A.3.5 Slower Reaction Controls [C]
5.2 GAS KINETICS
5.2.1 T
emperaTureandS
peedSofG
aSm
oleculeSIt is useful to begin by looking at the temperature dependence of the energies of atoms, ions, and molecules. Consider a box of with sides of equal length, L, with a monatomic gas atom having mass m and an x-component of velocity, vx, as shown in Figure 5.1. Assume that the walls of the box are impenetrable so the atom will hit a wall and rebound completely elastically in the negative x-direction. When it does so, the momentum change is ∆(mv) = 2 mvx—one for coming and one for going in the opposite direction with the same x-velocity. Therefore the force exerted by this single wall collision is simply
F ma d mv
= = (dt )
(5.1) or ∆F = ∆(mv)/∆t where ∆t = time between collisions of this atom with this wall and is just the time it takes for the atom to travel to the back wall and return; that is, ∆t = 2L/vx. Therefore, the force per atom on the wall is
∆ ∆
F mv∆ t
mv L v
mv L
x x
x
= ( )= = x
/ .
2 2
2
(5.2)
If N is the total number of atoms in the box, then they all strike the wall in the time ∆t so the total force, F, is just N∆F and the pressure, P, is given by P = F/L2, so
P F L
N F L
Nmv L
Nmv V
x x
= 2 = ∆2 = 3 2 = 2
(5.3) where V = L3 = volume of the box. Rearranging this, for an ideal gas
PV Nmv nRT N
N RT Nk T
x
A
= 2= = = B (5.4)
and therefore mvx2=k TB where NA is the Avogadro’s number and kB is the Boltzmann’s constant. The kinetic energy in the x-direction, KEx, for the atom is defined by
L
L L
X Z Y
VX m
FIGURE 5.1 Model of a gas molecule moving in a cubic box of side L in the x-direction with a velocity vx used to calculate the translational kinetic energy of molecules.
KEx=1mvx= k TB
2
1 2
2 . (5.5)
The velocity vector has three-components, one in each of the x, y, and z directions as shown in Figure 5.2. Therefore,
v2=vx2+vy2+vz2
and the total kinetic energy, KE, is given by the sum of the kinetic energies in each direction
KE mv mv mv mv
KE k T k T k T k T
x y z
B B B B
= + + =
= + + =
1 2
1 2
1 2
1 2 1
2 1 2
1 2
3 2
2 2 2 2
.
(5.6)
The second of these last two equations is a statement of the principle of equipartition of energy (Silbey and Alberty 2001) which essentially says that the microscopic modes—mechanisms—for absorbing energy will all have the same energy, 1/2 kBT, when things are in thermal equilibrium.
For a monatomic gas, the only energy-absorbing modes are the translational energies in the three directions leading to KE = 3/2 kBT. This assumes that the temperature is not high enough to excite electrons into higher energy states, which does not happen at temperatures of normal interest.* From above, the RMS—root mean square—speed of gas molecules is defined as
v k T
m
k T m
N N
RT
rms B B A M
A
= 3 = 3 = 3 (5.7)
where M is the molecular or atomic weight.
Take He gas as an example, M = 4 g/mol, at 300 K,
v 3(8.314 J mol K)(300 K)
4.00 10 kg mol 1368m s 1368m s
1
rms= / 3⋅
× / = /
= /
−
6609 m mi 3600s hour 3060 mph v 3060 mph
761.2 mph Mach 4.02M
rms
/ × / =
= / = aach
where Mach 1 = 761.2 mph, the speed of sound at sea level. Even at room temperature, the speed of gas atoms is indeed very fast.
5.2.2 m
axwell–B
olTzmannS
peedd
iSTriBuTionIn most cases, the interest is in the average speed of the atoms and not in their RMS speed. This can be calculated from the Maxwell–Boltzmann speed distribution. In Chapter 4, it was shown
* For example, at T = 1000 K: kBT = 1.38 × 10–23 J/K × 1000 K = 1.38 × 10–20 J = 1.28 × 10–20 J/1.602 × 10–19 J/eV = 0.08 eV which is considerably less than the several eV necessary for electron excitations. Note that at room temperature, T = 298 K, kBT ≅ 0.025 eV ≅ 1/40 eV.
x
y z
v
vy vz
vx
FIGURE 5.2 Schematic showing the x-, y-, and z-components of the velocity of a molecule that has a vector velocity v.
that the distribution of energies for NT total atoms or molecules, or their fraction in a given energy state is
1 1
N dN dE k T
T B
E k TB
= e− . (5.8)
This is an extension of the barometric formula that gives the number of molecules as a function of height above sea level, that is, their potential energy. The equivalence between the potential and kinetic energies of gas molecules is not surprising. For example, suppose an atom had a kinetic energy KEx=1 2/ mvz2 at sea level in the vertical direction, which is taken to be z. If it did not undergo any collisions with other gas molecules, then it would rise to a height where its initial kinetic energy would just reach its potential energy, its kinetic energy would go to zero, and then it would be accel- erated back to earth ending with its initial kinetic energy. Therefore, the number of atoms at any given height, or potential energy, says something about the Maxwell distribution of the kinetic energy of atoms at sea level moving with a vertical velocity, vz; that is,
1 2 2
N dN
dv A
z z z
mv k T
z
= ′e− B
which considers molecules moving in the z-direction only, a Gaussian distribution (Figure 5.3) where A′ is just some constant (Moore 1955).* To evaluate the constant, A′, for molecules moving in the x-direction, this expression needs to be integrated; that is,
A dv
dN N
mv
k T x
x
x x
′ e− B .
−∞
∞
−∞
∞
∫ ∫
= −
2
2 1
(5.9) Let mvx2/2kBT=y2, or vx= (2k T m yB / ) , and the integral becomes
A k T
mB e dyy
′ 2 − 2 1
−∞
∞
∫
= .Therefore, the key is evaluating the integral ∫−∞∞e dy−y2 . How this is done is shown in Appendix A.2, and everyone needs to see this integration at least once in his or her career. In any event, from Appendix A.2
e dy−y
−∞
∞
∫
2 = π. (5.10)So
A m
k TB
′ = 2π
* Of course, in reality, no single molecule or atom travels very far (only about 0.1 μm at 1 bar and 300 K) before colliding with another gas atom or molecule, as will be seen when diffu- sion in gases is discussed later. Nevertheless, there is simply an energy exchange between atoms or molecules and there must have been some of them that had this energy at sea level and transfer their energy through multiple collisions to other atoms that made it to the higher potential energies.
+x
−x Vx
1/NxdNx dvx
FIGURE 5.3 One-dimensional Gaussian distribution of gas molecule speeds.
and
dN N
m
k T dv
x
x B
mv
k T x
x
= − B
2
2 2
π e . (5.11)
This is only in one dimension and similar expressions exist for the other two directions (y and z) and what is of interest is the speed, v, which is v= vx2+v2y+vz2, as shown in Figure 5.2. The fraction of molecules in three dimensions is just the product of the fractions in each direction, namely
dN N
dN N
dN N
dN N m
k T
x x
y y
z z
B
m vx
=
=
−
2
3 2
π e
( 22 2 2 2
+v+v
k T x y z
y z
B dv dv dv
)
.
(5.12)
Make the substitution for the speed, v, above, and change the volume ele- ment from dV = dvxdvydvz to a spherical shell of radius v and thickness dv, so the volume element becomes dV= 4πv2dv = dvxdvydvz, Figure 5.4, with the change in coordinate system shown in Appendix A.3. This gives the Maxwell–Boltzmann distribution for the speeds of gas atoms and molecules in thermal equilibrium:
dN N
m
k T v dv
B
mv k TB
=
−
2 4
3 2
2 2
2
π π e (5.13)
shown in Figure 5.5. The high speed tail end of this distribution, the shaded region in Figure 5.5, is essentially an exponential function. Note that Equation 5.13 could also be written in terms of the kinetic energy of the atom or molecule, ε =(1 2/ mv) 2 as
dN
N k T e d
B
=
(
2)
3 2 1 2 −π
π ε ε ε.
5.2.3 m
eanm
olecularS
peedBy definition, the mean speed, v, is given by v vdN
N
m
k T v v dv
B
mv k TB
= =
∞ ∞
∫ ∫
−0
3 2
0
2 2
2 4
2
π π e (5.14)
and substitution of z mv= 2/2k TB (or v= (2k T m zB / ) 1 2) in the integral gives
v v dv k T
m z k T
m
mv
k TB B B
4 4 2 1
2 2
2 2
0
3 2
0
3 2 2 1
π e π
/ /
− /
∞ ∞
∫
=∫
22 1 2
2
0
4 2
2
z e dz
k T
m ze dz
z
B z
− −
−
∞
=
∫
( / )
π
ze dzz
0
∞ −2
∫ can be easily integrated by parts or found in most calculus books (Edwards and Penney 2002)
dV = 4πv2dv
dv
v x
y
FIGURE 5.4 The volume element in spherical coordinates depending only on the radius, v, the molecular speed.
v, speed 1 dN
N dv
EA
FIGURE 5.5 Maxwell–Boltzmann distribution for the molecular speed showing those molecules or atoms having energies above EA in the exponential tail of the distribution.
v v dv k T
m e z
k T m
mv
k T B z
B
4 B 4
2
2 1
4 2
2
2 2
0
2
0 2
π π
π e−
∞
− ∞
∫
= −
(
+)
=
− +2
(
0 1)
so
v m
k T
k T
B m
= B
2
4 2
3 2 2 2
π
π
and gives the desired result for the mean speed,
v k T
m
RT M
= 8 B = 8
π π (5.15)
which is about 7% smaller than the RMS speed, Equation 5.7.
5.2.4 f
luxofG
aSa
TomSi
mpinGinGonaS
urfaceThe number of gas atoms that strike a solid or liquid surface per unit area, per unit time is a quan- tity of interest. Figure 5.6 shows how this can be approximated. Consider a column of gas of height h v= RMS that contains a number of gas atoms per unit volume, η, determined by the pressure. At any point in time, one-sixth of the gas atoms are moving in the −z direction and in 1 second all of these atoms in the column will strike the surface, so the approximate number of atoms striking the surface per unit area per unit time, I, the rate of impingement is (Hudson 1998)
I≅1 vRMS
6η . (5.16)
A better approximation would use the mean velocity, v, but this still is only an approximation.
Equation 5.16 assumes that the atoms are only traveling in the three mutually perpendicular direc- tions. The reality is the gas atoms can strike the area A from any direction. When this is taken into consideration, along with the mean velocity, the exact relation—as shown in Appendix A.4—is
I= 1 v
4η . (5.17)
Substituting Equation 5.15 for v and η =N P RTA / gives
I N P
RT
RT M
I N P
RTM
A
A
=
= 1 4
8
2
π π
or as it normally found in the literature,
I P
mk TB
= 2π (5.18)
where Units(I) = atoms/m2-s, Units(m) = kg/atom, and Units(P) = Pa.
Not surprisingly, the impingement rate depends on the concentration of gas atoms, which is proportional to the pressure.
V
A η
1/6
FIGURE 5.6 A model to calculate the approximate number of gas molecules hitting a surface of area A per unit time with a mean speed v. Only 1/6 of the atoms per unit volume, η, are moving in the z-direction at any given time and only those strike the surface.
5.2.5 n
eedforH
iGHV
acuuminS
urfaceS
TudieSTake air with a mean molecular weight of M=28 84. g mol (Chapter 4) or 28.84 × 10−3 kg/mol (M = 28.84 at a temperature of 25°C or 298 K at a pressure of 1 atm or 1.0132 × 105 Pa). Then,
I P
mk TB
= = ×
( )
××
2 −
1 0132 10 2 3 1416 28 84 10
6 022 10 1 3
5 3 23
π
.
. .
. . 88 10 298
2 88 10 2 88 10
23
27 2 23 2
(
×) ( )
= × ⋅ = × ⋅
−
I . atoms m s . atoms cm s
(5.19)
Suppose that an experimentalist wanted to study the surface properties of pure nickel, Ni, without any adsorbed oxygen impurities. If the atmosphere over the nickel were air, it might be expected that all of the oxygen molecules hitting the surface will react to form at least a monolayer—one atom thick—of nickel oxide, NiO, since it is stable in air at room temperature. The number of oxygen atoms impinging on the surface would simply be
IO=2IO2= ×2 0 21 2 88 10× × atoms cm s⋅ =1 21 10× oxygen atoms cm
23 2 23 2
. . . ⋅⋅s
since air is 21% oxygen. For nickel, M = 58.693 g/mol and the density ρ = 8.90 g/cm3. So the number of Ni atoms per cm3 is
η ρ
η
Ni A
Ni
MN
atoms cm
= =
( )
( ) (
×)
= ×
8 90
58 693 6 022 10 9 13 10
23
22 3
.
. .
.
typical of most solids and liquids—something times 1022/cm3. The approximate number of nickel atoms per unit area is given by
Ni cm
Ni cm cm
/ Ni
/
2 2 3 22 2 3
2 15 2
9 13 10 2 03 10
≅ =
(
×)
≅ × −
η/ . /
. .
So the time it would take for the entire surface to be covered with a monolayer of NiO would simply be Ni cm
I s
O
/ .
. . !
2 15
23
2 03 10 8
1 21 10 1 68 10
= ×
× = × −
About 10 ns, this means that the experimentalist must be pretty quick in carrying out the experi- ment if it is to be done on a surface essentially free of oxygen atoms! A more reasonable pace to perform such an experiment might be about 1 hour during which time, say, only 1% of the surface gets covered with oxygen. In that case, IO must be substantially reduced and can be estimated by
I Ni cm time
I O cm s
O
O
= =
( )
×= × ⋅
/ 2 15
9 2
0 01 2 03 10 3600 5 64 10
. .
. .
Everything else being equal, this means that the pressure of air in the system would have to be reduced to about 5 64 10 1 21 10. × 9/ . × 23≅4 66 10. × −14 atm. As seen in Chapter 4, the experiment might be best done on the outside of the International Space Station where the vacuum has about this value. However, modern vacuum systems can reach vacuums on the order of 10−11 torr* or in the
* The quality of a vacuum is usually given in the units of torr = mm of mercury. One atmosphere = 760 torr.
order of 10−14 atm. Therefore, experiments on clean surfaces must be done in ultra-high vacuum on the order of 10−11 torr.
5.2.6 c
ondenSaTionande
VaporaTionorS
uBlimaTion5.2.6.1 The Model
Equation 5.18 also gives the rate of condensation or evaporation of atoms from the surface of a solid or liquid at equilibrium if the equilibrium pressure is pe,
I p
mk T
e B
= 2π . (5.20)
If the solid or liquid is not in equilibrium with the gas that has a partial pressure p of the solid or liquid molecules or atoms in the gas phase, the Equation 5.20 needs to be modified to (Pound 1972),
′ =
(
−)
J p p
mk T
e B
α
2π (5.21)
where J′ is the atom flux density to—or away from—the solid or liquid surface depending on the rela- tive values of p and pe: it p > pe, condensation will occur; if p < pe then evaporation or sublimation happens. Alpha, α, is called the accommodation coefficient or sticking coefficient (Adamson 1982) and is a positive quantity less than or equal to 1.0. If α ≠1 0. then the rate of condensation or evaporation is slower than if every atom or molecule that hit the surface stuck or remained there. Also, alpha less than one, implies that some kind of surface reaction must be occurring as well. An alpha of zero means that the atoms do not stick and just bounce off the surface. This situation will be explored in more detail later in this chapter. Also, the value of alpha may be different on evaporation than it is on condensation if the events are occurring far from equilibrium. However, for many substances, metals, inorganic compounds, and organics, α ≅1 0. when there is no chemical reaction but simply condensation or evaporation (sublimation) when near equilibrium (Pound 1972).
5.2.6.2 Example: Sublimation of 1,4-Dichlorobenzene
Today, mothballs consist of 1,4-dichlorobenzene (Figure 5.7) that has a melting point of 53.5°C, a molecular weight of 147.00 g/mol, and a vapor pressure of pe=3 27 10. × −3bar at 30°C (NIST Webbook). If there is no pressure of the compound in the gas phase, then dichlorobenzene will sublime. Assuming α = 1.0, then J′ becomes,
′ = =
(
×)( )
( )
×−
J p −
mk T
bar Pa bar
e
2 B
3 27 10 10 2 3 14 6 147 00 10
3 5
π 3
. /
. . ( . kkg mol atoms mol
J
) / ( . ) .
.
6 022 10 1 38 10 303
4 08 10
23 23
2
(
×) (
×) ( )
′ = ×
−
44molecules m s2⋅ =4 08 10. × 20molecules cm s2⋅ and
J J
N mol cm s
A
= ′
= ×
× = × − ⋅
4 08 10
6 022 10 6 78 10
20 23
4 2
.
. . .
Therefore, the dichlorobenzene is subliming as expected. Of practical interest is the rate of sublima- tion in terms of the thickness sublimed in a given amount of time. The density of dichlorobenzene ρ = 1.25 g/cm3 (Haynes 2013), so
J mol cm s g cm
M g mol A cm
dV cm d
2
3
2
1 3
(
⋅)
=( )
( )
( )
( )
ρ
/ tt s
( )
. (5.22)
Cl Cl
FIGURE 5.7 The chemical structure of 1,4-dichlorobenzene.
For a sphere or radius a, V=(4 3/ )πa3, dV dt/ =4πa da dt2( / ), and A=4πa2so J M
da
= ρ dt. Therefore, the rate of evaporation, da/dt, is
da dt
MJ da
dt cm s
= =
(
×)
= ×
−
−
ρ
147 02
1 25 6 78 10 7 97 10
4
2
.
. .
. .
This seems very fast compared to how long mothballs last, suggesting α ≠1 0. , so there is some kind of surface reaction taking place and/or diffusion through the gas phase is rate controlling, the latter being more likely case.
5.2.7 l
anGmuira
dSorpTioni
SoTHermIn Chapter 4, the role of catalysts in reactions was discussed: specifically, the adsorption of hydro- gen and deuterium onto platinum and the catalytic role of the platinum surface in the exchange of hydrogen and deuterium atoms. A reasonable question to ask is, “What might determine the surface concentration of the adsorbed gases?” A simple way to approach this is through the equilibrium between the adsorbed gas and the atmosphere such as
empty surface site, S + gas atom occupied surface site, S′; ∆Go (5.23) where ∆Go is the Gibbs energy of the adsorption reaction of the gas atoms (molecules) on the solid surface. An equilibrium constant for the reaction in Equation 5.23 can be written as
[ ]
′[ ]
= =S −
S P KL G
e RT
∆ °
(5.24) where [S′] =θ, the fraction of occupied surface sites by the adsorbed gas, [S] = 1 – θ, the fraction of unoccupied surface sites, and P = pressure. So Equation 5.24 becomes
θ θ
θ θ
θ 1
1
− =
= −
(
+)
=K P K P K P K P K P
L
L L
L L
with the final result, the Langmuir isotherm (Adamson 1982; Hudson 1998):
θ = + K P
K P
L
1 L . (5.25)
If P = 1 bar and KL= 1 at 298 K, then θ= 0.5 or one-half of the sites are occupied and ∆G° ≅ 0. Now,
∆Go=∆Ho−T S∆ o and, of course,
KL S R
H
=e e−RT
∆° ∆ °
where ∆So is the entropy difference between the vibrations and translations of the atoms or mole- cules on the surface and the gas phase and ∆Ho is the enthalpy of adsorption. If the adsorbed atoms
stick at one spot and do not translate on the surface, an estimate for the entropy change for water molecules adsorbing at 25°C and one bar is the entropy of sublimation of ice, ∆So298≅ −141 6. J mol K⋅ (Roine 2002). In order to have KL = 1.0, ∆H° would have to be about −42 200, Jmol. On the other hand, if there is 99% coverage (θ = 0.99), then ∆Go≅ −11 385, J mol or ∆Ho≅ −53 568, J mol.* If θ ≅0 01. , then ∆Ho= −11 435, J mol. All else being the same, as the pressure decreases, the fraction adsorbed decreases. Since the value of the bonding enthalpy, ∆Ho, determines what fraction of the surface is covered by an adsorbing gas, the boundary between physical adsorption and chemical adsorption or chemisorption becomes rather arbitrary and is around ∆H° = −10k J mol: less nega- tive for physical adsorption (Hudson 1998).