Kinetic Theory of Gases: Effusion and Collisions

5.62 Spring 2008 Lecture #29 Page 1 Kinetic Theory of Gases: Effusion and Collisions EFFUSION Consider the process by which molecules escape through a hole in a vessel and into a vacuum slit size “d” oven walls at definite T We assume that: (1) d is so small that the pressure in the vessel is unchanged; (2) the effusion does not perturb the velocity of the gas in the vessel; (3) there are no collisions when the molecules pass through the slit. Molecules that would have been incident on the portion of the wall where the hole is, now pass through the hole. This creates a flux of particles defined as the number of particles per unit area per unit time that leave the vessel. ! vdt dA Consider a square hole of area dA. A particle that is a distance vdt from the hole moves with speed v and at angle ! from the surface normal toward the hole. Draw a parallelepiped around the hole with length equal to vdt, at angle θ from the normal. All molecules within this volume moving toward the hole (i.e., with the correct !, " angle) with speed v will pass through the hole in time interval dt. ρ = density of gas Volume of parallelepiped = v cos ! dA dt (Note that, at grazing angles, θ ~ π/2, the volume is small) # of molecules crossing through dA in dt = ρvcos θdAdt (number density times volume) # molecules = ρvcos ! =FLUX. dAdt We must integrate this expression over the distribution of velocities of the gas to obtain the average flux J. The Maxwell-Boltzmann distribution (from Lecture #28) is revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 Page 2 3 2 $ m ' 2 * mv ! ! 2 F ( v) dv = F(v, !, ")v2 dvsin !d!d" = & ) e kT v dvsin!d!d" . % 2#kT ( Thus, for the average flux: !/2 rather than ! to obtain forward direction only 3 # m &2 J = !% ( $ 2"kT ' , +e 0 ) mv2 kT " 2 2" v3dv + cos* sin*d* + do 0 3 2 Factor of v : v from volume element, v from flux The result is: 3/2 2 1/2 # m & ) 1 # 2kT & ,) 1 , ! # 8kT & !v [ ] ( + % ( .+ . 2" = % ( = J = !% $ 2"kT ' * 2 $ m ' -* 2 4 $ "m ' 4 Note : This result could have been obtained in an alternative way. Consider a volume of gas behind the hole that contains ρdV molecules. Then J simply is: Volume dV=v zdt 3 J =!$ +# "# $ dvx dvy +# 0 2 & m ) 2 " mv dvz dvz ( + e 2kT ' 2%kT * Only the molecules with vz > 0 can exit the hole. Angular Distribution of Flux (or Effusing Molecules) The velocity distribution of particles exiting through the hole are the same as the velocity distribution of particles hitting the hole. As a result the angular distribution and speed distribution of flux j(v, !, ") leaving the hole is. 3 2 % m ( 2 3 + mv j(v, !, ")dvd# = $ ' * v e 2kT cos !dvd# & 2kT ) solid angle d#=sin!d!d" revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 Page 3 The angular distribution of the flux is cos ! # jangle (")d" = $ dv j(v, !,%) = 0 ' 0 E0 So σR(E) = π b 20 The reactive cross-section is a function of the relative translational energy because it includes all values of b ≤ b0 and b0 is explicitly dependent on E. E $ 2 " σR = π b 20 = πd AB # 1! 0 % E for E > E0 σR = 0 for E ≤ E0 2. Calculate k by averaging over relative velocities (energies). E k = σR –c = πd 2AB " 1! 0 $ c # E% revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 Page 11 But, we no longer want to multiply by the average relative velocity because σR is now dependent on the relative energy (velocity). Want to multiply σR by the relative velocity of the specific collision and then want to average over all possible relative energies. k = σR –c = ∫ σR c f(c) dc " µ % 3/2 E = πd 2AB " 1! 0 $ 4π $ c3 e–µc2/2kT dc ' ⌠ ⌡ # % E # 2!kT & Change variables to E instead of c because σR(E) 1 E = 2 µc2 dE = µcdc " µ % 3/2 dE E ! 2E $ –E/kT ! 2E $ 1/2 k = πd 2AB ! # 1" 0 % 4π $ &e ' # & 1/ 2 # $ & E # 2!kT & " µ % ! 2E$ " µ % µ# & " µ % " = 8! " µ % 3/2 2 πd AB # (E ! Eo ) e–E/kT dE ' 2 $ # & µ 2!kT E 0 lower limit is E0 because σR = 0 if E < Eo Let x = E ! E0 kT dE dx = kT then e–E/kT = e–E0/kT e–x ! 8! " µ % 3/2 2 –E /kT 2 0 k= 2$ πdAB (kT) e xe #xdx ' " µ # 2!kT & 0 !#"#$ 1 1/ 2 " 8kT % 2 (E 0 / kT k=$ !d e ' AB ! # !µ & % "$ sphere !#"# $hard cross section COLLISION THEORY RATE CONSTANT relative speed fraction of collisions having necessary component of energy along line of centers E0 Collision theory result looks very much like empirical Arrhenius expression (k = A e–Ea/kT) except that the collision theory preexponential factor has a T1/2 dependence revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 k = BTm e–E0/kT Page 12 where m =1/2 Most experimental measurements of k do not show a T dependent pre–exponential factor. This is because measurements of k are made over too small a temperature range to see the weak T1/2 dependence of the pre–exponential factor in the presence of the much stronger T dependence of the exponential term. This pre-exponential T1/2 dependence would show up as non-linearity in plot of lnk vs. 1/T. RELATIONSHIP BETWEEN E0 (critical line of centers energy) and Ea, EMPIRICAL ARRHENIUS ENERGY k = A e–Ea/kT d ln k d(ln AEa / kT) Ea = 2 dT = kT dT " 8kT % 1/2 2 kCT = $ πdAB e–E0/kT ' # !m & # & 1 1/2 2 d (ln(8k / !µ) !d + ln T " E / kT) CT AB 0 % (' d ln k $ 2 = dT dT 1 E = 2T + 02 kT Now we require that d ln k d ln kCT dT = dT Ea 1 E + 02 2 = kT 2T kT Thus kT Ea = 2 + E0 Ea must be larger than E0. WHY? FURTHER ANALYZE COLLISION THEORY RATE CONSTANT: Eo⎞ ⎛ –E/kT dE kCT ∝ ⌠ 1 – E⎠ Ee ⌡ ⎝ energy dependence of cross section, σR distribution of relative energies revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 Page 13 CALCULATE VALUES FOR " 8kT % 1/2 2 ! 8kT $ 1/2 2 kCT = $ πdAB e–E0/kT = # πdAB e1/2 e–Ea/kT ' & # !µ & " !µ % Get E0 from experimental Ea since E0 = Ea – kT/2 2 Get πdAB from transport data COMPARE TO EXPERIMENT ln kCT CALCULATED ln k OBSERVED CH3 + CH3 → C2H6 10.6 10.5 CF3 + CF3→ C2F6 10.3 10.4 REACTION H + CCl4 → HCl + CCl3 Cl + H2 → HCl + H 0.83 –1.4 –4.4 –4.7 Conclusions: works well for simple recombination steps where the Ea is small or non existent and where there is little steric hindrance. does not work well for reactions with steric hindrance revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 Page 14 — E0 dependence may not be isotropic — ignores effects of reactant vibrational and rotational energy in overcoming barrier. revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 Page 15 Part II. A "good" theory must take into account the internal degrees of freedom of the reactants and their angle of approach. An approach known as transition state theory = activated complex theory = absolute rate theory does so in an approximate way. POTENTIAL ENERGY SURFACE A correct theory must consider the internal structure of molecules and the forces acting on atoms in the molecules because bonds are being broken and formed during a reaction. During a reactive collision, the force on an atom depends on both intramolecular forces (forces between atoms in a molecule) and intermolecular forces (forces between molecules). Must treat the two colliding reactants as a single quantum mechanical system. This system exists only during collision process. The system's potential energy is calculated the same way the potential energy for nuclear vibrational motion is calculated. Within the Born-Oppenheimer approximation, solve Hel Ψel = Eel Ψel at fixed nuclear configuration. The resulting Eel is the potential energy at that nuclear configuration. Systematically vary the nuclear configuration (grid of points) to get potential energy as a function of nuclear coordinates. Problem: too many nuclear coordinates! Can't plot potential energy as a function of more than 2 coordinates. A plot of potential energy versus 2 coordinates is a 3-D plot where the potential energy is a SURFACE. Potential energy for more than 2 coordinates is still called a surface even though there are more than 2 coordinates. revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 Page 16 describes how potential energy changes with the positions of the nuclei EXAMPLE: H2 + F ! HF + H For collinear approach, there are 2 independent variables RHF and RHH on which the potential depends. revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 Page 17 The potential energy of this system can be represented as a 3-D surface as a function of RHF and RHH. The 3-D surface can be represented as a 2 D contour potential energy surface. HF + H B saddle point x F + H2 A revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 Page 18 Lines on a contour map represent EQUIPOTENTIALS. Valleys correspond to initial and final states. Section A-A Section B-B (cut through V(Q) along B) REACTION COORDINATE FOR F + H2 [H2F ] ‡ 1.4 34.7 REACTION COORDINATE — minimum energy path along deepest part of potential energy surface revised 4/22/08 5:38 PM 5.62 Spring 2008 Lecture #29 Page 19 — note reaction coordinate corresponds to an antisymmetric H2F vibration TRANSITION STATE — transitory [H2F]‡ complex with a definite structure — dissociates within one half antisymmetric vibration. TRANSITION STATE THEORY An approach to calculating a rate constant by reducing the dynamics of the reaction to an equilibrium between the reactants and the transition state along the reaction coordinate. A + B ↔ [AB]‡ → PRODUCTS Uses statistical mechanics to treat the equilibrium. The reaction coordinate is the 1-D antisymmetric vibration of the transition state. revised 4/22/08 5:38 PM