chapter 1411 copyright © by houghton mifflin company. all rights reserved. suggested problems ch 14...

Chapter 14 1Copyright © by Houghton Mifflin Company. All rights reserved. 1

Suggested Problems Ch 14

27, 28, 31, 33, 35, 41, 43, 45, 49, 53, 63,

67, 75, 77, 79, 91, 99, 101, 103, 109


Figure 14.21: Enzyme action (lock-and-key model).

Return to Slide 118


Concentration-Time Equations

First-Order Rate Law

Using calculus, you get the following equation.

kt- ]A[]A[

lno

t

ln 1/2 [A]o = kt1/2

[A]o

[A] = 1/2 [A]o


Half Life

First-Order Rate Law

You could write the rate law in the form

]A[k t

]A[ Rate

[A] = 1/2 {A]o 1/2 {A]o

k[A]o

= t


Half-life

The half-life of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value.

For a first-order reaction, the half-life is independent of the initial concentration of reactant.

21kt)ln(2

1

In one half-life the amount of reactant decreases by one-half. Substituting into the first-order concentration-time equation, we get:


Half-life

The half-life of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value.

Solving for t1/2 we obtain:

k693.0

t21

Figure 14.8 illustrates the half-life of a first-order reaction.


Figure 14.8: A graph illustrating that

the half-life of a first-order reaction is independent of initial concentration. Half life, t1/2, is the time it takes

for the [R] to decrease by 1/2.

This is exactly like radioactive decay.


Half-life

Sulfuryl chloride, SO2Cl2, decomposes in a first-order reaction to SO2 and Cl2.

At 320 oC, the rate constant is 2.2 x 10-5 s-1. What is the half-life of SO2Cl2 vapor at this temperature?

)g(Cl)g(SO)g(ClSO 2222

Substitute the value of k into the relationship between k and t1/2.

k693.0

t21


A model of SO2Cl2(g)


Hybridizations of S??


Half-life





1-5 s 1020.2

693.0t

21


Half-life





s1015.3t 421


Half-life

For a second-order reaction, half-life depends on the initial concentration and becomes larger as time goes on.

Again, assuming that [A]t = ½[A]o after one half-life, it can be shown that:

o]A[k1

t21

Each succeeding half-life is twice the length of its predecessor.


Graphing Kinetic Data

In addition to the method of initial rates, rate laws can be deduced by graphical methods.

If we rewrite the first-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.

ot ]Aln[kt]Aln[ y = mx + b




If we rewrite the first-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.

This means if you plot ln[A] versus time, you will get a straight line for a first-order reaction. (see Figure 14.9)


Figure 14.9: A plot of log [R] versus time.




If we rewrite the second-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.

ot ]A[1

kt]A[

1

y = mx + b




If we rewrite the second-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.

This means if you plot 1/[A] versus time, you will get a straight line for a second-order reaction.Figure 14.10 illustrates the graphical method of deducing the order of a reaction.


Collision Theory

Rate constants vary with temperature. Consequently, the actual rate of a reaction is very temperature dependent.

Why the rate depends on temperature can by explained by collision theory.


Collision Theory

Collision theory assumes that for a reaction to occur, reactant molecules must collide with sufficient energy and the proper orientation.

The minimum energy of collision required for two molecules to react is called the activation energy, Ea.


Figure 14.11: Two test tubes containing the same solution,one in water. Photo courtesy of American Color.

KMnO4 + H2C2O4 (oxalic acid)


Figure 14.11: One test tube shows a reaction while theother does not. Photo courtesy of American Color.

KMnO4 + H2C2O4 (oxalic acid)

CO2+ H2O


Transition-State Theory

Transition-state theory explains the reaction resulting from the collision of two molecules in terms of an activated complex.

An activated complex (transition state) is an unstable grouping of atoms that can break up to form products.

A simple analogy would be the collision of three billiard balls on a billiard table.




Suppose two balls are coated with a slightly stick adhesive.

We’ll take a third ball covered with an extremely sticky adhesive and collide it with our joined pair.




The “incoming” billiard ball would likely stick to one of the joined spheres and provide sufficient energy to dislodge the other, resulting in a new “pairing.”

At the instant of impact, when all three spheres are joined, we have an unstable transition-state complex.




If we repeated this scenario several times, some collisions would be successful and others (because of either insufficient energy or improper orientation) would not be successful.

We could compare the energy we provided to the billiard balls to the activation energy, Ea.


Figure 14.12: Importance of molecular orientation in the reaction of NO and CI2.


Potential-Energy Diagrams for Reactions

To illustrate graphically the formation of a transition state, we can plot the potential energy of a reaction versus time.

Figure 14.13 illustrates the endothermic reaction of nitric oxide and chlorine gas.

Note that the forward activation energy is the energy necessary to form the activated complex.

The H of the reaction is the net change in energy between reactants and products.

Figure 14.13 Potential-energy curve for the endothermic reaction of nitric oxide and chlorine.


Potential-Energy Diagrams for Reactions

The potential-energy diagram for an exothermic reaction shows that the products are more stable than the reactants.

Figure 14.14 illustrates the potential-energy diagram for an exothermic reaction.

We see again that the forward activation energy is required to form the transition-state complex.

In both of these graphs, the reverse reaction must still supply enough activation energy to form the activated complex.

Figure 14.14 Potential-energy curve for an exothermic reaction.


Collision Theory and the Arrhenius Equation

Collision theory maintains that the rate constant for a reaction is the product of three factors.

1. Z, the collision frequency

2. f, the fraction of collisions with sufficient energy to react

3. p, the fraction of collisions with the proper orientation to react

Zpfk



Z is only slightly temperature dependent.

This is illustrated using the kinetic theory of gases, which shows the relationship between the velocity of gas molecules and their absolute temperature.

m

abs

MRT3

velocity absT velocity or



Z is only slightly temperature dependent.

This alone does not account for the observed increases in rates with only small increases in temperature.

From kinetic theory, it can be shown that a 10 oC rise in temperature will produce only a 2% rise in collision frequency.



On the other hand, f, the fraction of molecules with sufficient activation energy, turns out to be very temperature dependent.

It can be shown that f is related to Ea by the following expression.

RTaE-

e f Here e = 2.718… , and R is the ideal gas constant, 8.31 J/(mol.K).



On the other hand, f, the fraction of molecules with sufficient activation energy turns out to be very temperature dependent.

From this relationship, as temperature increases, f increases.

RTaE-

e f Also, a decrease in the activation energy, Ea, increases the value of f.



On the other hand, f, the fraction of molecules with sufficient activation energy turns out to be very temperature dependent.

RTaE-

e f

This is the primary factor relating temperature increases to observed rate increases.



The reaction rate also depends on p, the fraction of collisions with the proper orientation.

This factor is independent of temperature changes.

So, with changes in temperature, Z and p remain fairly constant.

We can use that fact to derive a mathematical relationship between the rate constant, k, and the absolute temperature.


The Arrhenius Equation

If we were to combine the relatively constant terms, Z and p, into one constant, let’s call it A. We obtain the Arrhenius equation:

The Arrhenius equation expresses the dependence of the rate constant on absolute temperature and activation energy.

RTaE-

e A k



It is useful to recast the Arrhenius equation in logarithmic form.

Taking the natural logarithm of both sides of the equation, we get:

RTEa- Aln k ln



It is useful to recast the Arrhenius equation in logarithmic form.

We can relate this equation to the (somewhat rearranged) general formula for a straight line.

)(- Aln k ln T1

REa

y = b + m xA plot of ln k versus (1/T) should yield a straight line with a slope of (-Ea/R) and an intercept of ln A. (see Figure 14.15)


Figure 14.15: Plot of ln k versus 1/T.



A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k1, (1/T1)) and (k2, (1/T2)).

The two equations describing the relationship at each coordinate would be

)(- Aln k ln1

a

T1

RE

1

)(- Aln k ln2

a

T1

RE

2 and



A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k1, (1/T1)) and (k2, (1/T2)).

We can eliminate ln A by subtracting the two equations to obtain

)( ln21

a

1

2

T1

T1

RE

kk

chapter 1411 copyright © by houghton mifflin company. all rights reserved. suggested problems ch 14...

Documents

houghton mifflin company

o slide

order reaction

o ka o

halflife sulfuryl chloride

order rate law

reactant concentration

rate constant