Hydrogen-Iodine Reaction much faster than 2, we can take d[C

1st step: [Fast]

2nd step: [Slow]

Net:

- If the reversible step reaches a steady state condition:
- rate of disappearance of I2 = rate of formation of I2

ERT 108 – Physical Chemistry Semester II- 2010/2011

The Hydrogen-Iodine Reaction much faster than 2, we can take d[C

- For the rate-determining step:
Rate of formation of HI

= k3 [I]2[H2]

= K[H2][I2] (K=k1k3/k2)

ERT 108 – Physical Chemistry Semester II- 2010/2011

Example 6 much faster than 2, we can take d[C

- The thermal decomposition of ozone to oxygen:
2O3 (g) → 3O2 (g)

- The observed rate law:
Rate of disappearance of O3 =

- Show that the following mechanism is consistent with this experiment rate law.
1st:

2nd:

ERT 108 – Physical Chemistry Semester II- 2010/2011

Experimental methods for fast reactions much faster than 2, we can take d[C

- Many reactions are too fast to follow by the classical methods.
- Several ways to study fast reactions :

ERT 108 – Physical Chemistry Semester II- 2010/2011

Experimental methods for fast reactions much faster than 2, we can take d[C

- Continuous flow system
- Liquid phase:
- Reactant A & B are rapidly drive into the mixing chamber M by pushing in the plungers of the syringes.
- Mixing occurs in 0.5 – 1ms.
- The reaction mixture then flows through the narrow observation tube, where one measures the light absorption at a wavelength (at which one species absorbs to determine the concentration of that species).

ERT 108 – Physical Chemistry Semester II- 2010/2011

Experimental methods for fast reactions much faster than 2, we can take d[C

- For gas phase reaction, the syringes are replaced by bulbs of gases A & B.

Figure: A continuous flow system with rapid mixing of reactants.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Experimental methods for fast reactions much faster than 2, we can take d[C

- Stopped flow method:
- the reactants mixed at M & rapidly flow through the observation tube into the receiving syringe, driving its plunger against a barrier & thereby stopping the flow.
- this plunger hits a switch which stops the motor driven plungers & triggers the oscilloscope sweep.
- One observes the light absorption at P as a function of time.
- The continuous flow & stopped flow methods are applicable to reactions with half-lives in the range of 0.001 to 10s.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Experimental methods for fast reactions much faster than 2, we can take d[C

Figure:

A slopped-flow system

Figure:

A flash-photolysis experiment.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Experimental methods for fast reactions much faster than 2, we can take d[C

- Relaxation methods:
- Take a system in reaction equilibrium & suddenly change one of the variables that determine the equilibrium position.
- A limitation on relaxation methods – the reaction must be reversible, with detectable amounts of all species present in equilibrium.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Experimental methods for fast reactions much faster than 2, we can take d[C

- Rapid flow & relaxation method have been used to measure the rates of proton transfer (acid-base) reactions, complex-ion-formation reactions, ion-pair-formation reactions & enzyme-substrate-complex formation system.
- Relaxation methods apply rather small perturbations to a system & do not generate new chemical species.
- The flash-photolysis and shock tube methods apply a large perturbation to a system, thereby generating one or more reactive species whose reactions are then followed.
- NMR spectroscopy is used to measure the rates of certain rapid isomerization & exchange reactions.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Answer (Example 6a) much faster than 2, we can take d[C

- Tabulate the data as follows.
- Construct the Arrhenius plot of log10k versus 1/T for the reaction.
- Intercept (log10A)=13.5
A = 3x1013s-1

- Slope=-5500K,
Ea=25kcal/mol

=105 kJ/mol

Figure: Arrhenius plot of log10 k versus 1/T for this reaction. Note: the long extrapolation needed to find A.

ERT 108 – Physical Chemistry Semester II- 2010/2011

Answer (Example 6b) much faster than 2, we can take d[C

- Based on the given info:
- k2= 2k1 ,
- T1 = room temperature (298K), T2=298+10 = 308K,
- The Arrhenius equation:
- Substitute:
Ea= 53 kJ/mol

ERT 108 – Physical Chemistry Semester II- 2010/2011

Answer (Example 7) much faster than 2, we can take d[C

- Assume the 1st step reaches the steady state condition:
Rate of formation of O = Rate of disappearance of O

k1 [O3] = k2 [O2] [O]

- Assume the 2nd step is the rate-determining step:
Rate of disappearance of O3 = k3 [O][O3]

(where k = k1k3/k2)

ERT 108 – Physical Chemistry Semester II- 2010/2011

Apparatus of the determining the rate of decomposition of N much faster than 2, we can take d[C2O5

ERT 108 – Physical Chemistry Semester II- 2010/2011

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