Fraction of Molecular Collisions That Have Sufficient Energy to React. Now that we have established a barrier height, we need to

APPENDIX I HOW TO USE THE CRE WEB RESOURCES 931

Concept 3. Fraction of Molecular Collisions That Have Sufficient Energy to React. Now that we have established a barrier height, we need to

know what fraction of molecular collisions have sufficient energy to cross over the barrier and react. To discuss this issue, we consider reactions in the gas phase where the reacting molecules will not have only one velocity, U, but a distribution of velocities, f(U,T). Some will have high velocities and some will have low velocities as they move around and collide. These velocities are not defined with respect to a fixed coordinate system; these velocities are defined with respect to the other reactant molecules. The Maxwell-Boltzmann distribu- tion of relative velocities is given by the probability function, f(U, T)

k_B= Boltzmann’s constant = 3.29 10²⁴ cal/molecule/K m = Reduced mass, g

U = Relative velocity, m/s T = Absolute Temperature, K e = Energy kcal/molecule E = Kinetic energy kcal/mol

We usually interpret f(U, T) with dU, i.e., f(U, T) dU = fraction of reac- tant molecules with velocities between U and (U + dU).

Figure 3-3 A diagram of the orbital distortions during the reaction H • + CH₃CH₃→ H₂ + CH₂CH₃ •

The diagram shows only the interaction with the energy state of ethane (the C–H bond).

Other molecular orbitals of the ethane also distort. Courtesy of Richard I. Masel, Chemical Kinetics and Catalysis, p. 594 (Wiley, 2001).

f U T( , ) 4 m 2k_BT ---

⎝ ⎠

⎛ ⎞^{3 2} mU² 2k_BT --- U² exp

Rather than using velocities to discuss the fraction of molecules with suf- ficient energy to cross the barrier, we convert these velocities to energies using the equation for kinetic energy in making this conversion

Using this substitute, the Maxwell Boltzmann probability distribution of colli- sions with energy e (cal/molecule) at temperature T is

(3-19) In terms of energy per mole, E, instead of energy per molecule, e, we have

(3-20) where E is in (cal/mol), R is in (cal/mol/K), and f(E, T) is in mol/cal.

This function is plotted for two different temperatures in Figure 3-4. The distribution function f(E,T) is most easily interpreted by recognizing that [f(E,T) dE] is the fraction of collision with energies between E and E + dE.

(3-21) For example, the fraction of collisions with energies between 0.25 kcal/mol and 0.35 kcal/mol would be

This fraction is shown by the shaded area in Figure 3-4 and is approxi- mated by the average value of f(E,T) at E = 0.3 kcal/mole is 0.81 mol/kcal.

e 1

2---mU²

f e T( , ) 2 1 k_BT ---

⎝ ⎠

⎛ ⎞^{3 2} e^{1 2} e k_BT --- exp

f E T( , ) 2 1 RT ---

⎝ ⎠

⎛ ⎞^{3 2} E^{1 2} E ---RT exp

T₂ > T₁

T₂ = 500K T₁ = 300K

0 0 0.1 0.2 0.3 0.4 0.5

f(E,T) 0.49

-1 0.6 0.7 0.8 0.9 0.81

0.250.35 1 1.5 2 2.5 3

E

3.5 4

The fraction of collisions at T = 500K that have energy EA of 2.5 kcal/mol or greater

The fraction of collisions with energies between 0.25 and 0.35 kcal/mol is approximately f(E,T) dE = (0.81 kcal/mol)^–1 x (0.1 kcal/mol) = 0.081;

that is, 8.1% of the molecular collisions have energies between 0.25 and 0.35 kcal/mol.

4.5 5 5.5 6

(kcalmol) (kcalmol)

Figure 3-4 Energy distribution of reacting molecules.

π π

= ⎛

⎝⎜

⎞

⎠⎟ ⎡−

⎣⎢ ⎤

⎦⎥ f E T dE

k T E E

k T dE

( , ) 2 1

exp

B 3/ 2

1 2

B

∫

( )

⎛⎝⎜ ⎞

⎠⎟ = f E T dE Fraction with energies to between 0.25 and 0.35 kcal

mol ( , )

0.25 0.35

= Δ = ⎛ −

⎝⎜ ⎞

f E T dE( , ) f(0.3, 300 )K E 0.81 mol ⎠⎟ =

kcal 0.35kcal

mol 0.25kcal

mol 0.081

Thus 8.1% of the molecular collisions have energies between 0.25 and 0.35 kcal/mol.

We can also determine the fraction of collision that have energies greater than a certain value, E_A

(3-22)

This fraction is shown by the shaded area for E_A = 2.5 kcal/mole for T = 300 K (heavier shade) and for T = 500 K (lighter shade). One can easily see that for T = 500 K a greater fraction of the collisions that cross the barrier with E_A = 2.5 kcal/mol and react, which is consistent with our observation that the rates of reaction increase with increasing temperature.

For E_A > 3RT, we can obtain an analytical approximation for the fraction of molecules of collision with energies greater than E_A by combining Equa- tions (3-21) and (3-22) and integrating to get

(3-23)

Figure 3.5 shows the fraction of collisions that have energies greater than E_A as as function of the choose E_A at two different temperatures. One observes for an activation energy E_A of 20 kcal/mol and a temperature of 300 K the fraction of collisions with energies greater than 20 kcal/mol is 1.76 10¹⁴ while at 600 K, the fraction increases to 2.39 10⁷, which is a 7 orders of magnitude difference.

To summarize this discussion on the three concepts

Concept 1 The rate increases with increasing reactant concentration, Concept 2 The rate is related to the potential barrier height and to the

conversion of translational energy into potential energy, and Concept 3 The rate increases with the increasing fraction of collisions that have sufficient energy to cross over the barrier and form products.

∫

>

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟= > = ^∞

E E F E E T f E T dE

Fraction of Molecules

with

( , ) ( , )

A A E_A

π

⎡

⎣

⎢⎢

⎢

⎤

⎦

⎥⎥

⎥

= > ≅ ⎛

⎝⎜ ⎞

⎠⎟ ⎛−

⎝⎜ ⎞ E ⎠⎟

F E E T E

RT

E RT Fraction of collision

with energies greater than

( , ) 2

exp

A

A

A 1 2

A

T2 = 300K T1 = 600K

10⁻³⁰

0 5 10 15 20

E_A

F(E > EA,T)

25 30 35 40

10⁻²⁵ 10⁻²⁰ 10⁻¹⁵ 10⁻¹⁰ 10⁻⁵ 10⁰

(

kcal_mol

)

Figure 3-5 Fraction of collision with energies greater than E_A.

Thus 8.1% of the molecular collisions have energies between 0.25 and 0.35 kcal/mol.

We can also determine the fraction of collision that have energies greater than a certain value, E_A

(3-22)

This fraction is shown by the shaded area for E_A = 2.5 kcal/mole for T = 300 K (heavier shade) and for T = 500 K (lighter shade). One can easily see that for T = 500 K a greater fraction of the collisions that cross the barrier with E_A = 2.5 kcal/mol and react, which is consistent with our observation that the rates of reaction increase with increasing temperature.

For E_A > 3RT, we can obtain an analytical approximation for the fraction of molecules of collision with energies greater than E_A by combining Equa- tions (3-21) and (3-22) and integrating to get

(3-23)

Figure 3.5 shows the fraction of collisions that have energies greater than E_A as as function of the choose E_A at two different temperatures. One observes for an activation energy E_A of 20 kcal/mol and a temperature of 300 K the fraction of collisions with energies greater than 20 kcal/mol is 1.76 10¹⁴ while at 600 K, the fraction increases to 2.39 10⁷, which is a 7 orders of magnitude difference.

To summarize this discussion on the three concepts

Concept 1 The rate increases with increasing reactant concentration, Concept 2 The rate is related to the potential barrier height and to the

conversion of translational energy into potential energy, and Concept 3 The rate increases with the increasing fraction of collisions that have sufficient energy to cross over the barrier and form products.

∫

>

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟= > = ^∞

E E F E E T f E T dE

Fraction of Molecules

with

( , ) ( , )

A A E_A

π

⎡

⎣

⎢⎢

⎢

⎤

⎦

⎥⎥

⎥

= > ≅ ⎛

⎝⎜ ⎞

⎠⎟ ⎛−

⎝⎜ ⎞

E ⎠⎟

F E E T E

RT

E RT Fraction of collision

with energies greater than

( , ) 2

exp

A

A

A 1 2

A

T2 = 300K T1 = 600K

10⁻³⁰

0 5 10 15 20

E_A

F(E > EA,T)

25 30 35 40

10⁻²⁵ 10⁻²⁰ 10⁻¹⁵ 10⁻¹⁰ 10⁻⁵ 10⁰

(

kcal_mol

)

Figure 3-5 Fraction of collision with energies greater than E_A.

To carry this discussion to the next level is beyond the scope of this text as it involves averaging over a collection of pairs of molecules to find an average rate with which they pass over the transition state to become products.

Recapping the last section, the energy of the individual molecules falls within a distribution of energies some molecules have more energy than others.

One such distribution is the Boltzmann distribution, shown in Figure 3-4, where f(E, T) is the energy distribution function for the kinetic energies of the reacting molecules. It is interpreted most easily by recognizing the product [f(E, T) dE] as the fraction of molecular collisions that have an energy between E and (E + dE). For example, in Figure 3-4, the fraction of collisions that have energies between 0.25 kcal/mol and 0.35 kcal/mol is 0.081, as shown by the shaded area on the left. The activation energy has been equated with a minimum energy that must be possessed by reacting molecules before the reaction will occur. The fraction of the molecular collisions that have an energy E_A or greater is shown by the shaded areas at the right in Figure 3-4.

Figure 3-5 shows the fraction of collisions with energies greater than as a func- tion of E_A at two different temperatures.

3.3.2 The Arrhenius Plot

Postulation of the Arrhenius equation, Equation (3-18), remains the greatest single advancement in chemical kinetics, and retains its usefulness today, more than a century later. The activation energy, E, is determined experimentally by measuring the reaction rate at several different temperatures. After taking the natural logarithm of Equation (3-18), we obtain

(3-24) We see that the activation energy can be found from a plot of ln k_A as a func- tion of (1/T), which is called an Arrhenius plot. The larger the activation energy, the more temperature-sensitive the reaction. That is, for large E, an increase in just a few degrees in temperature can greatly increase k and thus increase the rate of reaction.

k_A=ln A

ln E

---R 1 T---

⎝ ⎠⎛ ⎞

Semilog Plot 10

1.0 0.1 0.01

0.0025 0.003 Low E Slope = – E

R 1

( )

T ^(K–1) (s^–1)

High E k

Figure 3-6 Calculation of the activation energy from an Arrhenius plot.

Calculation of the activation energy

Example 3–1 Determination of the Activation Energy

Calculate the activation energy for the decomposition of benzene diazonium chlo- ride to give chlorobenzene and nitrogen

using the information in Table E3-1.1 for this first-order reaction.

Solution

We start by recalling Equation (3-20)

(3-24) We can use the data in Table E3-1.1 to determine the activation energy, E, and fre- quency factor, A, in two different ways. One way is to make a semilog plot of k vs.

(1/T) and determine E from the slope (–E/R) of an Arrhenius plot. Another way is to use Excel or Polymath to regress the data. The data in Table E3-1.1 was entered in Excel and is shown in Figure E3-1.1, which was then used to obtain Figure E3-1.2.

A step-by-step tutorial to construct both an Excel and a Polymath spreadsheet is given in the Chapter 3 Summary Notes on the CRE Web site.

The equation for the best fit of the data

(E3-1.1) TABLE E3-1.1 DATA

k (s¹) 0.00043 0.00103 0.00180 0.00355 0.00717 T (K) 313.0 319.0 323.0 328.0 333.0

N₂ N——N

Cl

Cl

lnk _A ln A E R--- 1

T---

⎝ ⎠⎛ ⎞

Figure E3-1.1 Excel spreadsheet.

Tutorials

lnk 14,017 ---T 37.12

is also shown in Figure E3-1.2(b). From the slope of the line given in Figure 3-1.2(b) and Equation (3-20), we obtain

From Figure E3-1.2(b) and Equation (E3-1.1), we see

Taking the antilog, we find the frequency factor to be

(E3-1.2) Analysis: The activation energy, E, and frequency factor, A, can be calculated if we know the specific reaction rate, k, at two temperatures, T₁ and T₂. We can either use the Arrhenius equation (3-18) twice, once at T₁ and once at T₂, to solve two equa- tions for the two unknowns, A and E, or we can take the slope of a plot of (ln k) as a function of (1/T); the slope will be equal to (–E/R).

There is a rule of thumb that states that the rate of reaction doubles for every 10C increase in temperature. However, this rule is true only for specific combinations of activation energies and temperatures. For example, if the acti- vation energy is 53.6 kJ/mol, the rate will double only if the temperature is raised from 300 K to 310 K. If the activation energy is 147 kJ/mol, the rule will be valid only if the temperature is raised from 500 K to 510 K (see Prob- lem P3-7_B for the derivation of this relationship).

Figure E3-1.2 (a) Excel semilog plot; (b) Excel normal plot after taking ln(k).

(a) (b)

k = k1 exp E R--- 1

T₁ --- 1

T---

⎝ ⎠

⎛ ⎞

E R---

14 017 K,

E (14 017 K, )R (14 017 K, ) 8.314 J mol K ---

⎝ ⎠

⎛ ⎞

E 116.5 kJ mol---

ln A37.12

A1.3210¹⁶ s¹ k 1.32 10¹⁶ 14 017 K,

---T

exp

The rate does not always double for a temperature increase of 10C.

The larger the activation energy, the more temperature-sensitive is the rate of reaction. While there are no typical values of the frequency factor and activa- tion energy for a first-order gas-phase reaction, if one were forced to make a guess, values of A and E might be and 100 kJ/mol. However, for fami- lies of reactions (e.g., halogenation), a number of correlations can be used to estimate the activation energy. One such correlation is the Polanyi-Semenov equation, which relates activation energy to the heat of reaction (see Profes- sional Reference Shelf R3.1). Another correlation relates the activation energy to differences in bond strengths between products and reactants.⁴ While the activa- tion energy cannot be currently predicted a priori, significant research efforts are under way to calculate activation energies from first principles.⁵

One final comment on the Arrhenius equation, Equation (3-18). It can be put in a most useful form by finding the specific reaction rate at a temperature T₀; that is,

and at a temperature T

and taking the ratio to obtain

(3-25) This equation says that if we know the specific reaction rate k(T₀) at a tem- perature, T₀, and we know the activation energy, E, we can find the specific reac- tion rate k(T) at any other temperature, T, for that reaction.

3.4 Present Status of Our Approach to Reactor Sizing and Design In Chapter 2, we combined the different reactor mole balances with the defini- tion of conversion to arrive at the design equation for each of four types of reactors, as shown in Table 3-2. Next we showed that if the rate of disappear- ance is known as a function of the conversion X

then it is possible to size CSTRs, PFRs, and PBRs operated at the same con- ditions under which –r_A = g(X) was obtained.

4M. Boudart, Kinetics of Chemical Processes (Upper Saddle River, NJ: Prentice Hall, 1968), p. 168. J. W. Moore and R. G. Pearson, Kinetics and Mechanisms, 3rd ed.

(New York: Wiley, 1981), p. 199. S. W. Benson, Thermochemical Kinetics, 2nd ed.

(New York: Wiley, 1976).

5R. Masel, Chemical Kinetics and Catalysis, New York: Wiley, 2001, p. 594.

10¹³ s ¹

k T( )₀ Ae^E/RT0

k T( )Ae^E/RT

A most useful form

of k(T) k T( ) k T( )₀ e

E R--- 1

T₀ --- 1

T---

⎝ ⎠

⎛ ⎞

Where are we?

r_A

g X( )

In general, information in the form r_A g(X) is not available. However, we have seen in Section 3.2 that the rate of disappearance of A, , is normally expressed in terms of the concentration of the reacting species. This functionality (3-2) is called a rate law. In Chapter 4, we show how the concentration of the react- ing species may be written in terms of the conversion X

(3-26) With these additional relationships, one observes that if the rate law is given and the concentrations can be expressed as a function of conversion, then in fact we have r_A as a function of X and this is all that is needed to evaluate the isothermal design equations. One can use either the numerical techniques described in Chapter 2 or, as we shall see in Chapter 5, a table of integrals, and/or software programs (e.g., Polymath).

TABLE 3-2 DESIGN EQUATIONS

Differential Form

Algebraic Form

Integral Form

Batch (2-6) (2-9)

Backmix

(CSTR) (2-13)

Fluidized CSTR Tubular

(PFR) (2-15) (2-16)

Packed bed

(PBR) (2-17) (2-18)

Closure. Having completed this chapter, you should be able to write the rate law in terms of concentration and the Arrhenius temperature depen- dence. We have now completed the first two basic building blocks in our algorithm to study isothermal chemical reactions and reactors.

In Chapter 4, we focus on the third building block, stoichiometry, as we use the stoichiometric table to write the concentrations in terms of conversion to finally arrive at a relationship between the rate of reaction and conversion.

N_A0dX

---dt r_AV t NA0 dX

r_AV ---

0

X

The design

equations V F_A0X

rA

---

W FA0X rA ---

FA0dX

dV---rA V FA0 dX

rA

---

0

X

F_A0 dX

dW---rA W F_A0 dX

rA

---

0

X

r_A r_A

[k_A( )T ][fn(C_A,C_B, . . .)]

C_jh_j( )X

and then we can design isothermal reactors r_A

f C( )_j C_jh_j( )X

↓ r_A g X( )

The CRE Algorithm

• Mole Balance, Ch 1

• Rate Law, Ch 3

• Stoichiometry, Ch 4

• Combine, Ch 5

• Evaluate, Ch 5

• Energy Balance, Ch 11

Mole Balance Rate Law

S U M M A R Y

1. Relative rates of reaction for the generic reaction:

(S3-1) The relative rates of reaction can be written either as

or (S3-2)

2. Reaction order is determined from experimental observation:

A B C (S3-3)

r_A k

The reaction in Equation (S3-3) is order with respect to species A and order with respect to species B, whereas the overall order, n, is (). If 1 and 2, we would say that the reaction is first order with respect to A, second order with respect to B, and overall third order. We say a reaction fol- lows an elementary rate law if the reaction orders agree with the stoichiometric coefficients for the reaction as written.

Examples of reactions that follow an elementary rate law:

Irreversible reactions First order

Second order Reversible reactions

Second order

Examples of reactions that follow nonelementary rate laws:

Homogeneous

Heterogeneous reactions

3. The temperature dependence of a specific reaction rate is given by the Arrhenius equation

k Ae^E/RT (S3-4)

where A is the frequency factor and E the activation energy.

A b

a---B c a---C d

a---D

→

r_A

---a r_B ---b r_C

---c r_D ---d

r_A a --- r_B

b --- r_C

---c r_D ---d

⎯⎯→

C_AC_B

C₂H₆ ⎯⎯→ C₂H₄H₂ r_C

2H₆

kC_C

2H₆

CNBrCH₃NH₂ ⎯⎯→ CH₃BrNCNH₂ r_CNBr kC_CNBrC_CH

3NH₂

2C₆H₆ ⎯→←⎯ C₁₂H₁₀H₂ r_C

2H₆

k_C

2H₆ C²C₆H₆

C_C

12H₁₀C_H

2

K_C ---

⎝ ⎠

⎛ ⎞

CH₃CHO ⎯⎯→ CH₄CH₂ r_CH

3CHO

kC^{3 2}CH₃CHO

C₂H₄ H₂ ⎯⎯→ _cat C₂H₆ r_C

2H₄

k P_C

2H₄P_H

2

1 K_C

2H₄P_C

2H₄

---

If we know the specific reaction rate, k, at a temperature, T₀, and the activation energy, we can find k at any temperature, T

(S3-5) Concept 1 The rate increases with increasing reactant concentrations,

Concept 2 The rate is related to the potential barrier height and to the conversion of translational energy into potential energy, and

Concept 3 The rate increases as the fraction of collisions that have sufficient energy to cross over the barrier and form products increases.

Similarly from Appendix C, Equation (C-9), if we know the partial pressure equilibrium constant K_P at a temperature, T₁, and the heat of reaction, we can find the equilibrium constant at any other tempera- ture

(C-9) C R E W E B S I T E M A T E R I A L S

• Expanded Material 1. Collision Theory 2. Transition State 3. Molecular Simulation

4. How to Estimate the Activation Energy from the Polanyi Equation 5. Web P3-1_A Puzzle Problem “What’s Wrong with this Solution?”

6. Additional Homework Problems

• Learning Resources

1. Summary Notes for Chapter 3 2. Web Modules

A. Cooking a Potato

Chemical reaction engineering is applied to cooking a potato

with

k = Ae^–E/RT

B. Molecular Reaction Engineering

Molecular simulators (e.g., Gaussian) are used to make predictions of the activation energy.

k T( ) k T( )₀exp E R--- 1

T₀ --- 1

T---

⎝ ⎠

⎛ ⎞

K_P( )T K_P( )T₁ exp HRx

---R 1 T₁ --- 1

T---

⎝ ⎠

⎛ ⎞

Starch (crystalline)→k Starch amorphous

8 minutes at 400° F 12 minutes at 400° F 16 minutes at 400° F

• FAQs (Frequently Asked Questions)

• Professional Reference Shelf R3.1. Collision Theory

In this section, the fundamentals of collision theory

are applied to the reaction

A + B → C + D to arrive at the following rate law

The activation energy, E_A, can be estimated from the Polanyi equation E_A = Eº_A + γPΔH_Rx

R3.2. Transition-State Theory

In this section, the rate law and rate-law parameters are derived for the reaction A + BC ABC^#→ AB + C

using transition-state theory. Figure PW3B-1 shows the energy of the molecules along the reaction coor- dinate, which measures the progress of the reaction.

R3.3. Molecular Dynamics Simulations

The reaction trajectories are calculated to determine the reaction cross section of the reacting mole- cules. The reaction probability is found by counting up the number of reactive trajectories after Karplus.⁶

6M. Karplus, R.N. Porter, and R.D. Sharma, J. Chem. Phys., 43 (9), 3259 (1965).

U_R

A B

σAB σAB

Figure R3.1 Schematic of collision cross section.

rA AB 2 8k_BT

---

⎝ ⎠

⎛ ⎞^{1 2} N_Avoe^EARTC_AC_B Ae^EARTC_AC_B

⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

A

→←

Figure PW3B-1 Reaction coordinate for (a) S_N2reaction, and (b) generalized reaction. (c) 3-D energy surface for generalized reaction.

Energy

Reaction coordinate (a) E_b A – B – C

AB + C A + BC

Energy

Reaction coordinate (b) BC

+

A AB

+ C ABC

Reactants Products

Potential energy

(c)

Reaction path

A-B distance Initial state

A + BC Completely disassociated state A + B + C Final State

A – B + C B – C distance

Activated state

Nonreactive Trajectory

Reactive Trajectory

Q U E S T I O N S A N D P R O B L E M S The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.

A = ● B = ■ C = ◆ D = ◆◆

Questions

Q3-1_C (a) List the important concepts that you learned from this chapter. What concepts are you not clear about?

(b) Explain the strategy to evaluate reactor design equations and how this chapter expands on Chapter 2.

(c) Choose a FAQ from Chapters 1 through 3 and say why it was the most helpful.

(d) Listen to the audios on the CRE Web site. Select a topic and explain it.

(e) Read through the Self-Tests and Self-Assessments in Chapters 1 through 3 in the Summary Notes on the CRE Web site. Select one and critique it.

(f) Which example on the CRE Web site’s Summary Notes for Chapters 1 through 3 was most helpful?

Problems

P3-1_A (a) Example 3-1. Make a plot of k versus T and ln k versus (1/T) for E = 240 kJ/mol and for E = 60 kJ/mol. (1) Write a couple of sentences describing what you find. (2) Next, write a para- graph describing the activation, how it affects chemical reaction rates, and what its origins are.

(b) Collision Theory—Professional Reference Shelf. Make an outline of the steps that were used to derive

–r_A = Ae^–E/RT C_AC_B

C A

B RBC B

C

A

RAB

b

R

t

RAB

RA C

RBC

t₁ time R

R_AB

R_AB R_BC

R_BC A

B C

B C

b A

(c) The rate law for the reaction (2A + B → C) is –r_A = k_A C_B with k_A = 25(dm³/mol)²/s.

What are k_B and k_C?

P3-2_B Molecular collision energies—refer to Figure 3-3.

(a) What fraction of molecular collisions have energies less than or equal to 2.5 kcal at 300 K? At 500 K?

(b) What fraction of molecular collisions have energies between 3 and 4 kcal/mol at T = 300 K? At T = 500K?

(c) What fraction of molecular collisions have energies greater than the activation energy E_A = 25 kcal at T = 300 K? At T = 500K?

P3-3_B (a) Use Figure 3-1(b) to sketch the trajectory over the saddle point when the BC and AB molecules vibrate with the minimum separation distance being 0.20 Angstroms and the maximum separation being 0.4 Angstroms.

(b) At point Y, R_AB = 2.8 Angstroms, sketch the potential energy a s a function of the distance R_BC noting the minimum at valley floor.

(c) At Point X, R_BC = 2.8 Angstroms, sketch the potential energy as a function of R_AB noting the minimum on the valley floor.

P3-4_B Use Equation (3-20) to make a plot of f(E,T) as a function of E for T = 300, 500, 800, and 1200 K.

(a) What is the fraction of molecules that have sufficient energy to pass over a energy barrier of 25 kcal at 300, 500, 800, and 1200 K?

(b) For a temperature of 300 K, what is the ratio of the faction of energies between 15 and 25 kcal to the same energy range (15-25 kcal) at 1200 K?

(c) Make a plot of F(E > E_A,T) as a function of (E_A/RT) for (E_A/RT) > 3. What is the fraction of col- lisions that have energies greater than 25 kcal/mole at 700 K?

(d) What fraction of molecules have energies greater than 15 kcal/mol at 500 K?

(e) Construct a plot of F(E > E_A,T) versus T for E_A = 3, 10, 25, and 40 kcal/mole. Describe what you find. Hint: Recall the range of validity for T in Equation (3-23).

P3-5_A The following figures show the energy distribution function at 300 K for the reaction A + B → C

For each figure, determine the following:

(a) What fraction of the collisions have energies between 3 and 5 kcal/mol?

(b) What fraction of collisions have energies greater than 5 kcal/mol? (Ans.: Figure P3-5 (b) fraction = 0.28)

(c) What is the fraction with energies greater than 0 kcal/mol?

(d) What is the fraction with energies 8 kcal/mol or greater?

(e) If the activation energy for Figure P3-5(b) is 8 kcal/mol, what fraction of molecules have an energy greater than E_A?

(f) Guess what (sketch) the shape of the curve f(E,T) versus E shown in Figure P3-5(b) would look like if the temperature were increased to 400 K. (Remember: .)

C_A²

0.25 0.2 0.15 0.1 0.05

0 1 2 3 4 5 6 7 8

E(kcal)/mol T = 300 K kcal

mol

–1

f(E,T)

0.25 0.2 0.15 0.1 0.05

0 2 4 6 8 10

E(kcal/mol) T = 300 K kcal

mol f(E,T)–1

Figure P3-5(a) Figure P3-5(b)

f E T( , )dE

0

¹