Kinetic 2

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Second-Order Reactions
Many important biological reactions, such as the formation of double-stranded DNA from two complementary strands, can be described using second order kinetics. In a second-order reaction, the sum of the exponents in the rate law is equal to two. The two most common forms of second-order reactions will be discussed in detail in this section.
Reaction Rate: Integration of the second-order rate law
d[A]/dt=?k[A]2 (1.1)
(1.2)

Case 1: Identical Reactants
Two of the same reactants (A) combine in a single elementary step.
A+ A? (1.4)
2A?P (1.5)
The reaction rate for this step can be written as
Rate= ?1/2 d[A]/dt = +d[P]/dt (1.6)
and the rate of loss of reactant A:
d[A]/dt= ?k[A][A]= ?k[A]2 (1.7)

where k is a second order rate constant with units of M-1 min-1 or M-1 s-1.
Case 2: Different Reactants
Two different reactants (A and B) combine in a single elementary step:
A+B?P (1.8)
The reaction rate for this step can be written as:
Rate=?d[A]/dt =?d[B]/dt=+d[P]/dt (1.9)
and the rate of loss of reactant A :
-d[A]/dt = ?k[A][B] (1.10)
Where the reaction order with respect to each reactant is (1). This means that when the concentration of reactant A is doubled, the rate of the reaction will double, and quadrupling the concentration of reactant in a separate experiment will quadruple the rate.

Case 1: A + A ? P (Second Order Reaction with Single Reactant)
The rate at which A decreases can be expressed using the differential rate equation.
?d[A]/dt =k[A]2 (1.11)
The equation can then be rearranged:
d[A]/[A]2 = ?k dt (1.12)

Since we are interested in the change in concentration of A over a period of time, we integrate between t=0 and t the time of interest.


To solve this, we use the following rule of integration (power rule):

We then obtain the integrated rate equation.

Upon rearrangement of the integrated rate equation, we obtain an equation of the line:


It is important to understand how the equation directly relates to the graph which provides a linear relationship. In this case, and for all second order reactions, the linear plot of 1/[A]t versus time will yield the graph below:

This graph is useful in a variety of ways. If we only know the concentrations at specific times for a reaction, we can attempt to create a graph similar to the one above. If the graph yields a straight line, then the reaction in question must be second order. In addition, with this graph we can find the slope of the line and this slope is k , the reaction constant. The slope can be found be finding the "rise" and then dividing it by the "run" of the line. For an example of how to find the slope, please see the example section below. There are alternative graphs that could be drawn.
The plot of [A]t versus time would result in a straight line if the reaction were zeroth order. It does, however, yield less information for a second order graph. This is because both the graphs of a first or second order reaction would look like exponential decays. The only obvious difference, as seen in the graph below, is that the concentration of reactants approaches zero more slowly in a second-order, compared to that in a first order reaction.

Case 2: A + B ? P (Second Order Reaction with multiple reactants)
As before, the rate at which A decreases can be expressed using the differential rate equation:
d[A]/dt= ?k[A][B] (1.17)
Two situations can be identified.

Situation 2a: [A]0 ?[B]0
Situation 2a is the situation that the initial concentration of the two reactants are not equal. Let x be the concentration of each species reacted at time t
Let [A]0=a and [B]0=b, then [A]= (a?x) ;[B]=b?x The expression of rate law becomes:
?dx/dt=?k(a?x) (b?x) (1.18)
which can be rearranged to:
dx/(a?x)( b?x)= -kdt (1.19)

We integrate between t=0 (when x=0) and tt , the time of interest.
? dx/(a?x)(b?x)= k?dt (1.20)
To solve this integral, we use the method of partial fractions.

Evaluating the integral gives us:


Then for a?b k2 equal to:

Plotting Ln b(a-x)/a(b-x) versus time gives linear relationship as follows


This graph can be used in the same manner as the graph in the section above or written in the other way:
Situation 2b: [A]0=[B]0
Because A+B?P
Since A and B react with a 1 to 1 stoichiometry, [A]=[A]0?x and [B]=[B]0?x
at any time t , [A]=[B][A]=[B] and the rate law will be,
rate=k[A][B]=k[A][A]=k[A]2. (1.27)
Thus, it is assumed as the first case:

Upon integration we have:


Utilizing inigration method for this equation we obtain the following integrated form for second order reaction for case (a=b):

Or by rearrange equation 1-32


Example 1:
The following chemical equation reaction represents the thermal decomposition of gas E into K and G at 200° C?
5E(g )? 4K(g)+G(g) (1.28)

This reaction follows a second order rate law with regards to E . What is the initial rate of decomposition of E . For this reaction suppose that the rate constant at 200° C is equivalent to 4.0×10?2 M?1s?1 and the initial concentration is 0.050M0.050M ?
SOLUTION
Start by defining the reaction rate in terms of the loss of reactants
Rate (initial)= ?1/5d[E]/dt (1.29)
and then use the rate law to define the rate of loss of E
d[E]/dt= ?k[A]2 (1.30)

We already know k and [A]i but we need to figure out x . To do this look at the units of k and one sees it is M-1s-1 which means the overall reaction is a second order reaction with x=2 .
Initial rate=(4.0×10?2 M?1 s?1) (0.050M)2= 1×10?4 M -1s?1

Half-Life for 2nd order reaction
Another characteristic used to determine the order of a reaction from experimental data is the half-life (t1/2 ). By definition, the half life of any reaction is the amount of time it takes to consume half of the starting material. For a second-order reaction, the half-life is inversely related to the initial concentration of the reactant (A). For a second-order reaction each half-life is twice as long as the life span of the one before.
Consider the reaction 2A?P :
We can find an expression for the half-life of a second order reaction by using the previously derived integrated rate equation
We obtain the equation for the half-life of a second order reaction (X=1/2 a):
t1/2=1/k2[A]o
This inverse relationship suggests that as the initial concentration of reactant is increased, there is a higher probability of the two reactant molecules interacting to form product. Consequently, the reactant will be consumed in a shorter amount of time, i.e. the reaction will have a shorter half-life.

Example 1.1
If the only reactant is the initial concentration of A , and it is equivalent to [A]0=4.50×10?5M and the reaction is a second order with a rate constant k=0.89M?1s?1 what is the rate of the reaction?
SOLUTION
k[A]0=1/(4.50x10?5 M)(0.89M?1s?1)= 2.50×104s (1.39)
The graph below is the graph that tests if a reaction is second order. The reaction is second order if the graph has a straight line, as is in the example below.

Practice Problems
1. Given the following information, determine the order of the reaction and the value of k, the reaction constant.
Concentration (M) Time (s)
1.0 10
0.50 20
0.33 30

*Hint: Begin by graphing

2. Using the following information, determine the half life of this reaction, assuming there is a single reactant.
Concentration (M) Time (s)
2.0 0
1.3 10
0.9633 20

3. Given the information from the previous problem, what is the concentration after 5 minutes?
Solutions
a. Make graphs of concentration vs. time (zeroth order), natural log of concentration vs. time (first order), and one over concentration vs. time (second order).
b. Determine which graph results in a straight line. This graph reflects the order of the reaction. For this problem, the straight line should be in the 3rd graph, meaning the reaction is second order.
c. The numbers should have are:

1/Concentration(M-1) Time (s)
1 10
2 20
3 30
The slope can be found by taking the "rise" over the "run". This means taking two points, (10,1) and (20,2). The "rise" is the vertical distance between the points (2-1=1) and the "run" is the horizontal distance (20-10=10). Therefore the slope is 1/10=0.1. The value of k, therefore, is 0.1 M-2s-1.

2:
a. Determine the order of the reaction and the reaction constant, k, for the reaction using the tactics described in the previous problem. The order of the reaction is second, and the value of k is 0.0269 M-2s-1.
b. Since the reaction order is second, the formula for t1/2 = k-1[A]o-1. This means that the half life of the reaction is 0.0259 seconds.
3:
a. Convert the time (5 minutes) to seconds. This means the time is 300 seconds.
b. Use the integrated rate law to find the final concentration. The final concentration is .1167 M.

Example:
A solution of ethyl acetate 0.019 N at 293 K, this solution was saponificated with NaOH (0.002 N), within 23 minute 10% of the solution was saponificated. This reaction is a second order reaction how the time changes when reducing the concentrations of reactants by ten times?
Solution:

a= 0.019 N, b= 0.002 N, and x= 0.001 N,


When reduction of concentrations at the same temperature, rate constant(k) doesn t change and the time can be evaluated as follows:


Example 2: Reaction between thiosulphate and n-bropyl bromide was studied kinetically at 310 K,

Concentration of thiosulphate was estimated at different times via titration with iodine (0.0257) N for for each 10.02 mL of thio solution and the initial concentration of thio was (0.1)N. The obtained results are listed in table below, find rate constant (k2) for this reaction?



T time (sec.) 0 1110 2010 3192 5052 7 7380 11232 78840
V(ml) 37.63 35.20 33.63 31.90 29.86 28.04 26.01 22.24
Concentration of thio can be calculated as follows:
Thio= iodine
NxV= Nx V
Nx 10.02= 0.02572 x 37.63
Conce. Of thio, N= 0.0967.
Example2:
Rate constant for saponification of ethyl acetate with sodium hydroxide at 283 K equal to (2,83) , concentration of ethyl acetate 0.05 N .calculate the time required for saponification of 50% of ethyl acetate 1 m3 for the following cases:
1-mixing with 1 m3 of NaOH (0.05 N),
2-mixing with 1 m3 of NaOH (0.1 N),
Answer:
Upon mixing solutions together, concentrations would change and the new concs. Can be calculated from dilution law for each case as follows:
(N1xV1=N2 XV2)
Case a: in this case, the two concentrations are equal:
N1 x V1=N2 x V2 (0.05 x 1= N2 x 2), N2= a= 0.025
X= a x 50%= 0.025 x 50%= 0.0125
(k2= x/at(a-x)
(t=x/ak(a-x), t= 0.0125/2.38(0.025-0.0125)= 16.8 minute,
Case b: in this case we have different concentrations between substances a and b:

Time = 2.303/k2(a-b) Log b(a-x)/a(b-x), (a= 0.05, b=0.025)
Time= 2.303/2.38(0.025- 0.05)Log 0.05(0.025-0.05)/0.025(0.05-0.0125)= 6.81 min
Determination of reaction order
The order of a reaction cannot be deduced from the chemical equation of the reaction. It must be determined by experiment.
1-Trial method:
This method is based on applying 1st order reaction and evaluating rate constant for each substitution. If values of k1 are similar or the same in values then this means that reaction is 1st order in its kinetics. If not applying 2nd order reaction and so on …..
2-graphical methods
By plotting the obtained values according to zero order, 1st order, 2nd order, and 3rd order as shown below:












3-Isolation method
This method is applied for complex reactions, which involve more than two materials and it is based on separation of these substances each from other:
A+B+C P
According to this method, A can be separated from B and C and then finding order of reaction with respect to A only via using high concentrations of each of B and C in the reaction mixture . This can be achieved by following one of above methods. Then same way is applied for B and C separately. Then the order of the whole reaction will be summation of these orders as shown below:

Order of reaction= n1+n2+n3 ,
Whereas, n1 is the order of substance A, n2 is the order of B and n3 is the order of C and the order of the reaction will be summation of them.

Theories interesting of chemical kinetics
There are two main theories that interrupting of chemical kinetics: they are Collision theory, Activated complex theory, and transition state theor.
1- Collision theory,
Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates
Define the concepts of activation energy and transition state
Use the Arrhenius equation in calculations relating rate constants to temperature
We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates.
Collision theory is based on the following postulates:
1-The rate of a reaction is proportional to the rate of reactant collisions( rate ? Z, collision frequency, Z11 or Z12)
2-The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.
3-The collision must occur with adequate energy to permit mutual penetration of the reacting species’ valence shells so that the electrons can rearrange and form new bonds (and new chemical species) .
Reaction rate tends to increase with concentration phenomenon explained by collision theory
The collision theory states that when suitable particles of the reactant hit each other, only a certain fraction of the collisions cause any noticeable or significant chemical change; these successful changes are called successful collisions. The successful collisions must have enough energy, also known as activation energy, at the moment of impact to break the preexisting bonds and form all new bonds. This results in the products of the reaction. Increasing the concentration of the reactant particles or raising the temperature - which brings about more collisions and hence more successful collisions - therefore increases the rate of a reaction.
When a catalyst is involved in the collision between the reactant molecules, less energy is required for the chemical change to take place, and hence more collisions have sufficient energy for reaction to occur. The reaction rate therefore increases.Collision theory is closely related to chemical kinetics.
Rate constant
The rate constant for a bimolecular gas-phase reaction, as predicted by collision theory is:
.
Where: k(T), is a rate constant that is dependent on T ,Z is the collision frequency,,p is the steric factor, Ea is the activation energy of the reaction,T is the temperature, R is the gas constant.
The collision frequency is:
Z11= (2)1/2 ??2C-n
Z12=1/(2)1/2 ??2C-n2
The rate constant for a bimolecular gas phase reaction as predicted by collision theory is:

.
where: r, is the rate of reaction, Z is the collision frequency, is the steric factor, Ea is the activation energy of the reaction, T is the temperature., R is gas constant.
The collision frequency for A and B is:

whereas: NA is the Avogadro constant, ?AB is the reaction cross section, kB is Boltzmann s constant, ?AB is the reduced mass of the reactants.
2-Activated complex theory
Reaction intermediates are chemical species, often unstable and short-lived ,which are not reactants or products of the overall chemical reaction, but are temporary products and/or reactants in the mechanism s reaction steps. Reaction intermediates are often free radicals or ions.
The kinetics (relative rates of the reaction steps and the rate equation for the overall reaction) are explained in terms of the energy needed for the conversion of the reactants to the proposed transition states (molecular states that corresponds to maxima on the reaction coordinates, and to saddle points on the potential energy surface for the reaction).





The temperature dependence of the rate constant usually follows the Arrhenius equation:

where Ea is the activation energy and kB is the Boltzmann constant.
3-Transition state theory (TST):
The (TST) theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.






The basic ideas behind transition state theory are as follows:
1-Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface.
2-The details of how these complexes are formed are not important. The saddle point itself is called the transition state.
3-The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.
4-The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion
In general according to this theory:

Herein, (X*) is the activated complex which is less stable than both of reactants and products and it will be in a quasi-equilibrium between these materials.
If we suppose that, there is an equilibrium between AC(X*) and the reactants (A) and B. then equilibrium can be written:


Herein, CXx, CA and CB are the concentrations of activated complex, A and B.

Equilibrium constant, can be related to partion functions:

k*= (?X*)/ (?A.?B) . e –(Ea/RT) ---3
(?X*) is the partion function for X*, (?A. for A and ?B for B.

CX*= (?X*)/ (?A.?B) . e –(Ea/RT) . CA.CB ---4



Effect of temperature and pressure on the kinetics of reaction
1-Effect of temperature
To get to the other end of the road, an object must roll with enough speed to completely roll over the hill of a certain height. The faster the object moves, the more kinetic energy (Ek=1/2mV2) ,
V=(8RT/?m)1/2
If the object moves too slowly, it does not have enough kinetic energy necessary to overcome the barrier; as a result, it eventually rolls back down. In the same way, there is a minimum amount of energy needed in order for molecules to break existing bonds during a chemical reaction. If the kinetic energy of the molecules upon collision is greater than this minimum energy, then bond breaking and forming occur, forming a new product (provided that the molecules collide with the proper orientation.


The activation energy (Ea), labeled ?G‡?G‡ in Figure is the energy difference between the reactants and the activated complex, also known as transition state. In a chemical reaction, the transition state is defined as the highest-energy state of the system. If the molecules in the reactants collide with enough kinetic energy and this energy is higher than the transition state energy, then the reaction occurs and products form. In other words, the higher the activation energy, the harder it is for a reaction to occur and vice versa.

Arrhenius Equation
By 1890 it was common knowledge that higher temperatures speed up reactions, often doubling the rate for a 10-degree rise,. Finally, in 1899, the Swedish chemist Svante Arrhenius (1859-1927) combined the concepts of activation energy and the Boltzmann distribution law into one of the most important relationships in physical chemistry:


First, note that this is. What is "decaying" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent –Ea?/RT. And what is the significance of this quantity? Recalling that RT is the average kinetic energy, it becomes apparent that the exponent is just the ratio of the activation energy Ea to the average kinetic energy.
Determining the activation energy
The Arrhenius equation ( k=A e?Ea/RT), can be written in a non-exponential form that is often more convenient to use and to interpret graphically). Taking the logarithms of both sides and separating the exponential and pre-exponential terms yields
Lnk = ln(Ae?Ea/RT)
Lnk= lnA+ln(e?Ea/RT) = LnA+ (-Ea/R)(1/T)

which is the equation of a straight line whose slope is (–Ea?/R). This affords a simple way of determining the activation energy from values of k observed at different temperatures, by plotting (Lnk) as a function of (1/T).







- )Ea/R) = –3.27 × 104 K
Ea=– (8.314 J mol–1 K–1) (–3.27 × 104 K) = 273 kJ. mol–1

Calculating Ea without a plot
To calculate the value of Ea without plotting, We need to apply this equation at least at two different temperatures Because the ln k vs.-1/T plot yields a straight line, it is often convenient to estimate the activation energy from experiments at only two temperatures. The ln A term is eliminated by subtracting the expressions for the two ln-k terms via the following steps:p
Lnk1= LnA- (Ea /RT1) -----------------1
Lnk2= LnA-(Ea /RT2) --------------------2
By re-writing equation 2, and substituting for LnA,
Lnk1= Lnk2+ (Ea/RT2)-(Ea/RT1)
Lnk1-Lnk2= (Ea/RT2)- (Ea/RT1),
Lnk2/k1= -Ea/R(1/T2-1/T1)

The Role of Collisions
What would limit the rate constant if there were no activation energy requirements? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known as the frequency- or collision factor, Z:

In some reactions, the relative orientation of the molecules at the point of collision is important, so a geometrical or steric factor (commonly denoted by ? (Greek lower case rho) can be defined. In general, we can express A as the product of these two factors:
A=Z.p
Values of ? are generally very difficult to assess; they are sometime estimated by comparing the observed rate constant with the one in which A is assumed to be the same as Z. Usually, the more complex the reactant molecules, the lower the steric factors. The deviation from unity can have different causes: the molecules are not spherical, so different geometries are possible; not all the kinetic energy is delivered into the right spot; the presence of a solvent (when applied to solutions) and other factors




Figure: The Effect of Molecular Orientation on the Reaction of NO and O3.
Most collisions of NO and O3 molecules occur with an incorrect orientation for a reaction to occur. Only those collisions in which the N atom of NO collides with one of the terminal O atoms of O3 are likely to produce NO2 and O2, even if the molecules collide with E > Ea.

2-Effect of pressure on reaction kinetics
Increasing the pressure in a gaseous reaction will increase the number of collisions between reactants, increasing the rate of reaction. This is because the activity of a gas is directly proportional to the partial pressure of the gas. This is similar to the effect of increasing the concentration of a solution.
In addition to this straightforward mass-action effect, the rate coefficients themselves can change due to pressure.

The relationship between pressure and concentration
Increasing the pressure of a gas is exactly the same as increasing its concentration. If you have a given mass of gas, the way you increase its pressure is to squeeze it into a smaller volume. If you have the same mass in a smaller volume, then its concentration is higher.
You can also show this relationship mathematically if you have come across the ideal gas equation


Rearranging this gives:

Because "RT" is constant as long as the temperature is constant,
this shows that the pressure is directly proportional to the concentration. If you double one, you will also double the other.






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