Statistical Thermodynamic Treatment of Equilibrium Constants

The statistical thermodynamic expression for the free energy of an ideal gas can be reduced to a very simple form. Equation (6-79) reads

G = —RT(\n Q) + 7? Γ κ [ ^η ^ ] + RT ln N⁰ - RT.

However, only Q t r a n s depends on K, and for an ideal gas it is proportional to V by Eq. (4-68), so that [3(lnQ)/dK]^r is just l/V and the second and fourth terms of Eq. (6-79) cancel. Also, Eq. (6-79) was derived on the assumption that H⁰, the absolute enthalpy at 0 K, was set equal to zero. We now want to retain this reference point, and since for an ideal gas at 0 Κ H⁰ = E^Q , we obtain

G = E⁰ - RT ln Q + RTln N⁰ (14-102)

or

G = E⁰ - RT ln -5- . (14-103)

We now define a subsidiary partition function φ defined as Q = φν, that is, φ contains all the terms of Q except the volume factor. If the standard state is taken to be 1 atm pressure, then V = N^QkT, and Eq. (14-103) becomes

G° = E⁰ — RTlntykT). (14-104)

From our standard chemical reaction

aA + bB + = mM + « N + · · · ,

for which

AG⁰ = -RT In K^P [Eq. (7-11)].

If the standard state is 1 atm, substitution of Eq. (14-104) gives

K^P = (kT)^An Κ^φβ~^ΔΕ«'*^τ, (14-105) where An is the number of moles of products minus moles of reactants and Κ^φ

is the quantity

= Φ*Τφκ^η

-[Note the resemblance to Eq. (7-51).] Since by Eq. (7-14) K^c = K^P(RT)~^An, or Kⁿ = K^P(kT)-*ⁿ⁹ where Kⁿ is the concentration equilibrium constant using the concentration in molecules per unit volume n, Eq. (14-104) can be put in the final form

Kⁿ = K*e-*^E°^/RT. (14-106)

An equilibrium constant can thus be expressed as the product and quotient of the slightly modified partition functions φ multiplied by the term e-^AE°^/RT, which involves the change in zero-point energy accompanying the chemical reaction.

β. Application to Transition-State Theory

We first calculate the concentration of activated complex [AB]* as ([AB]*) = Kⁿ*(A)(B) = K<*e-*^Ei'*^T(A)(B).

A special assumption has been made regarding Q or φ for [AB]*, however, namely that one of its degrees of freedom is in translation along the reaction coordinate;

this particular component of φ^χ is then to be expressed separately with Eq. (4-67) for the translational partition function in one dimension, leaving φ^χ containing the usual three degrees of translational freedom but one less than the normal number of vibrational contributions. We therefore write

B 1,1 = ( i ^ ^ l e _ « , , „ ( A ) ( B ) >

2h φ^Αφ^Β

where a* must be some characteristic linear dimension associated with the saddle point of Fig. 14-10 and the factor of J is present since only forward translation along the reaction path is being considered.

The frequency with which molecules of [AB]* will pass over the saddle is next expressed as in terms of a velocity v*:

y*([AB]*) (27rm*k7y/² ώ* ^± rate = U = v*i —I— - f -^ * / « ( A ) ( B ) .

or 2n ΦΑΦΒ

The velocity v* is now supposed to have a Boltzmann distribution of values given by the treatment for a one-dimensional gas, Eq. (2-29). The average velocity for a three-dimensional gas is (SkT/nm)¹/² [Eq. (2-38)]; for a one-dimensional gas it

GENERAL REFERENCES 595

turns out to be (IkT/mn)¹/², and this result is inserted in place of v^t to obtain rate = ^ - f i j - e -^J^ ^ ( A ) ( B ) ,

η φ^Αφ^Β

from which the rate constant, expressed in molecules per cubic centimeter, is

k = π " e-^AE^'^RT . (14-107)

h ΦΑΦΒ

One may, of course, factor the partition function φ in the usual manner, writing Φ = ^trans^rot^vib , and evaluate the terms from the standard expressions of Chapter 4, remembering that V is dropped out of Qtrans and [AB]* has one less than the usual number of vibrations; otherwise the </>'s are the same as the Q's.

C. Discussion

The preceding derivation has led to an equation of the same form as Eq. (14-85) but shows more specifically the assumptions that are made in transition-state theory. There is a certain artificiality to the treatment of the rate of passage over the saddle point, but the more interesting aspect is the following. Although the <£'s are temperature-dependent, the major temperature term is εχρ(—ΑΕ⁰*/ΙΙΤ). The experimental activation energy, as obtained by means of the Arrhenius equation, is thus largely identified with the difference in zero-point energy between [AB]*

and the reactants A and B. Thus [AB]* is treated as a definite chemical species and not just as some special energized association of A and B. Its translational, rotational, and vibrational partition functions are treated as though [AB]* were in ordinary thermal equilibrium with its surroundings; that is, the ambient tempera­

ture is used in their evaluation.

The contrast with collision theory is especially striking in the case of a uni­

molecular reaction. Collision theory treats such a reaction in terms of a molecule A * which is essentially the same molecule as A but in a highly vibrationally excited state. Transition-state theory says that [A]* is a new chemical species of extra zero-point energy ΔΕ^0Χ whose vibrations are those corresponding to equi­

librium with the ambient temperature. The writer must confess a reluctance to accept this second picture.

The case of an elementary bimolecular reaction is somewhat different. Here we do rather expect that A and Β will form some new kind of molecule (as in the H² + I² reaction), and it makes sense to regard this as having a real ΔΕ^0Χ. There is still a question as to whether those molecules of [AB]* that pass on to reaction do, in reality, have vibrations (and rotations) corresponding to the ambient temperature. These questions are rather subtle ones and are brought up here mainly to give some perspective to the two theories of reaction rates.

G E N E R A L R E F E R E N C E S

BENSON, S. W. (1960). "The Foundations of Chemical Kinetics." McGraw-Hill, N e w York.

FROST, Α . Α . , A N D PEARSON, R. A. (1961). "Kinetics and Mechanism." Wiley, N e w York.

C I T E D R E F E R E N C E S BENSON, S. (1968). "Thermochemical Kinetics." Wiley, N e w York.

HARCOURT, Α . , A N D ESSEN, W. (1865). Proc. Roy. Soc. (London) 1 4 , 470.

HARCOURT, Α . , A N D ESSEN, W. (1866). Phil. Trans. Roy. Soc. (London) 1 5 6 , 193.

HARCOURT, Α . , A N D ESSEN, W . (1867). Phil. Trans. Roy. Soc. (London) 1 5 7 , 117 HERSCHBACH, D . , et al. (1956). / . Chem. Phys. 2 5 , 737.

K U N S M A N , C. H . (1928). / . Amer. Chem. Soc. 5 0 , 2100.

S U L L I V A N , J. (1967). / . Chem. Phys. 4 6 , 7 3 .

E X E R C I S E S

Take as exact numbers given t o o n e significant figure.

1 4 - 1 Calculate fc, τ, and t^{l l 2} for a first-order reaction for which t^xi^z = 25 min.

Ans. 0.044 m i n "¹, 22.8 m i n , 15.8 min.

1 4 - 2 The disappearance of a reactant is at the rate of 0.25 % m i n^{- 1}. Calculate the half-life if the reaction is first order.

Ans. 277 min.

1 4 - 3 A plot of 1/(A) versus t gives a straight line of intercept 150 a t m^{- 1} and slope 2 χ 1 0 ~³ a t m^{- 1} s e c^{- 1}. Calculate the half-life for this reaction.

Ans. 7.5 χ 1 0⁴s e c .

1 4 - 4 A gas-phase reaction has the stoichiometry A + 2 B - > products and obeys the rate law d(A)ldt = -A:(A)(B) with k = 2 χ 10"⁴ liter m o l e '¹ sec"¹ at 25°C. A reaction mixture at 25°C initially contains 2 0 % A and 80 % Β at a total pressure of 2 atm. Cal­

culate the percentage o f A and o f Β that have reacted after 1 hr.

Ans. 4.55% A , 2.27% B.

1 4 - 5 Calculate the half-life for the reaction o f Exercise 14-4 if (a) ( A )⁰ = 0.02 Μ and ( B )⁰ = 0.5 M , (b) if (A)o = 0.5 Μ and ( B )⁰ = 0.02 M , and (c) if ( A )⁰ = 0.02 Μ and ( B )⁰ = 0.04 M.

(Half-life is the time for half o f that reactant not i n excess t o have reacted.) A s s u m e i n cases (a) and (b) that the reactant i n excess is sufficiently s o that the reaction is p s e u d o first order.

Ans. (a) 6930 sec, (b) 3470 sec, (c) 1.25 χ 1 0⁵ sec.

1 4 - 6 Suppose that A and Β are volatile solids and that the reaction o f Exercise 14-4 is carried out in a five-liter flask at 25°C in the presence o f excess solid A and B. If the vapor pressures of A and Β are 0.1 atm and 0.02 atm, respectively, h o w long should it take for 0.5 mole of A t o be converted t o products?

Ans. 1.50 χ 1 08s e c . GARDINER, W. C , JR. (1969). "Rates and Mechanisms of Chemical Reactions." Benjamin, N e w

York.

R O B I N S O N , P. J., A N D H O L B R O O K , K . A . ( 1 9 7 2 ) . " U n i m o l e c u l a r R e a c t i o n s . " W i l e y (Interscience), N e w York.

W E S T O N , R. E., J R . , A N D S C H W A R Z , H . A . ( 1 9 7 2 ) . "Chemical Kinetics." Prentice-Hall, E n g l e w o o d Cliffs, N e w Jersey.

EXERCISES 5 9 7 14-7* A s a continuation o f the e x a m p l e o f Section 14-2B, rework part (a) assuming the reac­

tion stoichiometry t o b e 2 A + Β - > products. T h e rate constant, k, is still 0.02 M "¹ m i n " \ but is n o w defined by d(A)/dt = - 2 f c ( A ) ( B ) .

Ans. 16.26% A reacted.

14-8 Under certain conditions the reaction H² + B r² = 2HBr obeys a rate law of the form d(HBr)/dt = A:(H²)^a(Br²)⁶(HBr)^c. At a certain temperature Τ the reaction rate is found t o be R when (HBr) is 2 Μ and ( H²) and (Br²) are both 0.1 M. T h e rate varies with c o n ­ centration as follows:

Concentration ( M )

( H²) (Br²) (HBr) Rate

0.1 0.1 2 R

0.1 0.4 2 SR

0.2 0.4 2 16R

0.1 0.2 3 1.88Λ

Find the exponents a, b, and c.

Ans. 1, f, — 1.

14-9 Consider the reaction A² + B² 2 A B . Explain w h a U h e rate law for each of the following mechanisms should be: (a) A² - i 2 A (slow step), B² = 2B (rapid equilibrium highly

dis-k * Κ Κ

placed toward B2) and A + Β A B (fast), (b) A2 = 2 A , B2 = 2B (both rapid equili-bria, highly displaced toward A² and B², respectively), A 4- Β - * A B (slow step), (c) A² + B² A²B² (slow step), A²B² -i 2 A B (fast).

Ans. (a) R = ^ ( A^t) ; (b) R = k^&pp(A²)^lf2(B²yf\

where A:^{a p p} = Κΐ^/2κ1% ; (c) R = Αά(Α²)(Β²).

14-10 Derive the steady-state rate law for the mechanism

N O + H² % N O · H²

N O · H² + N O ^> N²0 + H²0 . S h o w under what condition this reduces t o the form of Eq. (14-39).

Ans. R = ^2( Ν Ο )2( Η2) / [ ^ + A:2(NO)].

14-11 Verify the expressions for k^t that follow Eq. (14-68).

14-12 Calculate the activation energy for the reaction 2 H I H² + I² using the first and last entries of Table 14-3. Obtain also the frequency factor A.

Ans. 44.6 kcal m o l e^{- 1}, 1.19 χ 1 0^{1 1} liter m o l e^{- 1} s e c^{- 1}.

14-13 Calculate the rate constant for the reaction C H³ + H² - > C H⁴ + Η at 100°C.

Ans. 478 liter m o l e ~¹ sec~ *.

14-14 Calculate log A' for the reaction of Exercise 14-13, assuming a collision diameter of 3 A.

Ans. 10.27.

14-15 The reaction cyclopropane propylene conforms (approximately) to the simple L i n d e m a n n mechanism. T h e limiting rate constant has the parameters l o g Λ = 15.17 (A in sec - x) and E* = 65 kcal m o l e "1 a n d at 765 K, knni = 1 0 "4 s e c "1 at a pressure o f 1 Torr. Calculate &uni,a> at 765°C and the ratio k^xjk_².

Ans. 4.0 χ 1 0 -4 sec"1, 3.0 Torr.

14-16 Calculate and ΔΗ^η for the decomposition of Ν²Ο^δ. At high pressures the reaction is first order with k = 3.35 χ 10~5 sec"1 at 25°C and 0.0048 sec"1 at 65°C.

Ans. 24.9 kcal m o l e " \ 4.39 cal K "1 m o l e "1.

14-17 The entropy of activation is - 2 0 cal K r1 m o l e "1 for the reaction C H3 + N H8 C H4 + N H2, thought to be a simple reaction. Estimate the translational contribution to this entropy and hence, by difference, that from rotation and vibration. A s s u m e 25°C.

Ans. J 5 ? *a n B = - 3 2 . 2 cal K ~1 m o l e " \ ( J S J o t + ^JS*l b) = 12.2 cal K "1 m o l e "1.

PROBLEMS

14-1 The reaction 4A(g) - * B(g) + 6C(g) is studied, and the following data are obtained, where Ρ is the total pressure. Assume that n o products are present initially and that the reaction goes to completion. S h o w what the order of the reaction is and calculate the rate constant.

Ρ (Torr) 500 687.5 781.3 ((sec) 0 60 120 14-2 The decomposition C l C O O C C l3 -> 2 C O C l2 goes to completion, and the total pressure

is found to vary with time (at 280°C) as follows:

r(sec) 0 500 800 1300 1800 Ρ (Torr) 15.0 18.9 20.7 23.0 24.8 Show what the order of the reaction is and calculate the rate constant.

14-3 The reaction 2NOC1 - • 2 N O + C l2 is studied at 200°C. The concentration of a sample initially consisting of NOC1 only changes with time as follows:

/ ( s e c ) 0 200 300 500 CNOCI(M) 0.02 0.0159 0.0144 0.0121 Assuming that the reaction goes to completion as stated, show what the order is and calculate the rate constant.

14-4 A reaction between the substances A and Β is represented stoichiometrically by the equa­

tion A + Β C. Observations on the rate of this reaction are obtained as follows (figures are "rounded" values):

Initial concentration (M) Concentration of A at end of

A Β i ( h r ) interval (M) 0.1 1.0 0.5 0.095 0.1 2.0 0.5 0.080 0.1 0.1 1000 0.05 0.2 0.2 500 0.10

PROBLEMS 599 Formulate the differential equation that expresses the rate of this reaction and calculate the specific rate constant. Indicate your reasoning.

14-5 The half-life for the decomposition of N H³ as catalyzed by a hot tungsten filament (at about 1100°C) is found to vary with the initial N H³ pressure as follows:

Ρ (Torr) 300 100 30

i^{l / 2}( m i n ) 8.4 2.8 0.84

[see Kunsman (1928)]. S h o w what the order of the reaction is (in Ρ^{Ν Η} ) and calculate the rate constant.

14-6 A reaction obeys the rate law of Eq. (14-31). H o w should (A) be plotted as a function of time so as to give a straight line plot? Relate the slope of this plot to k> ί^χ/², and n.

14-7 S h o w that Eq. (14-23) c a n be put in the form

where F^A = ( A ) / ( A )⁰ and ρ = ( B )⁰/ ( A )⁰ . M a k e plots o f F^A versus (B)⁰kt for ρ = 0 . 1 , 0.5, 1.0, 2.0, and o o .

14-8 Derive the integrated rate expression for a reaction of order i and the equation relating k t o half-life.

14-9 The following hypothetical rate data are f o u n d for the reaction 3 H² + N² - + 2 N H⁸ at 450°C.

Initial pressures (Torr)

Initial rate, —dPtotldt

Experiment ^{p o p o} ( T o r r h r -¹)

1 100 1 0.01

2 2 0 0 1 0.04

3 4 0 0 0.5 0.08

(a) T h e rate law is of the form R a t e = & ΡΗ, ΡΝ, . S h o w what the values o f χ a n d y are.

(b) Calculate the rate constant, (c) Calculate the initial rate for the conditions of Experiment 1, but for 500°C, taking the activation energy t o be 45 kcal m o l e "¹. 14-10 T h e reaction A + Β - > products o b e y s the rate law d(A)ldt = —k(A)(B). In Experiment

1 the initial concentrations are 0.001 Μ a n d 2 Μ for A and B , respectively, and in Ex­

periment 2 the initial concentrations are b o t h C0. F i n d the value for C0 such that the time for half reaction will b e the s a m e for the t w o experiments.

14-11 The d e c o m p o s i t i o n o f o z o n e , 2 0⁸ - > 3 0², is observed t o o b e y the rate law R = k(Oz)2l(02). Suggest a mechanism.

14-12 T^{n e} following reaction sequence is proposed:

N 0²C 1 % N 0² + CI,

*-l

N 0²C 1 + CI ^ N 0² + C l².

Show what the overall reaction is and derive the stationary-state rate law.

14-13 R. Ogg proposed the following mechanism for the reaction N²0⁶ - > N²0⁴ + ^ 0²:

N205 f c^ N Oa + N 03,

*-i

N 0² + N 0³ - i N O + O^z + N 0², N O + N 0³ ^ N^a0⁴ .

Derive the stationary-state rate law for this mechanism. (Both N 0³ and N O are treated as stationary-state intermediates.)

14-14 Sullivan (1967) reported that k² of reaction (14-66) is 1.12 χ 10⁵ liter² m o l e "² sec"¹ at 418°K. Calculate K^x and k.².

14-15 For the decomposition of N²0⁵ the following is observed:

t (°C) 25 35 45 55 65

l O ^ i s e c "¹) 1.72 6.65 24.95 75 240

Calculate AH⁰* and the frequency factor. Calculate AG⁰* and AS⁰* for the reaction at 50°C.

14-16 For unimolecular gas reactions the activation entropy can often be neglected. Assuming the activation entropy to be exactly zero, calculate the rate constant and half-life at r o o m temperature (25°C) for a reaction with a heat of activation of (a) 15 kcal and (b) 20 kcal.

14-17* A s k e y a n d H i n s h e l w o o d obtained the f o l l o w i n g data for the pyrolysis o f dimethyl The system was believed to conform to the general Lindemann mechanism. Plot the data in a linear form s o as to obtain values for k² and k^²/k^x from the best least-squares slope and intercept. The observed activation energy for &uni,«> is 65.5 kcal m o l e^{- 1}. Estimate the value of fin Eq. (14-89) that will account for the experimental k ^².

14-18 Estimate the value of k^xK² and k-²\k^x for the isomerization of cyclopropane (using Fig.

14-9) and also the best value of / i n Eq. (14-89).

14-19 The rate constants for the second-order reaction C H³C H O + I² - * C H⁴ + C O + I² vary with temperature as follows:

Γ ( Κ ) 630 645 660 675 695 A: (liter m o l e "¹ sec"¹) 12 22 41 76 140 Calculate A and E*⁹ and, using transition-state theory, AS⁰*. (Optional: make a least-squares fit of the data to an Arrhenius plot in evaluating the above quantities.) Assume 630 Κ in the calculations.

14-20 Before s o m e recent new developments (see Section 14-5) the H² + I² = 2 H I reaction was thought to be a simple one, the probable transition state being

0.97 H - H

1.75 / \ 1.75 I 1

2.95

(distances in angstroms). Calculate the value of AS⁰* at 575 Κ assuming that the three moments of inertia are 921.5, 6.9, and 928.4 (in units of 1 0^{- 4 0} g c m²) [note that σ = 2;

see Eq. (4-90)], and that the vibrational frequencies are 994, 86, 1280, 1400, and 1730 (all in c m^{- 1}) .

14-21* Calculate k for activation enthalpies varying f r o m 10 t o 60 kcal m o l e "¹ in increments o f 10 kcal m o l e " \ and activation entropies varying from —20 t o 2 0 cal K "¹ m o l e "¹ in

In document 14-2 Rate Laws and Simple Mechanisms (Pldal 51-59)