Chapter 20  -  Statistical Thermodynamics: the machinery

(pp 615-617)  E20.1(a),  E20.3(b), E20.13(a),  E20.15(a)

notes

E20.1(a)   The equipartition theorem (pp4-5) predicts the maximum value of Cv which is approached at higher temperatures, assuming the molecule doesn't decompose.  The translational contribution is always fully realized (3/2 R).  Each fully active rotational degree of freedom (v_R* =2, linear; 3, nonlinear) contributes 1/2 R to the molar Cv.  Each fully active vibrational degree of freedom (v_V* = 3N-6) contributes R (1/2 R for each: P.E. and K.E) to the molar Cv.  The Solutions Manual estimates Cv assuming the vibrational contribution will be negligible at 25^oC -- justifiable for CH₄ , less so for C₆H₆ and I₂ as can be seen by looking at the experimental Cv values, e.g., Cv for I₂ is almost 7/2 R indicating that the vibrational contribution is almost fully realized.  The maximum vibrational contributions are:  R for I₂,  (3N-6)R = 9R for CH₄, and (3N-6)R = 30R for C₆H₆.  (Low frequency vibrational modes, due to weak bonds and/or heavy atoms, increase the vibrational contribution.)   Why might benzene be expected to show the greatest deviation from the equipartition prediction? (It has many, 3N-6=30, vibrational modes, some of which are low frequency.)  Why might I₂ show a deviation?  (Heavy atoms, so the single frequency is low - see E20.15(a))  Which would you expect to show the greatest Cv deviation from the equipartition prediction: CH₄ or CF₄?  (CF₄ because it has more low frequency vibrational modes.)

The general form of Fig. 20.11, p 605, although for diatomic molecules, is applicable to more complex gas molecules if the numerical values on the vertical axis are ignored.

E20.3(b)  The rotational constant, B, for O₂ is in Table 16.2, p942.  The symmetry number, sigma, for a homonuclear diatomic molecule is 2.
   q^R = kT / [(sigma) (hc) (B cm^-1)]
   q^R(298K)  =  (1.381x10^-23J/K)(298K) / [2(6.626x10^-34Js)(2.998x10¹⁰cm/s)(1.446cm^-1)] = 71.6
   q^R(523K)  =  126

Alternatively, the equation for q^R in Table 20.5, p602 can be used:
    q^R = (0.6950/sigma) (T / B)
    q^R(298K)  =  (0.6950 / 2) (298K / 1.446 cm^-1)  = 71.6

E20.15(a)      I₂(g) --->  2 I(g)
The K is determined by eqn(59), p610

delta Eo = De - 1/2 hc(wavenumber)

De is the energy from the bottom of the potential well to the dissociation limit; delta Eo is the energy from the ground (v=0)  vibrational level to the dissociation limit.

De = (1.5422 eV/m'cule) (1.602x10^-19J/eV)* = 2.471x10^-19J/m'cule
* inside front cover of Atkins

delta Eo = (2.471x10^-19) - (0.5) (6.626x10^-34 Js) (2.998x10¹⁰cm/s) (214.36cm^-1)
             =  2.45x10^-19 J/m'cule
delta Eo(J/mol)  =  (2.45x10^-19J/m'cule) (6.02x10²³m'cules/mol)
              =  1.47x10⁵J/mol

Now the partition functions . . .

I atoms (only translational and electronic contributions)
  q_total(I) = q^T(I) q^E(I);

  q^E(I) = 4 ;   ²P_3/2  implies a 4-fold degenerate ground electronic state; g_o=2J+1= 2(3/2)+1 = 4, see pp375-8 (Chem 461).
  q^T(I) = [(2PI m k T)^3/2 / h³] V
    Since the partition functions are standard state (1 bar, 10⁵Pa) and molar (n=1), the volume can be determined from the perfect gas equation, and q^T computed with the above formula.  Alternatively, and simpler, Table 20.4, p602 condenses the partition functions formulas into forms more amenable to computation.

  q^T = N_A (2.561x10^-2) T^5/2 M^3/2      (with the molar mass, M, in units of g/mol)
at 1000K,
  q^T(I)  = N_A(2.561x10^-2) (1000K)^5/2 (126.9 g/mol)^3/2  =  6.97x10³²
then,
  q_total(I) =  q^T(I) q^E(I)  =  (6.97x10³²) (4)  =  2.79x10³³

I₂ molecules (translational, rotational, vibrational contributions)
   q_total(I₂) = q^T(I₂) q^R(I₂) q^V(I₂);

   q^E(I₂) = 1  (so doesn't contribute)
at 1000K, using equations in Table 20.4, p602
   q^T(I₂)  =  N_A(2.561x10^-2) (1000K)^5/2 (253.8 g/mol)^3/2 = 1.97x10³³
   q^R(I₂)  =  (0.6950 / 2) (1000K / 0.0373 cm^-1) = 9316
   q^V(I₂)  =  1 / [1 - EXP{-1.4388(214.36 cm^-1) / 1000K}] = 3.768
then,
   q_total(I₂)  =  (1.97x10³³) (9316) (3.768) = 6.92x10³⁷

Now the equilibrium constant, K, from eqn(59), p610 . . .

K =  ([2.79x10³³]² / {[6.92x10³⁷] N_A})  EXP[-1.47x10⁵J mol^-1/ {(8.314 J mol^-1K^-1)(1000K)}]
    =  3.9x10^-3
 

Chapter 19  -  Statistical Thermodynamics: the concepts

(pp 588-589)  E19.1(b) 623K, E19.3(a), E19.4(a), E19.6(a)  There may be more after Exam 2 (see below).    Exam 2 will  include sections 19.1 (Boltzmann distribution) and 19.2 (molecular partition function).

Another additional problem from Chapter 19:

     From expressions for the thermal energy and entropy as functions of the partition function (see Atkins, Table 19.1, p587, or the equivalent expressions from class, Wednesday, May 24) determine the thermal energy (energy above that at T=0K) and absolute entropy (third law entropy, see Atkins, pp 111-112) of 1.00 mol helium at 1.00 bar (10⁵Pa) and 25^oC .

     The energy was worked out in class:  U_m = 3/2 RT = 3/2(8.314 J/K mol) (298K) = 3.72x10³J/mol.   This result is that predicted from the classical equipartition of energy "law" (see Chapter 0, p 4-5;  note especially the last paragraph in section 0.3, p5).

     The entropy can be worked out from the general entropy expression (last key equation in Table 19.1, p587 with Q=q^N/N! for indistinguishable particles, i.e., gas).  More simply, for a monatomic (one-atom) perfect gas, so only translational energy need be considered, the Sackur-Tetrode equation is equivalent to the last equation in Table 19.1.

     When you have calculated the molar entropy of He at 25^oC and 1.00 bar from the Sackur-Tetrode equation, compare your result from statistical thermodynamics with that given in Table 2.6, p925.   And while you're at Table 2.6, compare the standard molar entropy of argon calculated in Example 19.5, p586 in Atkins with the value in Table 2.6.  Most of the standard entropy values in Table 2.6 were determined experimentally as described in Atkins, pp109-110.

Additional problem:  Using eqn(1), p 569, Atkins, confirm the number of microstates associated with each configuration of the systems discussed in class on Monday, May 15.  These systems were:
--------
system I:   N = SUM n_i  = 5;     E = SUM e_i n_i  = 2 units
available energy levels:  e_o (zero energy),  e₁(one energy unit higher)
Using Atkin's nomenclature to specify the configurations (see bottom of p 568):
system I has one configuration:
   {3, 2}configuration   (10 microstates)  [see Fig 19.1, p 569]

system II:   N = SUM n_i = 3;     E = SUM e_i n_i = 3 units
available energy levels:  4 equally spaced levels, ground level at zero energy, next level one unit higher, etc.
three configurations:
   {2, 0, 0, 1}  (3 microstates)
   {1, 1, 1, 0}  (6 microstates)
   {0, 3, 0, 0}  (1 microstate)

system III:    N = SUM n_i = 5;     E = SUM e_i n_i = 5 units
available energy levels:  5 equally spaced levels, ground at zero energy, next level one unit higher, etc.
seven configurations
   {4, 0, 0, 0, 0, 1}  (5 microstates)
   {3, 1, 0, 0, 1, 0}  (20 microstates)
   {3, 0, 1, 1, 0, 0}  (20 microstates)
   {2, 2, 0, 1, 0, 0}  (30 microstates)
   {2, 1, 2, 0, 0, 0}  (30 microstates}
   {1, 3, 1, 0, 0, 0}  (20 microstates)
   {0, 5, 0, 0, 0, 0}  (1 microstate)

system IV:    N = SUM n_i = 10;     E = SUM e_i n_i = 5 units
available energy levels: as system III
seven configurations
   {9, 0, 0, 0, 0, 1}  (10 microstates)
   {8, 1, 0, 0, 1, 0}  (90 microstates)
   {8, 0, 1, 1, 0, 0}  (90 microstates)
   {7, 2, 0, 1, 0, 0}  (360 microstates)
   {7, 1, 2, 0, 0, 0}  (360 microstates)
   {6, 3, 1, 0, 0, 0}  (840 microstates)
   {5, 5, 0, 0, 0, 0}  (252 microstates)

Chapter 28  -  Processes at Solid Surfaces

(pp 873-876)  E28.2(b) 7.3x10² Pa ,  E28.3(a)*, E28.5(a),  E28.7(b)** 3.7x10³ J/mol , P28.4***Voo =1.44 mL, area=5.0 m²,  P28.5***,  P28.15***

* One face (square) of a face-centered cubic has an atom on each of the 4 corners, each atom in contact with the central face atom.  The diagonal of this face (square) is 4x(atomic radius) and is also 361 x sqrt(2) pm, from which the radius and cross sectional area of one atom can be determined.  The Solutions Manual determines the area of the atom differently.

** Activated means the rate is controlled at least partially by an activation energy barrier (See Fig. 28.29, p863), so Arrhenius-like behavior is expected.

*** Think spreadsheet on these.  For P28.5 the data can be plotted according to the Langmuir isotherm and according to the BET isotherm (as done in Example 28.3, p 861).  Which isotherm equation gives the best fit?  Plotting the data according to each isotherm to check for best linear fit, or doing a complete linear regression and seeing which has the best  regression statistics (in Excel, Multiple R closer to 1) will determine which isotherm fits best.  Since the problem asks for a value of c, (a BET parameter), you can probably guess which isotherm fits the data best.  (V_mon=75 cm³, c=3.98)

P28.4  After the number of molecules (N = nN_A) have been determined from Voo using PV = nRT, and after the volume of a single molecule has been determined from the density, V= (molar mass) /(DN_A), some geometric assumption concerning how to treat the molecules on the surface must be made.  One approach is to imagine the molecules in cubic boxes and determine the area of one face of these boxes as the "molecular footprint."   Another approach is to treat the molecules as spheres and determine the cross-sectional area of the spheres as the "molecular footprint" as the Solutions Manual does for this problem.  The results do not depend significantly on which route one takes.
 

Chapter 27  -  Molecular Reaction Dynamics

(pp 844-848)  E27.1(b) 6.6x10⁹s^-1, 8.1x10³⁴m^-3s^-1, 1.6%, E27.2(a), E27.4(a),  E27.5(b), 3.2x10¹⁰L mol^-1s^-1,  E27.6(b) 2.0x10⁶m³mol^-1s^-1, 2.4x10⁵m³mol^-1s^-1,  E27.8(b) 1.4x10^-3(unitless), E27.10(a),  E27.15(b) k_CT/k_CH=0.06, k_CO-18/k_CO-16=0.89, P27.2, P27.5

notes on exercises/problems

E27.1(b)  R is the collision radius (R = d/2,  sigma = PI d² = 4PI R²).
collision cross section (see Fig 1.20, p30): sigma = 4 PI (180x10^-12m)² = 4.07x10^-19m²
molecular mass: m  = M/Na = (0.028kg/mol) (1mol/6.02x10²³ m'cules) = 4.65x10^-26kg
molecular density: N = N/V = N_A[A] = p/kT = 10⁵Pa / [(1.38x10^-23) 298K] = 2.43x10²⁵m'cules/m³
collision density (total molecular collisions per second per m³), eqn(6), p821:
Z = sigma [(4 kT / PI m)^1/2 N²       NOTE:  N  =  N_A[A]
Z = (4.07x10^-19m²) (336m/s) (2.43x10²⁵m'cules/m³)²  = 8.07x10³⁴collisions/(s m³)
NOTE:   Using eqn(6) for E27.1(a) gives:  Z = 1.15x10³⁵collisions/(s m³)

The collision frequency (collisions per molecule per second), from eqn(10), p821:
 z = 2 Z / N = 2(8.07x10³⁴ ) / 2.43x10²⁵ = 6.64x10⁹ collisions/s

At constant volume, the molecular density, N=N/V, doesn't change with temperature.
Z is proportional to T^1/2 so Z_308K = Z_298K (308/298)^1/2 = 8.07x10³⁴ (1.017) = 8.20x10³⁴collisions/(s m³)
(8.20 - 8.07)x10³⁴ / 8.07x10³⁴ =  0.016  (1.6%)

E27.6(b)  See comment on centipoise (cP) units at the bottom of Example 24.3, p 732.

E27.10(a)  The relationship between the activation energy, Ea, and the delta H* (read: delta H double dagger) depends on whether the reaction is gas phase or solution.  If it's a gas reaction the relation is:  Ea=delta H* +2RT (see p 835).  If it's a solution reaction the relation is: Ea=delta H* +RT (see Solutions Manual).  This difference leads to a slightly different formula  for k₂ for a gas reaction (see eqn65, p 835) than for a solution reaction as in this problem.  In evaluating B in this problem, keep in mind standard pressure, p^o, is 1 bar (standard pressure is no longer defined as 1 atm - close but not identical), so p^o is 10⁵ Pa.

E27.15(b)  298K is assumed.  For part (a) the reduced masses are essentially the mass of the hydrogen atom because  m_H, m_T << m_C so, for example,  m_Hm_C/(m_H+m_C) ~ m_H.   This is the reason CH stretching frequencies are relatively independent of the rest of the molecule.  This is not true for CO stretching frequencies, where the reduced mass associated with the CO stretch will vary somewhat depending on the rest of the molecule.  The answer to the second part of E27.15(b) reflect a reduced mass determined from the carbon and oxygen atoms alone.

P27.5  As can be seen in the Solutions Manual, the fourth point is suspect.  Plotting only the first three points gives the "limiting slope" as the solution approaches very dilute concentrations.  Linear regression of the first three points, i.e., log₁₀(k) vs I^1/2, gives a slope of  +2.54 --  not as close to an integer as one might hope.  The Solutions Manual concludes Z_B is +2, although +3 would have been equally justified from this data.   This indicates the approximate nature of equation (73), p836.  When seeking integer results approximate equations are often sufficient, although results roughly half way between integers can be problematic.   By contrast, Example 27.3 (p836) is unambigous: Z_B=+1.00.   Self-test 27.3 (p837) results in Z_B= -0.42 which like P27.5 is not close to an integer.  Here, however, unlike P27.5, a charge of -1 can be reasonably assumed since a charge of zero is probably not reasonable.   The message is clear and not surprising: data don't always yield definite results.

Both problems P27.2 and P27.5 are conveniently done on a spreadsheet.  In P27.2, columns containing T/K and k are entered, converted in adjacent columns to 1/T and ln k, and plotted to check for linearity after which the intercept can be determined from  the Excel formula:  =INTERCEPT(Y data block, X data block).  Or a complete linear regression. can be carried out.  The equation is different for P27.5, and the slope rather than the intercept is needed, but the approach is the same.

Chapter 26  -  The Kinetics of Complex Reactions

(pp 815-818)  -  E26.1(b)*,  E26.3(b)**,  E26.4(a),  E26.5(a)***,  P26.1,  P26.5,  P26.7(solve for the rate law in terms of  of A_o and P_o)****

* Assume the steady state approximation for the intermediates, [NO₃^.] and [NO^. ].  This gives the two equations in Example 25.7 (p 781).  Solve these two simultaneous equations for [NO^.] and [NO₃^.] in terms of the rate constants and species that are not intermediates.  Then put these into the rate expression for [N₂O₅].   In this case there is an algebraic shortcut to the final results, e.g., see Example 25.7 (p 781).   The final rate expression:
   d[N₂O₅]/dt = - {2k₁k₂/(k' + k₂)}[N₂O₅]

**  Figure 26.3 suggests that at 700K there will not be a chain-branching explosion because the system does not exist between the first and second explosion limits.  What about above 10⁶ Pa (~10 atm)?   What is expected at 900K?

*** As mentioned in Example 26.3 (p 800) reviewing Example 11.1 (p 290) is useful.

**** The rate law in differential form is d[P]/dt = k [A][P]²  or  dx/dt = k[A_o- x ][P_o+ x]².  To solve for the integrated rate law (as in Table 25.3), an integral is needed which can be worked out by partial fractions.  Alternatively, from integral tables the following general integral form can be obtained where a and b are constants:

S dx / [(a - x) (b + x)² ]  =  -{1/(a+b)}{1/(b+x) + [1/(a+b)] ln[(a-x) / (b+x)]}

The answer to P26.7:

[1/(A_o+P_o)] [x /{P_o(P_o+x)} + {1/(A_o+P_o)}ln[A_o(P_o+x) /{P_o(A_o-x)}]  =  kt

Chapter 25  -  The Rates of Chemical Reactions
  (pp 788-791)  -  E25.2(a), E25.4(a), E25.5(a)^*,  E25.10(b),  E25.12(b), P25.1^**, P25.14, P25.17

^* The pressures given in E25.5(a) are partial pressures of acetaldehyde, not total pressures of the system.

^** Problems involving tabular data, e.g., P25.1, are easily done on a spreadsheet.  Columns of times (t) and m(urea)/g can be entered.  Adjacent columns are calculated for m(amm.cya.)/g and amount(amm. cya.)/mol which is also the molar concentration of ammonium cyanate here since the volume is one liter.  Then columns for various functions of concentrations assuming different reaction orders (see Table 25.3) and plots of  various functions vs. time to search for linearity.   Once the order is found, linear regression on the appropriate spreadsheet columns will yield reactions constants, etc.

The answers to E25.10(b): 0.64 microgram and 0.18 microgram.  The answer to E25.12(b): 1.5x10⁶ s.

solution to E25.12(b):   2A  --->  P

   v  =  -(1/2) d[A] / dt = k [A]³

   d[A] / dt  =  -2k [A]³

   Separating variables, [A] and t, and integrating between limits ([A]_o, t=0) and ([A], t) gives the third order integrated rate law:
   Sd[A] / [A]³   =   -2 k Sdt
   [ Sdx / x³  =  - (1/2) (1/x²) + constant ]

   (1/2) (1/[A]² - 1/[A]_o ²)  =  2 kt

    The 3-order integrated rate law (above) can also be obtained from the general n-order rate law at the bottom of Table 25.3, p772.   For the problem here:  [A] = 0.021 mol/L,  [A]_o = 0.077 mol/L,  k  =  3.50x10^-4  L²/(mol²s).  Solving for time, t, gives: 1.5x10⁶s

Chapter 24  -  Molecules in Motion

     In Chapter 24, we will not deal with sections 24.7, 24.8, and 24.9 on ionic conduction and electrochemistry.  Below are the assigned exercises/problems at the end of Chapter 24.   In addition to the assigned problems, it's also useful to attempt the Self-test problems in the chapters.

(pp 757-760)  -  E24.1(a), E24.2(a), E24.7(a), E24.10(a), E24.13(a),  E24.22(b),  P24.4, P24.17

    The answer to E24.22(b) is 2.1x10^-10 m.  The answers to P24.4:  (a) 2.7x10²³collisions/cm²/s , 2.7x10¹⁸collisions/cm²/s and (b) 1.8x10⁸collisions/Ti atom/s , 1.8x10³ collisions/Ti atom/s. Answers to P24.4(b) may vary slightly depending on how you deal with the area of a single titanium surface atom determined from the nearest-neighbor distance.

Notes on Chapter 24

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Determining the mass in kilograms of a single molecule or atom, using argon (39.95 g/mol , 39.95 amu/atom) as an example:

from molar mass, (M g/mol),
(39.95 g/mol) (1 kg / 10³ g) (1 mole / 6.022x10²³ atoms)  =  6.634x10^-26 kg/atom

or, alternatively, from atomic mass, (amu/atom)

(39.95 amu/atom) (1.6605x10^-27 kg/amu)  -=  6.634x10^-26 kg/atom

The conversion factor in the second expression is, numerically, the product of the two conversion factors in the first expression.

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The SI units of viscosity are kg m^-1s^-1, although tabulated viscosities are often given in poise (P), centipoise (cP), or micropoise (uP) units, named after Jean Poiseuille, a physiologist interested in the flow of blood in capillaries.  His formula (see pp.731-2), published in 1840, is for the rate of flow of fluids through pipes and is well known to chemical engineers who deal with such flow.  The poise unit originates from the earlier centimeter-gram-second (cgs) unit system: 1 poise (P) = 1 g cm^-1s^-1, so 10 P = 1 kg m^-1s^-1 and 1 cP =10^-3 kg m^-1s^-1.

While on poise units and the Poiseuille formula, it has been pointed out by a perceptive 463 student that Example 24.3 (p. 732) Using the Poiseuille formula to measure a viscositymay have numerical errors in the last step or two.  For a 100 cm (0.100 m) tube length and 1.00 mm tube diameter (0.0005 m radius), the viscosity of air calculates to 1.8x10^-4 kg m^-1s^-1.  An earlier edition of Atkins text has this identical example problem except that the tube length is 1.0 m.  With a 1.0 m tube length, the viscosity of air calculates to 1.8x10^-5 kg m^-1s^-1, a result which does agree well with 1.4x10^-5 kg m^-1s^-1 from the kinetic theory expression.

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The answers for P24.4 make clear the need for very high vacuum systems when studying catalytic properties on metal surfaces.  The vacuum systems in Dr. Bussell's research equipment on the 4th floor (CB 450) permit vacuums at pressures well below 10^-7torr (mm Hg).  At 10^-7 torr pressure, what approximately would be the number of O₂ collisions per second on a single surface titanium atom (use data in P24.4)?   ANSWER:  0.024 collisions/s.  The reciprocal is seconds per collision, so about one collision every 42 seconds.   At 10^-7 torr pressure and 300K, how many O₂ molecules would be in one cubic centimeter (cm³) assuming only O₂ present?  HINT:  the perfect gas equation.  ANSWER:  about 3x10⁹ molecules(3 billion molecules in one cm³) in a very high vacuum.

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Some perspective on the answer for P24.17(b) can be obtained by comparing the calculated concentration of sucrose at 5.0 cm after one year of diffusion with the concentration of the sucrose (1) before diffusion begins and (2) after the concentration gradient drops to zero (uniform concentration).  These initial and final concentrations can be obtained from the total mass of sucrose (10.0 g) and the volumes involved.  Keep in mind that the diffusion process slows with time due to the decrease in the concentration gradient.

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Self-test 24.7 (p. 753)   This derivation is done as Example 24.7 right above the Self-test, except that the sum (integral) is taken over each x² (rather than each x) weighted by the probability of its occurrence.  Then the square root of the result is taken to get the root-mean-square distance.  The definite integral (0 to infinity) of  x² EXP(-x² / (4Dt) dx  where EXP is the base of natural logs is needed here.  Standard integral tables or programs such as Mathcad can be used to get such integrals.  For example, the Handbook of Chemistry and Physics (57th ed., CRC Press, 1976-77) in the chemistry student seminar room (CB200 ) has this as definite integral #666, page A-165.  (In other editions the page and integral # may differ.)  The general form of integral #666 can be made specific to this problem by recognizing that n=1 and a=1/(4Dt).

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