Chemical Kinetics By Prof. Sanaa Taher Arab Physical Chemistry

Chemical Kinetics By Prof. Sanaa Taher Arab Physical Chemistry

Reaction Kinetics Thermodynamics: spontaneous reactions (possible to take place) Thermodynamics: can’t answer (1) how to make them happen (2) how fast they will take place (3) the reaction mechanism For example

Reaction Kinetics -16.63

Reaction Kinetics T, P, catalyst T, catalyst
Chemical Kinetics Studies the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). A reacting system is not in equilibrium T, P, catalyst T, catalyst

Chemical Kinetics Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = - D[A] Dt D[A] = change in concentration of A over time period Dt rate = D[B] Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative.

Reaction Rates Rxn Movie Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. [A] vs t

A B time rate = - D[A] Dt rate = D[B] Dt

Outline: Kinetics Reaction Rates How we measure rates. Rate Laws
How the rate depends on amounts of reactants. Integrated Rate Laws How to calc. amount left or time to reach a given amount. Half-life How long it takes to react 50% of reactants. Arrhenius Equation How rate constant changes with T. Mechanisms Link between rate and molecular scale processes.

Reaction Rates High reaction speed :
1. Ba Cl2 + Na2 SO Ba SO Na Cl 2. OH H H2O Low reaction speed : O H H2O Also, Iron roast needs very long time to occurred.

Measurement of Reaction Rates
The concentration of a reactant or product as a function of time Cooling the sample removing a catalyst Diluting the mixture Adding a species Chemical Method At constant T At intervals Slows down or stop the reaction Reaction vessels Rapidly analyze chemical compositions of the mixture

Measurement of Reaction Rates
Physical Method (More accurate and less tedious) Measures a physical property of the reaction system as a function of time ( change in : absorbance , gas evolution , density ,pH , conductance and pressure . This allows the reaction to be followed continuously as it proceeds.

Chemical Kinetics The rates of reactions are affected by
Chemical kinetics - speed or rate at which a reaction occurs: The rates of reactions are affected by 1- Reactant concentration. 2- Temperature. 3- Reactant states. 4- Catalysts.

Factors affecting the reaction rates
1- Reactant concentration. The dependence of the rate of the reaction on concentration, called the order of the reaction, the rate equation summarizes the dependence of the rate on the concentrations of substances that affecting the rate of reaction , this expression involves the rate constant, k which is a constant of proportionality linking the rate terms. The order of any reaction with the various substances concentration, can be calculated from the

Following equation : R α CAm. CBn R = K CAm
Following equation : R α CAm . CBn R = K CAm . CBn Where the order will be (n + m ) ,the order can be integrals (1,2,3….) or part of it as (1/2,2/3 ,3/4 ) . The reaction equation does not express the reaction order as can be seen from the following examples : 1- The decomposing of N2O5 2N2O5 4NO2 + O2 The order equation will be first order as :

-dC(N2O5)/dt = K[N2O5] 1 2- The decomposing of NO2 is a second order inspite of that the decomposing of both N2O5 & NO2 have the same shape 2NO NO +O2 -dC(NO2)/dt = K[NO2]2 3- The reaction between triethayl amine and ethyl bromide is a second order reaction ,first for each : (C2H5)3N C2H5 Br (C2H5)4NBr The reaction rate equation is as follows : -dC (C2H5Br)/ dt = K[C2H5Br] [(C2H5)3N]

The reaction order will be ( 3/2 ) .
4- The decomposing of acytaldhyde occurred as follows : CH3CHO CH4 + CO The reaction rate equation is : -dC(CH3CHO)/dt = K[CH3CHO]3/2 The reaction order will be ( 3/2 ) . If the order of the reaction = 0 ,the concentration of the reactants does not effect the reaction rate .

2- Temperature. Increase in temperature generally increases the rate of reaction. Knowledge of just exactly how temperature affects the rate constant can give information leading to a deeper understanding of how reactions occur. Chemical reaction speed up when the temperature is increased. －molecules must collide to react －an increase in temperature increases the frequency of intermolecular collisions.

The following (Arrhenius equation) will explain the effect of temperature :
k = A • exp( -Ea/RT ) R : is the gas constant(8.314J/Kmol) E : is the activation energy (J/mol) A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation T : is the temperature in Kelvin for reaction. k :is the rate constant , k is determined experimentally at several temperatures

Activation energy طاقة التنشيط A + B C + D Endothermic reaction Exothermic reaction The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.

Activation energy

Arrhenius Equation y = mx + b
Taking the natural logarithm of both sides, the equation becomes y = mx + b When k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. 1/T.

The effect of temperature on the rate of the reaction and activation energy
Arrhenius Equation ln k = ln A - Ea/RT ln k2 k1 = Ea R - 1 T2 T1

ln k2 k1 = Ea R - 1 T2 T1 ln k1 = 50.2 R - 1 T2 298

3- Reactant states In order to react, molecules must come in contact with each other. The more homogeneous the mixture of reactants, the faster the molecules can react. The nature and kind of the reactants will produce different rates to the reaction, as an example the reduction of MnO4– by Fe +2 occurred very fast ,while the reduction of MnO4– by H2C2O4 is low according to the following : Fe MnO Fe Mn+2 2H2C2O4 + 2MnO4- + 6H H2O + 4CO Mn+2

4- Catalysts Catalysts increase the rate of a reaction by decreasing
the activation energy of the reaction. Catalysts speed up reactions by changing the mechanism of the reaction, by which the process occurs. Catalysts are not consumed during the course of the reaction. Homogenous and heterogeneous catalysts are defend as a catalyst in the same phase (gases and solutions) as the reactants is a homogeneous catalyst.

When the catalyst is in a different phase than reactants (and products), the process involve heterogeneous catalysis. Chemisorption, absorption, and adsorption cause reactions to take place via different pathways. Platinum is often used to catalyze hydrogenation in heterogeneous catalysis. Molecules catalyzed by enzymes are called substrates. They are held by various sites (together called the active site) of the enzyme molecules and just before and during the reaction.

Molecularity and order of a reaction
The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process.

The molecularity of a process tells how many molecules are involved in the process.
The rate law of elementary reaction is derived from the equation. The order is the number of reacting molecules because they must collide to react. A molecule decomposes by itself is a unimolecular reaction (step); two molecules collide and react is a bimolecular reaction (step); & three molecules collide and react is a termolecular reaction (step).

NO2 + NO2 NO3 + NO rate = k [NO2]2
O O2 + O rate = k [O3] NO2 + NO NO3 + NO rate = k [NO2]2 Br + Br + Ar Br2 + Ar* rate = k [Br]2[Ar] Caution: Derive rate laws this way only for elementary reactions 1. Molecularity and order some-times are the same as in the decomposing of hydrogen iodide ( equal 2 , Bi-molecular & Second- Order ) i) 2HI H2 + I2 2. Molecule decomposes or inter- conversion by itself is a unimolecular reaction as : ii a) Br Br

ii-b) The inter-conversion of Malic acid to Fumiric acid :
iii) In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step. NO2 (g) + CO (g) NO (g) + CO2 (g) The rate law for this reaction is found experimentally to be Rate = k [NO2]2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps

A proposed mechanism for this reaction is
- Slow Initial Step A proposed mechanism for this reaction is Step 1: NO2 + NO NO3 + NO (slow) Step 2: NO3 + CO NO2 + CO2 (fast) The NO3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. For the reaction :

A proposed mechanism is
The rate law for this reaction is found (experimentally) to be Because ter-molecular processes are rare, this rate law suggests a two-step mechanism. A proposed mechanism is Step 1 is an equilibrium- it includes the forward and reverse reactions.

The rate law for the slow step (the rate determining step) should be The rate of the overall reaction depends upon the rate of the slow step. Distinguish between order of a reaction and molecularity of a reaction. The main differences between the order of a reaction and molecularity of a reaction are summarized as follows:

MOLECULARITY OF A REACTION
ORDER OF A REACTION MOLECULARITY OF A REACTION It is the sum of the exponents of the concentrations in the rate law equation. It is the number of atoms, ions or molecules that must collide with one another simultaneously so as to result into a chemical reaction. It need not be a whole number (integrals) i.e. it can be fractional as well as zero. It is always a whole number. It can be determined experimentally only and cannot be calculated. It can be calculated by simply adding the molecules of the slowest step. It is for the overall reaction and no separate steps are written to obtain it. The overall molecularity of a complex reaction has no significance. It is only slowest step whose molecularity has significance for the overall reaction. Even the order of a simple reaction may not be equal to the number of molecules of the reactants as seen from the unbalance equation. For simple reactions, the molecularity can usually be obtained from the Stoichiometry of the equation.

Rate Laws For many reactions
k: rate constant (rate coefficient), k = f (T, P) , , : order or partial order + +  +  n : overall order shows the relation between the order and K

Reaction order The unites reaction constant (K) ML-1 Sec-1 n = 0
L1/2 M-1/2 Sec-1 n = 3/2 L M-1 Sec-1 n = 2 L2 M-2 Sec-1 n= 3

Concentration and Rate
To determine the rate law we measure the rate at different starting concentrations. Compare Experiments 1 and 2: when [NH4+] doubles, the initial rate doubles.

Concentration and Rate
Likewise, compare Experiments 5 and 6: when [NO2-] doubles, the initial rate doubles.

Rate Laws A rate law shows the relationship between the reaction rate and the concentrations of reactants. For gas-phase reactants use PA instead of [A]. k is a constant that has a specific value for each reaction. The value of k is determined experimentally. “Constant” is relative here k is unique for each rxn

Rate Laws Exponents tell the order of the reaction with respect to each reactant. This reaction is First-order in [NH4+] First-order in [NO2−] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall.

Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) [C4H9Cl] M In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times, t.

Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) Average Rate, M/s The average rate of the reaction over each interval is the change in concentration divided by the change in time:

Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) Note that the average rate decreases as the reaction proceeds. This is because as the reaction goes forward, there are fewer collisions between reactant molecules.

Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) A plot of concentration vs. time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time.

Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1. Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH. Rate = -[C4H9Cl] t = [C4H9OH]

Chemical Kinetics Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82- (aq) + 3I- (aq) 2 SO42- (aq) + I32- (aq) rate = k [S2O82-]x[I-]y Initial Rate (M/s) [I-] [S2O82-] Experiment No 2.2 x 104- 0.034 0.08 1 1.1 x 104- 0.017 2 0.16 3

Integrated Rate Laws 1- First-Order: i) Due to reactants
Consider a simple 1st order rxn: A B Differential form: How much A is left after time t? Integrate:

The integrated form of first order rate law:
Can be rearranged to give: [A]0 is the initial concentration of A (t=0). [A]t is the concentration of A at some time, t, during the course of the reaction.

which is in the form y = mx + b If a reaction is first-order, a plot of ln [A]t vs. t will yield a straight line with a slope of -k. So, use graphs to determine rxn order.

1- First-Order: ii) Due to products
Consider a simple 1st order rxn where (a) is the initial concentration and X the concentration of the products at time : R = dx/dt = K(a-x) dx /(a-x) = Kdt The integrated form will be : -ln(a-x) = Kt + K- Where K΄ integration constant ,it calculated as (at t=o , X=0 ) - ln a = K΄ and the following equation will obtained: ln (a/a-x) = Kt Which is the same for the first order . Differential form is :

Half life & k of First Order
The time required for half of A to decompose is called half life t1/2. Since [A] = [A]o e – k t or ln [A] = ln [A]o – k t When t = t1/2, [A] = ½ [A]o Thus ln ½ [A]o = ln [A]o – k t1/2 – ln 2 = – k t1/2 k t1/2 = ln 2 =  relationship between k and t1/2 Radioactive decay usually follow 1st order kinetics, and half life of an isotope is used to indicate its stability. Evaluate t½ from k or k from t½

N2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s. Solution: Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t = 30 s or = 1.0 e – k t apply [A]o = [A] e– k t ln 0.9 = ln 1.0 – k 30 s – = 0 – k * k = s – t½ = / k = 197 s apply k t ½ = ln 2 [A] = 1.0 e – *500 = Percent decomposed: 1.0 – = or 82.7 % After 2 t½ (2*197=394 s), [A] = (½)2 =¼, 75% decomposed. After 3 t½ (3*197=591 s), [A] = (½)3 =1/8, 87.5% decomposed.

First-Order Processes ,Examples
1- Consider the process in which methyl isonitrile is converted to acetonitrile. CH3NC CH3CN How do we know this is a first order rxn? This data was collected for this reaction at 198.9°C. Does rate=k[CH3NC] for all time intervals?

When ln P is plotted as a function of time, a straight line results
When ln P is plotted as a function of time, a straight line results. The process is first-order. k is the negative slope: 5.1  10-5 s-1.

2- N- chloro acetanilide is converted to P- chloro acetanilide according to first order reaction .
The proceeding of the reaction is followed by adding Iodine solution to specific amount of the studied reaction and the obtained solution is calibrated by standard solution of Na2S2O3 (Iodine can replace the chloride only in compound 1) Then the application of eq. log (a/a-x) = [Kt ]/ to calculate K of the reaction .

3-Radioactive Decay Radioactive Samples decay according to first order kinetics. This is the half-life of samples containing e.g. 14C , 239Pu, 99Tc. Example The half-life of 39Sr 38 is 20 years ,what is the time for 90% of Sr to decay . t1/2 = 0/693/K K= 0.693/20= Year-1 K = 2.303/t log a/(a-x) t = 2.303/ log 100/(100-90) = Year.

H2O2 → H2O + 1/2 O2 Time (min. ) 5 10 20 30 50 37.1 29.8 19.6 12.3 5.0

Typical Problem the 1st Order Reaction
The decomposition of A is first order, and [A] is monitored. The following data are recorded: t / min [A]/[M] Calculate k (What is the rate constant? k = ) Calculate the half life (What is the half life? Half life = 13.89) Calculate [A] when t = 5 min. (What is the concentration when t = 5 min?) Calculate t when [A] = (Estimate the time required for 90% of A to decompose.)

A Pseudo-first Order Reaction
Example : 1) hydration of alkyl hyliede ; methyl iodide CH3I(aq) + H2O(l) CH3OH(aq) + H+(aq) + I-(aq) Rate = k [CH3I] [H2O] Since the reaction is carried out in aqueous solution [H2O] >>>> [CH3I] \ [H2O] doesn’t change by a lot; The molarity of water in aqueous solution will be = 1000/18 = M , so the concentration of H2O is essentially constant R = k [CH3I] [constant] = k` [CH3I] where k` = k [H2O] The reaction is pseudo first order since it appears to be first order, but it is actually a second order process.

Second-Order Processes
For a reaction A + B products If the rate of interaction proportional to the first exponential of concentration of both interacting substances A&B , the interaction is of the second degree, so ,supposed there is an interaction with the primary concentration of substance is (a) and for the other is (b) so for the equation of rate of interaction there are two possibilities : 1- Both interacting substances of the same concentration which means a = b and the rate of the reaction proportional to result of this interaction. 2- Both interacting substances of different concentration which means a  b and rate of interaction is proportional to result of this interaction

1) For the first case a=b, integrating the rate law for a process that is second-order in reactant A: Rearrange, integrate: also in the form y = mx + b So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line with a slope of k.

Determining rxn order The decomposition of NO2 at 300°C is described by the equation NO2 (g) NO (g) + 1/2 O2 (g) and yields these data: Time (s) [NO2], M 0.0 50.0 100.0 200.0 300.0

Graphing ln [NO2] vs. t yields:
The plot is not a straight line, so the process is not first-order in [A]. Time (s) [NO2], M ln [NO2] 0.0 -4.610 50.0 -4.845 100.0 -5.038 200.0 -5.337 300.0 -5.573 Does not fit:

A graph of 1/[NO2] vs. t gives this plot.
Time (s) [NO2], M 1/[NO2] 0.0 100 50.0 127 100.0 154 200.0 208 300.0 263 This is a straight line. Therefore, the process is second-order in [NO2].

A 2nd Order Example Work out the formulas and then evaluate values
—— – —— = k t [B] [B]o —— – —— [B] [B]o t = ———————— k [B]o [B] = —————— [B]o k t + 1 t = [B] = Work out the formulas and then evaluate values

Half-Life Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t1/2 is one-half of the original [A], [A]t = 0.5 [A]0 For a second-order process, set

Half life of 2nd Order Chemical Kinetics
1/[B] t = [B] = How does half life vary in 2nd order reactions?

2) If concentration of substance A is a, and concentration of substance B is b and as a b ,If the resulting part of interaction at is x  The remaining part of the substance A is : (a – x) and the remaining part of substance B is : (b – x ) By replacement in equation of rate of interaction of the second order at a b ,then : dx/dt = K (a-x) (b-x) ordering the equation : dx/(a-x) (b-x) = K dt using partial integration

Considering (b – x) = constant = B, the equation is interpreted to
dx/dt = KB(a-x) dx/(a-x) = KB dt By integration : ∫dx/(a-x) = ∫KB dt -ln(a-x) = KBt + K- When x = 0 so t = 0 K = ln a ln a/(a-x) = KBt Considering (a- x) constant = A the equation is interpreted to the form dx/dt = KA (b-x) dx/(b-x) = KAdt ln b/(b-x) = Kat By subtracting this equation from equation ( ) and replacement in A, B we found : ln b/(b-x) – ln a/(a-x) = Kat – KBt ln b(a-x)/ a(b-x) = (a-b) Kt LM-1 Sec-1

Deriving a rate law from a mechanism - 1
The decomposition of H2O2 in the presence of I– follow this mechanism, i H2O2 + I–  k1® H2O + IO– slow ii H2O2 + IO–  k2® H2O + O2 + I– fast What is the rate law? Solution The slow step determines the rate, and the rate law is: rate = k1 [H2O2] [I –] Since both [H2O2] and [I –] are measurable in the system, this is the rate law ,the reaction order will be 2 .

Examples for second order reaction :
(1) Saponification of ethyl acetate This interaction could be done in present of hydroxyl ions (OH) in small amounts to affect is concentration, so the degree of interaction is the second degree, while presence of large amounts of hydroxyl so its concentration is not affected will be first degree interaction and the interaction equation could be written as follows :

CH3COOC2H5+OH- → CH3COO- + C2H5OH
Following the progress of interaction by taking known amounts of the interaction mix in successive periods and their calibration with standard acid, estimation of interaction constant by drawing the relationship between : x/(a-x) # t . K is the slope of the resulting straight line.

The interaction equation could be written : 2Hl + H2O2 l2 + 2H2O
(2) Oxidation of Iodide hydrogen acid using hydrogen peroxide in acidic media : The interaction equation could be written : 2Hl + H2O l2 + 2H2O This second degree interaction occurred in many steps as follows : l H2O H2O + lO- H lO HIO HlO + H+ + l H2O + l2 The first step is the slow step which determine the speed of interaction, degree and so the equation of speed of interaction could be written as follows R= K [ H2O2 [[HI [= Rate of reaction

A Pseudo-Second Order Reaction :
Interactions which are not really of the second degree but being that under special conditions and their degrees changes with changes of these conditions are termed false double interactions, for example the effect of alcohol on organic acids in excess amounts of alcohol, as in the following equation : R – COOH + R΄ OH R - COOR΄ + H2O In this interaction the speed of interaction depends on concentration of the organic acid (RCOOH) and this acid perform the role of catalyst for this reaction and its concentration changed during the interaction and so change the speed of interaction as follows :

Rate of interaction constant x [R-COOH]2
The interaction of second degree although participation of more than two molecules in the interaction at each moment. When the metal acid is added to perform the role of catalyst and its concentration remains constant with change during the interaction the degree of this interaction become of the first degree.

Integrated Rate Law - zero order
If the order of the reaction = 0 ,the concentration of the reactants does not effect the reaction rate . rate = k [A]0 = k Where [A]0 is the initial concentration at time t=0 and [A] the concentration at time = t .The application of the following eq. will produce a straight line. [A] - [A]0 = -kt

To calculate the half life time at n= 0
The zero reaction order depends usually on other factors such pressure ,the catalysis used or as on the photo reactions, To calculate the half life time at n= 0 t½ = t when [A] = [A]0/2 ,so [A]0 t½ = 2k

Integrated Rate Law – Third-Order Reactions
For the reaction: A + B + D products So, R = K CA . AB . CD Three cases will be found : i- All interacting substances of the same concentration which means a = b = d and the rate of the reaction proportional to result of this interaction. R= dx/dt = K (a-x)3 ii- Two of the interacting substances of same concentration and the third one has different concentration which means : a =b  d and rate of interaction is proportional to result of this interaction R= dx/dt = K (a-x)2(d-x)

R= dx/dt = K (a-x) (b-x) (d-x)
iii. The three interacting substances of different concentrations which means a  b  d and rate of interaction is proportional to result of this interaction R= dx/dt = K (a-x) (b-x) (d-x) Concerning the first case: all interacting substances of the same concentration ,then R = K CA . AB . CD CA = CB = CD = a dx/dt = K (a-x)3 by integrating the following equation : ∫dx/(a-x)3 = ∫ K dt 1/2(a-x)2 = Kt + K-

where K΄ is the interacting constant , to find the value of K΄ at, t = o , x = o ,so,
By substituting with K΄ value in the integrated equation, so, 1/2(a-x)2 = Kt + 1/2a2 Kt = 1/2(a-x)2 – 1/2a2 K = 1/2t (1/(a-x)2 – 1/a2) L2M-2Sec-1 The oxidation of nitric oxide with X2 , where X2 = O2 ; Br2 ; or Cl2 is a good example for the third order reaction as in the following equations :

The rate equation for the previous reactions will be as follows
2NO + X2 → 2NOX 1) 2NO + O2 → 2NO2 2) 2NO + Br2 → 2NOBr 3) 2NO + Cl2 → 2NOCl The rate equation for the previous reactions will be as follows -d(NO) / dt = K(NO)2(X2) Half-life for the third can be determined from the following equation when : a = b = d : K=1/2 t1/2 [1/(a-a/2)2 – 1/a2] = 1/2 t1/2.3/a2 t1/2 =3/2 1/Ka2 Sec.

Reaction Orders Zero-order First-order Second-order Third-order
No simple order

Outline: Kinetics t1/2=(3/2)1/Ka2 First order Second order Third order
Rate Laws Integrated Rate Laws complicated Half-life t1/2=(3/2)1/Ka2

nth-Order Reaction For any reaction of the order n ,the rate equation can be illustrated as: R α Cn R = K (a – x )n or dx/dt = K (a – x )n To integrate this equation two conditions must be presented : i) The order of the reaction must be n  1 ii) All reacting substances have the same concentration which means a = b = d etc… and the rate of the reaction proportional to result of this concentration . dx/(a-x)n = K dt

1/ (n-1)(a-x)(n-1) = Kt + (n-1) a(n-1)
So ,by the separation of the variables will give ∫ dx/ (a- x)n = ∫K dt By integration concerning that n  1 1/ (n -1)(a-x)(n-1) = Kt + K΄ at to fined the value of the integral constant K΄ when t = o , x =o  K΄ = 1/ (n-1)(a-o)(n-1) = 1/ (n-1) a(n-1) By substituting with the value of K΄ : 1/ (n-1)(a-x)(n-1) = Kt + (n-1) a(n-1) K = 1/ t(n-1) [1/ (a-x)(n-1) - 1/a(n-1)]

K = 2/t [ a½ - (a –x) ½]M½. L-½ .Sec-1
When, n=2 , then K = [1/t ] x/a(a-x) LM-1 Sec-1 at n=3 K = 1/2t [1/(a-x)2 – 1/a2] L2 M-2 Sec-1 at n= 3/2 K = 2/t [1/ (a -x)1/2 - 1/a1/2] L1/2 M-1/2 Sec-1 Finally , at, n=1/2 K = 2/t [ a½ - (a –x) ½]M½. L-½ .Sec-1

, nth-Order Reaction Half-life: Half-life: For n  1 For n = 1

Determination of the order of a reaction
Three methods may be used for the determination of the reaction order , these are : 1- Differential method : The experimental data table generally contains concentration and time data. The use of the variation of rate value R with time t . Van ’t Hoff method : For any reaction the following equation is written : R = dc/dt = KCn , where dC the change in the concentration of reactants , dt ,the change in time ,n is the reaction order . At two different concentrations ,the following equations is obtained :

So , by divided the two equations :
R1 =-dc1/dt1 = KC1n R2 = -dc2/dt2 = KC2n So , by divided the two equations : R1/R2 = K C1n/K C2n = (C1/C2)n  logR1 – logR2 = n(logC1-logC2) n = logR1- logR2/ logC1-logC2 n= logC2/t2 – logC1/t1 / logC1-logC2 ii) Instantaneous method : The simplest means of determining the order of a reaction is to measure the rate R at different concentrations a of the reactants.

log R = log k + n log Ca log R
A plot of ln R or , log R against ln or log a is then a straight line with slope equal to the order. The rates are taken from : 1- a single plot of concentration vs. time, at each of several times during the reaction, the measure of the slope of the concentration-versus-time curve, and then make a plot of log of rate against log of concentration to obtain the order. 2- A plot of several concentrations against time, the concentrations of all the reactants change with time. R = K Can log R = log k + n log Ca log Ca log R

2- Integrated method (CONCENTRATION/TIME RELATIONSHIPS)
In this method the integrated form of the rate low is used : 1- Method of Try and fail : The equations of the integrated form of the rate low of each order is used to obtain the order of any reaction . The substitution of the obtained results from the experimental data will give the reaction order as follows : The transformation of ammonium cyocyanat to thiouora give the following results: 600 312 157 72 3 Time (min. ) 0.0228 0.0348 0.0512 0.0656 0.0916 Concentration (a – x) M

By applying the equations :
i. for the first order it does not give constant value for the reaction constant, k . ii. When the application of the second order is continue , a constant value for k is found ,their-for , the reaction is 2nd ordered n=2 K=1/t x/a(a-x) n= 1 K=2.303/t log a/(a-x) X Concentration (a – x) M Time (min. ) 0.601 0.026 0.0656 72 0.0558 0.0037 0.0404 0.0512 157 0.0551 0.0568 0.0348 312 0.0549 0.0688 0.0228 600

2- Method of plots: It is easily to find the order of the reaction when time and concentration are known , since the equation : y = mx + b , can be applied .By choosing the set of variables that gave you the best straight line and insert them in place of x and y in the generalized equation for a straight line. where “A” is reactant A and Ao is the initial concentration of reactant A at time zero [the y-intercept]. The following equations will be obtained ,and the plots in the form : y = mx + b , is applied , zero order [A] = −kt + [Ao] first order ln[A] = −kt + ln [Ao] second order 1/[A] = kt + 1/[Ao] Also recognize that slope = k, since the rate constant is NEVER negative.

Plots to determined the reaction order
ln[A]t = -kt + ln[A]0 [A]t = -kt + [A]0 1/[A]t = kt + 1/[A]0

3- Half life time t1/2 method :
The time required for one half of one of the reactants to disappear is called t1/2. -For the first order reaction : t1/2 = / k -For the second order reaction : t1/2 = 1/ka -For the third order reaction : t1/2 =(3/2)1/Ka2 i)Half –life period method: When two half –life times are recorded at two initial concentrations then the following equations are used to calculate the order of the reaction : then ,

Problems : ii) Mutual half life method :
In this method the following equation is applied -For the first order reaction : the value of t1/2 is constant (independent of the initial concentration a) . -For the second order reaction : t1/2 a = constant . -For the third order reaction : t1/2 a2 = constant . Problems : 1- The decomposing of PO3 at different initial pressure give the following recorded half –life times. Find the reaction order (n).As can be seen ,this reaction is 1st order reaction (t1/2 is constant), n=1 P (initial) 77 79 37.5 t1/2 84 83

The following of a reaction at different initial pressure give the following recorded half –life times. Find the reaction order (n) P m.m (initial) 250 300 350 400 450 t1/2 136 112.5 97 85 75.5 As can be seen ,in this reaction (t1/2 is not constant , so this reaction is not 1st order . The calculation of the value t1/2 P give almost a constant value : t1/2 P : , , , = constant . The reaction order is 2 .

4- Isolation method The isolation method is a technique for simplifying the rate law in order to determine its dependence on the concentration of a single reactant. Once the rate law has been simplified, the differential or integral methods may be used to determine the reaction orders. The dependence of the reaction rate on the chosen reactant concentration is isolated by having all other reactants present in a large excess, so that their concentration remains essentially constant throughout the course of the reaction. As an example, consider a reaction A + B → P, in which B is present at a concentration 1000 times greater than A. When all of species A has been used up , the concentration of B will only have changed by 1/1000, or 0.1%, and so 99.9% of the original B will still be present. It is

Chapter two Complex Reactions.

3- the two reactions are of the second order :
1- the two reactions of the first ordered ,so, R = R1 – R-1 dx / dt = K1 [A] – K-1 [X] or , dx / dt = K1 (a – x) – K-1 X 2- the two reactions one them is of the first order and the other of the second order dx / dt = K1 [A] 1 – K-1 [X] 2 dx / dt = K1 (a – x) – K-1 X 2 or , dx / dt = K1 [a – x] 2– K-1 X 3- the two reactions are of the second order : dx / dt = K1 [A] 2 – K-1 [X] 2

Here the first type will be considered :
The rate of the reaction will be R = R1 – R-1 a) At initial concentration of A : a and B =0 ,so, (a-x) and X are the concentrations of A and B at t time : dx / dt = K1 [A] – K-1 [X] or , dx / dt = K1 (a – x) – K-1 X At equilibrium R1 = R-1 , R = 0 , and X = Xe; so, dx/dt = K1 (a – Xe ) – K-1 Xe = 0 K-1 = K1 (a – Xe ) / Xe

By substitution with the value of K-1
dx / dt = K1 (a – X ) – K-1(a – Xe) / Xe .X dx / dt = [K1 a/Xe] (Xe – X ) Separation of variable and integration at t=0 and X=0 , the following equation is obtained : K1 = (Xe / a t) ln [Xe /( Xe – X)] This equation can be written in the form : ln [Xe /( Xe – X)] = [(K1a t) / Xe ] by substitution with the value of K-1 K-1 = K1 (a – Xe ) / Xe

then K-1 .Xe = K1 ( a – Xe) = K1a – K1. Xe
( K-1 + K1) Xe = K1. A Xe / a = K1 / (K-1 + K 1 ) So, the following equation is obtained : K1 = (K1 / K1 + K-1 ) / t ln [ Xe / ( Xe – X)] Or K1 + K-1 = 1/t ln [Xe / (Xe – X ) ] This equation is the same of the simple first order, with a = Xe and (K1 + K-1 ) instead of K .

then  Xe = -A / B so the equation will be in the form : dx / dt = A + BX Separation of variable and integration at t=0 and X=0 , the following equation is obtained : dx / (A + BX ) = dt ( 1/B) ln ( A + BX) = t + K΄ The value of K΄ is found when t = 0 , x = 0 K΄ = (1/B ) ln A (1 / B ) ln (A + BX) = t + (1 / B ) ln A (1 / B ) ln (A + BX) – (1 / B) ln A = t

(1 / B ) ln (A + BX) = t + (1 / B ) ln A
(1 / B ) ln (A + BX) – (1 / B) ln A = t By substitution with the value of (1 / B) then , ( -1 / K1 + K-1 ) ln ( A + BX ) + (1 / K1 + K-1 ) ln A = t  (1 / t) ln A/ ( A + BX) = K1 + K-1 by dividing the two equations and substitution with the value of Xe = A/B : ( 1 / t) ln A/B /(A/B + B/B X) = K1 + K-1 1/t ln [ - Xe/ (-Xe +X) ] = K1 + K-1 1/t ln (+ Xe/ (Xe – X) ) = K1 + K-1 This equation is the same of the simple first order ,with a =Xe and (K1 + K-1 ) instead of K .

1- The transformation of ᴽ hydroxyl butyric acid to lactone :
This reaction is reversible from the first order in the two directions , at equilibrium Xe = M/l. , then using the obtained data in the following equation : K1 + K-1 = 1/t ln (+ Xe/ (Xe – X) ) K1 and K-1 can be calculated. ᴽ hydroxyl butyric acid lactone

2- The transformation of α glucose to β glucose :
This reaction is reversible from the first order in the two directions ,the concentration of β glucose in water is followed by polarization . The initial concentration of β glucose = 0 ,the values of (K1 and K-1 ) is then calculated .

The reaction rate equation is written as :
The reaction between hydrogen and iodine is proceed through second order reaction as : The reaction rate equation is written as : R = R1 – R-1 R1= K1 CH2 .Cl2 , R-1 = K-1 C2Hl dCHl / dt = K1 CH2 .Cl2 –K-1 C2Hl The analysis of [HI ] will used to calculate (K1 + K-1 ) .

From the first order - d CA / dt = K1 CA = K1 [A] K1 = 1/t ln (a/ Ca) , Or K1 t = ln a – ln CA the equation in natural exponential form ,so, CA = a. ē K1t The rate equation for compound D for the first order : dD / dt = K2 CB = K2 [B] The rate of the formation of B will be d [B] / dt = K1 CA – K2 CB By substitution with the value of CA , the following equation will be obtained:

On multiply the equation by the value e K2t
d [B] / dt = K1ae –K1t – K2 CB On multiply the equation by the value e K2t d [B] / dt e K2t = K1ae (K2- K1)t – K2 CB. e K2t :.d [B] / dt e K2t + K2 CB. eK2 = K1 a e (K2 - K1)t this equation is for two multiple equations ,so, it may written as : d (e K2t CB) / dt = K1 a e (K2 - K1)t By integration ∫d( e K2t CB ) = ∫ K1ae (K2 – K1) t dt :. e K2t CB = (K1 / K2 – K1 ) a e (K2 – K1) t + K΄ To fined the value of K΄ ,the equation is divided by e K2t , at the start of the reaction, t= 0 and CB = 0 ,so, CB = K1a / ( K2 – K1) ē K1t + K΄ ē K2t

CB = K1a / ( K2 – K1 ) e –K1xo + K ᷆ e-K2 xo = o
:. K ᷆ = - K1a / ( K2 – K1 ) By substitution with the obtain value of K ᷆ the following equation will be obtained : CB = K1a / ( K2 – K1 ) e-K1t - K1a / ( K2 – K1 ) e-K2t CB = K1a / ( K2 – K 1 ) (e-K1t - e-K2t ) Knowing the values of K2 , K1 , the values of CA , CB , CD at time = t , is available , where : a = CA + CB + CD ,[D][A] , [B] : the following figure shows the relation between

dx / dt = R = R1 + R2 = K1 ( a – x ) + K2 ( a – x )
 dx / dt = ( K1 + K2 ) ( a – x ) Separation of variable and integration at t=0 and X=0 , the following equations are obtained :  dx / ( a – X ) = ( K1 + K2 ) dt and , ln ( a – x ) = ( K1 + K2 ) t + K΄ K΄ = - ln a  K1 + K2 = 1 / t ln a / ( a – x) This equation the same of the first order of simple reaction only the rate constant k become (K1 + K2 ),or ,

ln a / ( a – x) = 1 / t (K1 + K2) a plot of ln [a / ( a – x ]t vs. t will yield a straight line with a slope of (K1 + K2). The rate of formation of [B] and [D] will be as : d [B] / dt = [B] = K1 ( a – x ) d [D] / dt = [D] = K2 ( a – x ) d [B] / d [D] = K1 / K2 = [B] / [D] Their concentrations are determined by the individual rate constants, such that [B]/[D] = k1/k2 always. Such systems are convenient to study experimentally.

I) The time for 99% of acetic acid to decomposed
At high temperatures( ) ⁰C , acetic acid CH3COOH decomposed into CO2 and CH4 and in the same time into CH2 CO and H2O . CH3COOH CO CH4 CH2 CO H2O The value of K1, for the first reaction is Sec -1 and for K Sec -1 , calculate : I) The time for 99% of acetic acid to decomposed ii) The maximum amount of CH2CO to form by % to acetic acid.

Chain Reactions: The chain reactions one of the complex reactions ,it usually when started continues until all or at least one of the reactants disappeared. The reaction proceed through the formation of activated species and affected by catalysis . Many factors may affect this type of reaction, these are : The concentration C : The reaction rate is proportional directly to the concentration ; R α F1 (C) The surface area S : The reaction rate is proportional

iii) Inversely to the surface area S : R α 1 / F2 (S)
iv) The number of collusions : The reaction rate is proportional inversely to the number of collusions between the reactants ; R α 1 / F3 (g) The Chain Reaction between Hydrogen and Brome: The reaction proceeds as follows : Initiation of the reaction : Br K Br (1) ii) Propagation of the reaction : Br + H K HBr + H (2)

iii) Inversely to the surface area S : R α 1 / F2 (S)
iv) The number of collusions : The reaction rate is proportional inversely to the number of collusions between the reactants ; R α 1 / F3 (g) The Chain Reaction between Hydrogen and Brome: The reaction proceeds as follows : Initiation of the reaction : Br K Br (1) ii) Propagation of the reaction : Br + H K HBr + H (2) H Br K HBr + Br (3)

iii) Inhibition of the reaction :
H HBr K H2 + Br (4) iv) Stopping of the reaction : 2Br K Br (5) the intermediate species [H] and [Br] formed during the reaction are d [H] / dt = , d [Br] / dt = 0 So, d[Br] /dt = K1[Br2] - K2[Br][H2]+ K3[H][Br2]+ K4 [H][HBr] – K5[Br]2 = 0 d[Br]/dt = K1[Br2]–{K2[Br][H2] - K3[H][Br2]- K4 [H][HBr]} – K5[Br]2 = 0

Also, d[H] /dt = K2[Br] [H2] - K3[H][Br2] - K4 [H][HBr] = 0
then , d[Br] /dt = K1[Br2] – 0 – K5 [Br]2 = 0 [Br˙]2 = K1 / K5 [Br2] and , [Br] = ( K1 / K5 )½ [Br2]½ By replacing with the value of (Br) in the following equation : [H] = K2 ( K1 / K5 )½ [H2] [Br2]½ / K5 [Br1] + K4 [HBr] So, the formation of HBr will be as follows :

d[HBr] / dt = K2 [Br][H2] + K3[H][Br2] – K4[H][HBr]
By substitution with the value of (Br) and (H) , then : d[HBr] /dt = 2K2 (K1 / K5)½ [H2][Br2]½ / K4 [HBr] / K3 [Br2] The equation can be written as follows : d[HBr] /dt = K [H2][Br2]½ / 1+ [HBr] / Ḱ [Br2] Where k = 2K2 ( K1 / K5 )½ and Ḱ = K3 / K4

Theories of chemical Reactions
Chapter three Theories of chemical Reactions

Division of Arrhenius equation
From the earliest studies of reaction rates, it has been evident that they are profoundly influenced by temperature. The most elementary consequence of this is that the temperature must always be controlled if meaningful results are to be obtained from kinetic experiments. However, with care, one can use temperature much more positively and, by carrying out measurements at several temperatures, deduce important information about reaction mechanisms.

The studies of van ’t Hoff and Arrhenius form the starting point for all modern theories of the temperature dependence of rate constants. Harcourt had earlier noted that the rates of many reactions approximately doubled for each 10⁰C rise in temperature, but van ’t Hoff and Arrhenius attempted to find a more exact relationship by comparing kinetic observations with the known properties of equilibrium constants. Any equilibrium constant K varies with the absolute temperature T in accordance with the van ’t Hoff equation:

d (ln K)/dT = ΔH⁰/RT2 where R is the gas constant and ΔH⁰ is the standard enthalpy change in the reaction. But K is regarded as the ratio k1/k-1 of the rate constants k1 and k-1 for the forward and reverse reactions (because the net rate of any reaction is zero at equilibrium). On the separation of the variables d (ln K) = ΔH⁰/RT 2 dT Integration at the absolute temperature the following equation is found : ln K = - ΔH ⁰ /RT ln Kα

By separation the integrated equation can be written as:
Where : ΔH⁰ = H⁰P - H⁰R H⁰P the enthalpy for the products and H⁰R the enthalpy for the reactants . By separation the integrated equation can be written as: ln K1/K-1 = -(H⁰P-H⁰R/RT) + ln[(K1) α/(K-1) α Or ln K1/(K1) α - H⁰R/RT = ln K-1/(K-1) α - H⁰P/RT = Y where Y is a quantity about which nothing can be said a priori except that it must be the same in both equations because otherwise it would not vanish when one equation is subtracted from the other), so, Y = –ΔH*/RT .

ln K1/(K1) α = -(H* - H⁰R) /RT
ln K-1/(K-1) α = - (H *- H⁰P) /RT So, the equations may written as : K1 = (K1)α e - (H*- H⁰R /RT) K-1 = (K-1)α e - (H*- H ⁰P/RT) The values of (H*-H⁰R) or (H*-H⁰P) always positive which mains that H* is greater than H⁰R or H⁰P , H* value will be corresponds to the activation energy Ea . The rate equation for the forward and reverse reactions can be written as :

and for the reverse reaction : K-1 = A -1 e - (H*- HºP /RT) ,
For the forward reaction : K1 = A e - (H*- H⁰ R /RT) ln k = ln A - Ea/RT and for the reverse reaction : K-1 = A -1 e - (H*- HºP /RT) , ln k-1 = ln A-1 - E-1a/RT where Ea is the activation energy , corresponds to the standard enthalpy of reaction ΔH0 in the van ’t Hoff equation.

Taking the natural logarithm of both sides, the equation becomes
y = mx + b When k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. 1/T.

Elementary collision theory
It is instructive to relate the rates of reactions in the gas phase with the frequencies of collisions between the reactant molecules. According to the Maxwell–Boltzmann distribution of energies among molecules, the number of molecules in a mixture that have energies in excess of -Ea is proportional to e-Ea/RT. Therefore interpret the Arrhenius equation to mean that molecules can take part in a reaction only if their energy exceeds some threshold value, the activation energy. A simplified approach for describing chemical reactions is the “hard-sphere collision model ”, which considers a chemical reaction as a collision of rigid spheres and

-dnA/dt = -dnB /dt = ZAB e-E/RT
does not take into account any other intrinsic information about the reactants . In this interpretation, for the following reaction: A B AB Molecules can only react if they collide with each other. The rate of reaction will be equal to the number of active collision , so, the rate of the reaction R can be written as -dnA/dt = -dnB /dt = ZAB e-E/RT Where nA , nB number of A and B molecules in cm3 ; the concentration of A& B , ZAB = the number of collisions between the reactant molecules A&B in a second of time in cm3.

then , the rate of reaction is : ZABe-Ea/RT = K nA nB
-dn/dt = the decrease in the number of reactants molecules A&B per time ; the rate of reaction = R ,so the rate of reaction may written as : -dn/dt = K nA nB then , the rate of reaction is : ZABe-Ea/RT = K nA nB K= ZAB/nA nB e-Ea/RT so, K= Z- e-Ea/RT in which , Z- collision number is defined as Z- = ZAB/ nA nB This equation is the same equation of Arrhenius : K = A e -∆Ea/RT

where σA σ B , is the radii of the molecules A and B.
in which Z- instead of A Arrhenius constant . To fined the value of Z-, by substitution with the value of ZAB from the kinetic theory of gases. ZAB = [(σA+ σB/2)2 √8π(mA+ mB) kT/mA mB ]nA nB where σA σ B , is the radii of the molecules A and B. mB , mA represents the mass of the molecules A and B: k is the Boltzmann constant = R/N , Nav = 6.022×1023 mol-1. Z- = (σA+ σB /2)2 √8π(mA + mB) kT/ mA mB = A to calculate A , by substitution with the values of σ , mA mB and k the Boltzmann constant :

A = (10-8)2 (4x104) = 4x10-12 cm3/atom/sec
A = 4x10-12 x N/1000 = 4x10-12 x 6x1023/1000 = 2.4x109 L/mol/sec The obtained value for A is acceptable . For some simple reactions in the gas phase, such as the decomposition of hydrogen iodide, A is equal to Z, but in general for complicated reactions such as the reaction between tre-ethyl amines and ethyl amine : 0r , the reaction between butadiene with acriladhyde :

it is necessary to introduce a factor P,
k = P Z e-Ea/RT assume that, in addition to colliding with sufficient energy, molecules must also be correctly oriented if they are to react. The factor P is then taken to be a measure of the probability that the correct orientation will be adopted spontaneously, so modify the interpretation above that at infinite temperature

every collision is productive if the orientation is correct.
However, virtually all of the reactions that interest biochemists concern complicated molecules in the liquid phase, and collision frequencies have little relevance for these. The application of collision theory on the tre-molecules reactions: For the reaction between NO and O2 , collision of three molecules

at the same time is rare , for every one of tre-collision ,ten thousand collisions of the bimolecular reaction -collision ,so, it is necessary to introduce a mechanism of two molecules collision instead of three molecules collisions . For the reaction: 2NO + O NO2 The rate of the reaction can be written as : -dC[O2 ] /dt = K[NO]2[O2] where [NO] and [O2] are the concentrations of nitric oxide and oxygen respectively , k is the rate reaction constant of the reaction .

for this step the rate of the reaction is : d(NOX2)/dt = K1[NO] [X2]
It was suggested that the reaction proceed through the formation of a complex molecule as follows ,where X 2 represents oxygen or any molecule such as Br2 ,I 2 NO + X NOX2 for this step the rate of the reaction is : d(NOX2)/dt = K1[NO] [X2] this step will followed by : NOX2 + NO NOX The rate of the reaction for this step is : R = K2[NO][NOX2]

The overall rate reaction will be :
R = K2K1[NO]2[X2] = K[NO]2 [X2] this suggested that the following equilibrium is occurred 2NO N2O2 N2O2 + X2 2NOX So ,a di-collision step is decisive for the occurrence of the reaction , also, it may conceder that the two molecules in contact , then a collision between these two molecules and the third one will occurred .

Transition-state theory
The transition-state theory (sometimes called the theory of absolute reaction rates). It is so called the transition state , because it relates the rates of chemical complex in chemical reactions to the thermodynamic properties of a particular high-energy state of the reacting molecules, (The term activated complex is also sometimes used). As a reacting system passes along a notional “reaction coordinate”, it

must pass through a continuum of energy states, and at some stage it must pass through a state of maximum energy. This maximum energy state is the transition state, and should be clearly distinguished from an intermediate, which represents not a maximum but a metastable minimum on the reaction profile. A bimolecular reaction can be represented as in which the reactants molecules are in equilibrium with the active formed compound as follows: A+ B (AB) Products Or,

A+B Q+P X‡

K≠ = f≠/fA fB e-(Єo≠ -ЄoA- ЄoB/kT)
The values of ν and K≠ can be calculated from the equation of the partition function f- as : K≠ = f-≠/f-A f-B where f- is defined as f- = fe-Є0/kT by substitution with the values f≠, fA and f B , the following equation is found : K≠ = f≠/fA fB e-(Єo≠ -ЄoA- ЄoB/kT) where e-Єo the zero point energy of the molecule ,so, K≠ = f≠/fA fB eEo/RT in which Eo=No(Єo≠-ЄoA-ЄoB)

When ν is too small the value hv/kT will be too small ˂˂ 1 then the
The partition function f- can be written in the form of its components : Translation , Rotation and Vibration .For the activated compounded X‡ ,this function may written as combination of two functions : f≠ = fv f≠ Where f≠ is the partition function as in the stable molecules ,it is the remaining function after removing fv ,which equal to : fv= kT/hν ehv/kT When ν is too small the value hv/kT will be too small ˂˂ 1 then the fv= kT/hν ehv/kT fv= kT/hν , and f≠ = fv f≠ = f≠ = kT/hν f≠ by substitution with the value of fv :

K≠ = kT/hν (f≠/fA fB) e-Eo/RT
K = kT/h (f≠/fA fB) e-Eo/RT This equation is the same equation of Arrhenius : K = A e -∆Ea/RT in which [kT/h (f≠/fA fB)] is instead of A Arrhenius constant. The calculation of the partition function f is important for the comparison . The partition function f is consist of : Translation , Rotation and Vibration ,the total partition is 3N , it is divided to : translation =3 , rotation depending if the molecule for linear

molecules=2 , for non linear =3
molecules=2 , for non linear =3 . The vibration will be the reminder , so, = (3N- 5 ) or (3N-6) . The following table shows the degree of freedom of any molecule ,where N = the number of atoms in the molecule. Freedom No. Mono atom Di -atom Linear molecules Non Linear molecules Total No (3N) 3 6 3N Translation ft Rotation fr - 2 Vibration fv 1 (3N - 5) (3N - 6)

For the reaction : H2 + I2 (HI )2 2HI the calculation of the frequency factor from the point view of the transition-state theory , will be given as : A = kT/h f≠/ fH2 fI2 For this reaction 1. the activated spices (HI )2 ,3N = 3X4 =12 , (HI )2 : Translation ft = 3 , Rotation fr = 3 ,Vibration fv = 6 f≠ = ft3. fr3. fv6 2. for H2 and I2 , 3N = 3 X 2 =6 Translation ft = 3 Rotation fr = 2 ,Vibration fv = 1

A =kT/h (ft3 fr3 fv6/(ft3 fr2 fv ft3 fr2 fv ) = kT/h (fv4/ ft3 fr)
fI2 = fH2 = ft3.fr2.fv A =kT/h (ft3 fr3 fv6/(ft3 fr2 fv ft3 fr2 fv ) = kT/h (fv4/ ft3 fr) by substitution with the values of Translation ,Rotation and Vibration at ordinary : ft = 108 , fr = 10, fv  1, kT/h  1013 the obtained value of A will be : A = 1013/ (108)3 10 = cm2/atom/sec To calculate the A by mol /L the following value is found A = 10-12x 6x1023/1000 = L/mol/Sec Which is in good agreement with the measured one for any reaction .

Given that the total number of degrees of freedom for a polyatomic molecule is 3N,
calculate the number of vibrational modes open to (a) an atom, (b) a diatomic molecule, (c) a non-linear polyatomic molecule with N atoms and (d) a non-linear activated complex with N atoms. Answer : (a) an atom has three translations, but no rotations and no vibrations; (b) a diatomic molecule has three translations, two rotations

and (3 x 2) = 1 vibration; c) a non-linear molecule has three translations, three rotations and ; 3N - 6 vibrations; (d) a non-linear activated complex has three translations, three rotations, one internal free translation and ; 3N 7 vibrations. Note. Compare the answers to points (c) and (d). The activated complex has one vibration less open to it compared with the polyatomic molecule with the same number of atoms.

The relation between the free energy ,entropy with
the rate reaction constant From the point view of the transition-state theory ,the rate constant K will be given as K = (kT/h) (f≠ / fA fB) e-Eo/RT also ,for the equilibrium constant K≠ the following equation can be written : K≠ = (f≠/ fA fB )e-Eo/RT So, K = kT/h K≠ But the equilibrium constant K≠ : K≠ = e-∆G≠/RT ∆G≠ = ∆H≠ - T∆S≠

Where ∆S is Entropy (S) which is defined as a measure of the randomness or disorder of a system and ∆H is the enthalpy and ∆G is the free energy of the reaction . From the second law of thermodynamic : ∆G≠ = ∆H≠ - T∆S≠ K≠ = e-(∆H*-T∆S≠ / RT) K≠ =∆S≠/RT e-∆H≠/RT K= kT/h e∆S≠/R e-∆H≠/RT ∆H≠ K = kT/h e∆S≠/R This equation is the same equation of Arrhenius ,only ∆H≠ instead of ∆E , so : A = kT/h e∆S≠/R also, from the collision theory A = PZ- PZ- = kT/h e(∆S≠/R)

Reactions in Solutions
The rate constant of a reaction usually calculated in the gas phase , so, its value may be has a different value . The collisions between reacting molecules is given as A + A Products is : collisions The number of ZII = Zn2A if there is other molecules B ,the collisions between A&B will be the same .so, for the reaction : A + B (M ) ≠ Products By applying the transition-state theory ,the equilibrium constant K will be given as : K≠ = a≠/aA aB = v≠c≠ /νAcA vBcB

where aA , aB , and a≠ the activity of A,B and (M ) ≠
where aA , aB , and a≠ the activity of A,B and (M ) ≠ . νA , vB and v≠ activity co-efficient of A,B and (M ) ≠, also, cA, cB , and c≠ concentration of A,B and (M ) ≠ respectively . From the rate of reaction form the point view kinetics of may written as : R = KCA.C B = v c≠ K = vK≠ (vA.vB/v≠) but , K≠ is the equilibrium constant ,so, K≠ = (kT/ vh) K≠ K = (kT/h) K≠ (vA.vB/v≠)

This equation is for reactions in the solutions
This equation is for reactions in the solutions . For comparison ,in the gas state ,the following equation is used : Kg= (kT/h ) K≠ so, K = Kg(vA. vB/v≠) the variation of the rate constant value between the gas and solution states depend on v ,if the solvent decrease the free energy of reacted substances more than the activated compound ,then vB,vA values will be smaller than v≠ ,then K for solution will be smaller than in gases . If the reaction does not proceeds in gas state ,then the reaction will be studied in infinite dilutions ; (v=1), and Kg will be K0 ,

the rate equation is : K = K0(vA.vB/v≠)0 K = Ko(vA.vB/v≠)
K0 = (kT/h )K≠ K = K0(vA.vB/v≠)0 K = Ko(vA.vB/v≠) This equation was obtained by Boroniched and Beirm . By taking the log of the this equation : log K = logK0 + log vA + log vB - log v≠ From Dibie Hikel equation the vi is given to be equal to :

log K = logK0 – 0.5√I [ZA2 + ZB2-(ZA+ZB)2]
log vi = - 0.5Zi2√I Where Z is the ion charge and I is the ionic strength given as : I = 1/2 ∑i Zi2 Ci by substitution with the values of vi the following equation will be obtained : log K = logK0 – 0.5√I [ZA2 + ZB2-(ZA+ZB)2] Where ZAB = (ZA + ZB) , log K = logK √I [-2ZA.ZB] = logK0 + ZA.ZB√I log K /K0 = ZA.ZB√I

the plots of log(K/K0) and √I will give straight line with a slope of ZA ZB , the value of K will depends on the charge of the reacting ions : 1- the value of ZA ZB positive ,then K will increased with increase of I. 2- the value of ZA ZB negative ,then K will decreased with increase of I. 3- the value of ZA ZB =0 ,then K will not affected with increase of I. The following plots shows the relation of log(K/K0) and √I

logK/K0 The effect of √I on K

ZAZB=4 [Co (NH3)5Br]2+ + Hg2+ I ZAZB=2 S2O82- + I- II ZAZB=1 NO2NCO2C2H5-+I- III ZAZB=0 C12H22O11+OH- IV ZAZB=-1 H2O2+H+ +Br- V ZAZB=-2 [Co(NH3)5Br]2+ +OH- VI

Solve Problems :

5. For the following reaction:
[Cr Cl (NH3)5] OH [Cr (NH3)5 OH ] Cl- the rate constant of the reaction k0 is 89 min -1 .L , when the concentration of [Cr Cl (NH3)5]+2 = 3x10-3 M presented as Br- and OH- = 3.5x10-3 M , calculate : the rate constant k of the reaction in absence and presence 0.2 M NaCl .

6. What is the activation energy obtained from an Arrhenius equation for the following reaction?
2NOCl → 2NO + Cl2 1/T (T, K) lnk (k, 1/M.s) 2.5×10-3 2.79 1.7×10-3 -7.32

Factors affecting the reactions mechanisms
Chapter four Factors affecting the reactions mechanisms

Factors affecting the reactions mechanisms:
All the reactions in chemistry are affected by factors that may influence the reaction mechanisms , the role by which these factors is not clear ,but in gives some suggestions that may helps in understanding how the reactions proceeds . 1. Free Energy . 2.Activation of Chemical Reactions 3.Rate of Reactions and Free Energy of Activation. 4.Kinetics Energy and the Rate Determining Step of Reactions 5. Expectations of Reaction Mechanisms.

equilibrium . The value of (ΔE) of the reaction is not enough to deicide the reaction direction .
The change in the free energy (ΔG ) associated with a chemical reaction is a more useful indicator of whether the reaction will proceed spontaneously, since the change in free energy ΔG for a reaction can be expressed in terms of the equilibrium concentrations of the reactants and products. This is illustrated by the following equation : ΔG = -R T ln K Also, Free Energy change (ΔG ) is a combination of enthalpy

change ΔH or, the activation energy , ΔE , Entropy change ΔS, and temperature T.
This relation is given by the following equation : ΔG = ΔH - TΔS when ΔG is negative, the reaction is possibly spontaneous, this usually obtained when the sum of the two terms in the equation is negative (ΔH is negative for endothermic reactions and is positive for increase ΔS in the reactions ) when the degree of freedom decreases as in combination reactions A + B C

also, as in cyclization of chain compounds
ΔG = ΔH - (+) TΔS = negative value In the case when ΔG is positive, the reaction is possibly not spontaneous ( ΔH is negative for the exothermic reactions and ΔS is negative for the reactions in which the degree of freedom increases as in the decomposition reactions . A B C ΔG = ΔH - (-) TΔS = positive value When ΔG is zero, there is no equilibrium and therefore, there is no reaction proceeds .

2.Activation of Chemical Reactions
When ΔG is negative, the reaction is spontaneous, and the reaction proceed in the direction to give the products but , this is usually not enough to till about the reaction speed of the reaction. The reaction does not proceeds as fowling from up to down it must go through a maximum value , so ,the gain of some energy is necessary for the start of any reaction.

there is a minimum amount of energy required for reaction: the activation energy, Ea. Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier. The following reaction possess a high negative ΔG value , and the formation of the products is predicted but , the reaction does not start unless some of heat (or ,any kind of energy) is given to it : (C6H10O5)n + 6(n)O (n)CO2 + 5(n)H2O

3.Rate of Reaction and Free Energy of Activation:
For any reaction the transfer of reactants to products , an activation energy necessary must provide for the reaction occurrence .This energy is consumed in braking and building up the bonds in reacting and products compounds respectively . The formation of intermediate which is not stable compound , at the maximum value of energy is necessary for the reaction occurrence, it is, also, called the

activated compound in which the formation of new bonds occurred before braking the required bonds .The following example explains that : In the alkali hydrolysis of methyl bromide ,the reaction is suggested to as the formation of HO ……. C bond is found before the braking of C…..Br bond :

4.Kinetics Energy and the Rate Determining Step of Reactions :
The new bonds formation need to supply with energy ,this energy comes from the reactant particles collusions which is help in getting over the energy barrier of the reaction . 4.Kinetics Energy and the Rate Determining Step of Reactions : The determination of the reaction rate can be carried out by measuring the change in the concentration of either the reactants disappearance nor the production

of products during the proceeding of the reaction at constant temperature .This can be done by chemical analyses of reactants or the products , or by following the change in the absorbance or by titration ,but this is not enough to suggest the reaction mechanism . As can be seen from the following reaction : CH3Br + OH- CH3OH + Br- This reaction is first order for the two reactants and the rate equation will be written as : R = K [CH3 – Br] [OH-]

But for the reaction: CH3 – CO – CH3 + Br2 CH3COCH2Br + HBr The rate equation of this reaction is written as : R = K[CH3COCH3][OH-] in which Br2 does not appear in the rate . For this reaction as can be seen the alkaline of the solution is affecting the rate of the reaction ,the role of Br2 in the reaction found in the final product of the reaction the production of CH3COCH2Br.

ΔG Reactants Products Time Time

5. Expectations of Reaction Mechanisms:
It usually difficult to have a full and perfect data about the reaction mechanisms .The following factors will give some light for the understanding of the reaction mechanisms . i. Kinetics Energy ii. The effect of Isotopes iii. Detection of the Intermediates of Reactions iv. The effect of Geometric Structure of the Reactants molecules . v. Study the Differences between the expected and real Results.

Kinetics Energy : 1. C6H6 + HNO3 C6H5NO2 + H2O
The information supplied from the studied reaction is some times not accurate enough to explain the reaction mechanism . The measured concentration of reactants or the products is not the effective substance that included in the slow step ,this caused deviations in the collected information to explain the reaction mechanism . In the following two reactions the concentration measured is not that of the effective substance in the rate equation : 1. C6H6 + HNO C6H5NO H2O the effective substance which must involve in the slow step is

NO2+, but the rate equation involve only the concentration of HNO3 which is can be measured . Also, in the reaction : RX H2O ROH + XH the rate equation of this reaction is : R = K [R-X] A first order reaction ,in which the concentration of H2O is not involved in rate equation due to its high concentration. Inspite that water molecules are must involved in the reaction . If this reaction was carried on non aqueous solution a second order reaction . The deviations in the collected information cause errs on the kinetics energy which is calculated from E= 1/2mR2 , if the value of R is not correct the obtained kinetics energy is not accrue .

ii. The effect of Isotopes :
The data obtained from the kinetic study does not till about the broken or the formation of bonds of the reaction . The use of isotopes through some light in the study of the reaction mechanism .The energy required to brake bonds in a molecule is not the same if two isotopes are used for the study of a reaction . The following examples show the effect of isotopes on the reaction mechanism : The oxidation of diphenilmethanol in the following reaction by KMnO4 Ph2CHOH Ph2C=O + H2O

can be done using oxygen isotope in Ph2CHOH, the bond energy required to brake 7 times greater than the energy required to brake the bond in the isotope Ph2CDOH , the measured reaction rate in Ph2CDOH was found to be less 7 than that of the bond in Ph2CHOH ,so, it is suggested that the rate determining step will contain the broken of CD bond . While in the nitration of benzene in the two isotopes (C6H6) and (C6D6) by HNO3 acid the rate of the two reactions was the same . can be done using oxygen isotope in Ph2CHOH, the bond energy required to brake 7 times greater than the energy required to brake the bond in the isotope Ph2CDOH , the measured reaction rate in Ph2CDOH was found to be less 7 than that of the bond in Ph2CHOH ,so, it is suggested that the rate determining step will contain the broken of CD bond . While in the nitration of benzene in the two isotopes (C6H6) and (C6D6) by HNO3 acid the rate of the two reactions was the same .

It was found that R1= R2 , this suggested that the rate determining step will not contain the broken of CD bond . In Kanizaro reaction :

iii. Detection of the Intermediates of Reactions :
Oxidation – redaction reaction of benzalhyde is carried out in aqueous solution . The interference of water molecules H2O in the reaction mechanism can be found by using the isotope of water D2O instead of H2O in the reaction . The obtained data were the same which confirmed that H2O does not shear in the reaction mechanism , and the formation of the suggested intermediate PhCHDOH is not found ,the hydrogen atom in benzalhyde alcohol comes from another benzalhyde molecule . iii. Detection of the Intermediates of Reactions : In the Hofman Reaction : R – CO – NH RNH2

The following intermediates were found : (R-CONHBr), (RCONBr),
and (RNCO), so the following mechanism is suggested : If the separate intermediates are actually intermediate and not a side products the suggestion of the mechanism will be accurate . In the study of the formation of oxyimate from ketones ,the absorbance band of C=O disappeared before that of C=N- band in the following reaction : Rearrangement

It was suggested the reaction proceeds through the formation of
the following intermediate: Some times when it is not easy to separate an intermediates in a reaction ,and a confirmation of the mechanism may carried out by the addition of a compound to react with the suggested compound as an intermediate ,then identification of the products is done. For example to detect the formation of C=Cl2 in

As can be seen from this reaction the reactants are photo-active species ,while the products are inactive . It is suggested that the reaction goes through the formation of two intermediate species have opposite photo-active directions . Also, in the addition of brome Br2to cyclopenteen , give only trance- cyclopenteen , the cis- cyclopenteen isomer does not appear ,which indicates that the reaction proceeds in one direction .

In the following reaction the elimination of acetic acid , it usually that the reaction in which the hydrogen atom and acetyl part in one direction occurred faster than the in the opposite direction , but as can be seen ,this does not happen ,due to the formation of hydrogen bond between CO of the acetyl and the hydrogen : v. Study the Differences between the expected and real Results. For any reaction ,there is usually expectation for the reaction , how it supposed to proceed in the studied condition . When the

obtained results comes different ,this mains that the suggested mechanisms does not proceed , so , alternative mechanism must be suggested . In ammonia solution ᴩ- chlorotoloine usually reacts with NH3 to give ᴩ- aminotoloine , but the obtained gave m-aminotoloine . So, the expected mechanism is a substation reaction in ᴩ- positions ,but the found mechanism is adding NH3 on the m-position then elimination reaction of HCl from ᴩ- and m- positions as :

Chemical Kinetics By Prof. Sanaa Taher Arab Physical Chemistry

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Chemical Kinetics By Prof. Sanaa Taher Arab Physical Chemistry

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