Laboratory Measurement Values and Kinetic Theory

Transcription

1 Laboratory Measurement alues and Kinetic Theory Kinetic Theory is an attempt to explain macroscopic phenomena such as pressure, reaction rates diffusion rates, etc by using basic principles of motion. There are a number of complicating factors in attempting such an endeavor, however, properties of substances can be explained by looking at systems of particles in a manner analagous to the ideal gas law. As in the ideal gas law, assumptions are made. For kinetic theory, the assumptions are as follows: Gas particles are far apart from each other relative to their size. (olume effects do not play a role until high pressures are achieved, so this is reasonable.) The molecules are in free motion, unperturbed by any external forces (i.e. magnetic fields) All collisions between particles and with the walls of a container are completely elastic. There are no interparticle interactions. (familiar?) With these assumptions, we may proceed to develop theories based on the molecular motion. Surface Collision Frequency and Effusion We wish to develop a molecular foundation for the macroscopic lab measurement of pressure. If we note that pressure is the force exerted by an ensemble of particles on the walls of a vessel per unit area, we need to know two things: What is the force exerted by a single collision How many collisions per second occur. If we multriply these two, we will get the total force with which we can then calculate the pressure. Your Responsibilities.. Followiing is a kinetic theory development of relationships between molecular motion, velocity, temperature, pressure, volume and more. This is not empirical, but a development based on classical physics. It is here to show how one can use the mechanics of physics to produce an underlying theory that connects macroscopic measurements to microscopic (or particle motion) properties. You will not be asked to reproduce it. (Whew!) BUT, you're responsible for understanding, conceptually, how the results are produced and you are responsible to being able to use the final results in order to calculate molecular properties.

2 How many collisions occur on a wall per second? Consider a system in which gas particles are incident upon a surface We would like to determine how many molecules will collide within a given area on a single wall surface per unit of time. The total number of particles colliding with the wall per second per unit area will be given by the total number of particles per unit volume multiplied by the fraction of them moving with a velocity in the x-direction, or Defining Z "collisions unit area per unit time", we have Z AΔt v x_pos_avg The item in parenthesis is the average velocity of particles traveling towards the wall. or rearranging... Z A Δt v x_pos_avg Now, this is the TOTAL number of collisions with a single wall in the y-z plane. What is the force exerted by a particle for each collision? In order to calculate the pressure, we need to multiply the total number of collisions on the wall by the force exerted by each. Therefore, we need, first, to calculate the force. As a particle approaches and strikes a wall, the momentum changes from +mv to -mv if the collision is supposed to be perfectly elastic. The force, then, is the change in the momentum per unit time as defined by Newton's second law. As such, we have.. Δp mv x_pos Mv x_pos M v x_pos From physics, we know that Δp F thus... Δt M v x_pos F per collision. Δt

3 What is the TOTAL force exerted by all particles then? Now, as described above, the TOTAL force is the product of these two quantities, or... F total ZF A Δt v x_pos_avg Mv x_pos Δt or, simplifying... F total N a MA vx_pos_avg We have the total force, now what is the pressure and how is it related to particle mass, velocity, temperature..etc.?? i.e. Bring on the answers!!! We know from physics that P F A so we now have.. P N a MA vx_pos_avg A or P N a M v x_avg_pos vx_avg_pos P N a M 3

4 It is important to note that we have talked about collisions against wall in the +x direction. We now need to consider both walls. If we do we see that v x_avg v x_avg_pos For both walls in the x-direction... For all three dimensions (we do live in a 3-D world after all!),then, we have v avg vx_avg 3 3 v x_avg_pos 6 v x_avg_pos vx_avg_pos vavg 6 Now combining into the pressure expression, we now have P N a M vx_avg_pos vavg 6 Solving for P... P M v avg 3 Now we see a connection between the average velocity of the gas and the pressure of the container. But what is the average velocity? Let us turn to that. 4

5 Important Note before going on: k b is a very important fundamental constant which we will see more of in a little bit and is known as the "Boltzmann's Constant". It is the constant that connects Energy to the macroscopic measurement of Temperature as we will see mol and R J molk R then k b k b J K Now, where were we... oh yes.. From experiments on ideal-behaving gases, we know that.. P RT k b T But also, from above.. Continuing, we see that P M v avg 3 M v avg 3 k b T Solving for the average velocity.. vavg 3k b T 3k b T or v avg M M We will find in a moment that this average is the root mean square average of particles. With these values, we are able to calculate many average properties. It would be of much greater use if we knew how the particles were distributed in velocity or energy. To do this, we further the Kinetic Theory assumptions to produce such a distribution. When done, we will have a powerful tool indeed! 5

6 Boltzmann Distribution In the previous development, I attempted to show how one can use basic physics in order to explain what we see in the laboratory. The results presented so far connects macroscopic values of pressure, volume, mass, temperature, etc. to microscopi properties of particle mass, number and average velocity. Note the word "average". This gets us part of the way. What we now need is to know how many particles have any given velocity. Knowing this will allow us to find out even more! The velocity of particles is given in terms of a "distribution" which is an expression that says to us.. "Tell me the mass and the temperature and I'll tell you how many particles are traveling at any given velocity" The development of this expression is more involved and beyond the scope of this course. It is developed from the same basic physics as used above. The result is called the Maxwell-Boltzmann distribution and is given here. f( v) dv M 4π k b T e v dv for infinitesimal ranges of velocity π T k b 3 Mv An approximate measure of this is given as... f( v) Δv 4π M π k b T 3 Mv k b T e v Δv approximate expression becomes more exact as v becomes infinitely small. Note: This expression as any probability expression is read and interpreted as follows: The fraction of particles having a velocity in the range of v to v + dv (or v) is given as f(v)dv (or f(v)v ). From this expression we can obtain values for average velocities, energies, fractions of particles having energies in a given range, etc. 6

7 Example: Calculate the fraction of Nitrogen molecules having velocities in the range of 300 to 30 m/s at 300K. k b J mol T 300K K M.08 kg v 300 m Δv 0 m mol s s Substituting these values in gives.. f( vt) 4π M π k b T 3 Mv k b T e v Δv f 305 m 300K s Note the lack of units. Probability is a unitless value! A plot of this function produces a "probability distribution" which is a measure of the relative probability of a particle having a given velocity. An example is shown here... v 0 m s 5 m s 800 m s Maxwell-Boltzmann Distribution 00 K 300 K 800 K velocity (m/s) Note how the distribution of partcles "spreads" as the temperature rises! In order to be good, the theory must fall in line with what is observed experimentally. To wit: 7

8 Experimental erification of the Maxwell-Boltzmann Distribution The Maxwell-Boltzmann distribution has been verified experimentally, most straightforwardly by Kusch and co-workers at Columbia University in the 950 s. The experimental setup that they used, illustrated here, consists of a furnace with an exit hole allowing atoms to escape. The atoms are collimated and passed through a velocity selector and to a detector. The velocity selector consists of a set of rotating disks, each with a narrow groove cut into it. The grooves are offset from each other at known positions. As the collection of disks rotate, only the particles possessing the precise velocity will pass through all slits as they make their way to the detector. Experimental Setup of Kusch et. al. Diagrammatic representation of a velocity selector. Comparison of the predicted velocity distribution of particles from the Maxwell-Boltzmann distribution along with the data points obtained by Kusch et. al. (circles) is presented below. Note the excellent agreement. Reference: McQuarrie & Simon Physical Chemistry, A Molecular Approach 8

9 Average alues: For a function as described above, average values for particles can be obtained by carrying out integrations. In general, for a function of some variable, x... v avg vf ( v) dv v avg v f( v) dv v rms v Using this to calculate quantities from the Boltzmann distribution gives. avg v avg and so on. v avg 8k b T k b T 3k b T v mp v rms πm M M Example: Calculate the average, most probable and root mean square values of the velocity of Argon atoms at 500K. T 500K M kg mol v rms 3k b T 8k b T k b T v avg v mp M πm M v rms m v avg m v mp m s s s Compare these values with those in the graphic below... Probability Distribution RMS Average v Most Probable v Average v elocity (m/s) 9

10 An important Point... Consider the v rms result given above. We know that the kinetic energy is given as.. E M v Substituting E M 3k b T M or 3 E k bt!!!!!!! The kinetic energy is independent of mass!! Indeed, when we convert the Maxwell-Boltzmann distribution into mass instead of velocity, we get.. PET ( ) π πrt 3 exp E RT E Note the absence of a mass!! Plotting... E 0 kj 0.0 kj 5 kj mol mol mol Probability ( x 0^4) Distribution in Energy 300 K 500 K 800 K Energy (kj/mol) We see the same variation in temperature but all curves will be identical regardless of the mass of the particle. 0

11 Some Additional Quantities... Now that we have a solid means by which to determine averages, we can extract a few more items. Consider, first the pressure again. We found earlier that the number of collisions per unit tme per area was given as: Z AΔt v x_pos_avg From the ideal gas law, we now know that.. P k b T or P k b T The average velocity in the +x direction can be obtained from the same Boltzmann distribution as the other average values. Without proof, the result is.. v x_pos_avg k b T πm Combining these two expressions into the collision equation,we get Z AΔt P k b T k b T πm P πm k b T This final expression is useful for a myriad of applications and is called the Knudsen equation. Here is one example. See the homework for more.. Collisions_per_second_per_area_wall P πm k b T

12 Example: How many collisions are there against the wall of a vessel containing bar and Pa of Argon at 300 K? kg mol M Ar M Ar kg T 300K k b J mol P bar or P 0 5 molk P Coll Coll π T M Ar k b m s kg ms For P kg ms P Coll Coll π T M Ar k b m s Grahams Law of Effusion We have seen that the velocities of particles depend on T and M. If two gases are at the same temperature, then the velocity depends only on mass and, as we will see, in a simple relation! Consider two gases of Mass M and M both at the same temperature, T. The velocities are given for each as.. v avg_ 8k b T 8k b T and v avg_ πm πm If we take the ratio of the velocities, we get.. 8k b T v avg_ v avg_ πm 8k b T v avg_ M v avg_ M πm

13 The previous expression is called "Graham's Law of Effusion" and has many applications. Consider the following examples. Example: Consider that the two rare gases, He and Ar, are released at one end of a long tube. a.) How much faster does He travel as compared to Ar? b.) If the tube were 00 m long, what would be the average distance between the gases by the time He reaches the end? kg mol M He M He kg kg mol M Ar M Ar kg M Ar elocity_ratio elocity_ratio 3.59 M He or over 3 times faster! b.) From physics, we have d vt T 300K d 8k b T 00m t v He t He t He s v πm He v He d Ar v He v Ar t d Ar t He d Ar m elocity_ratio So, by the the time the average He molecule arrives at 00 m, Ar has only reached m Important Note: Keep in mind that gases are a large number of particles all traveling at different velocities! So even though the AERAGE distance is as given above, the gases are far from separated as the distances are very much spread by the time they travel 00 m! 3

14 Distribution of Energy Now, let us really put this to use! Consider, for example, a chemcal reaction. You know from general chemistry that a reaction must have an "Activation Energy" or minimum energy necessary in order to react. The Boltzmann distribution will tell us this! We know that velocity and kinetic energy are related, therefore, it is possible to present the distribution in velocity as given above as a distribution in kinetic energy. Again, this involves some math, so I'll simply present the result as follows:.. PE ( ) de π πkt 3 exp E kt EdE or PE ( ) ΔE π πkt 3 exp E kt EΔE As in the velocity expression, this expression is interpreted and read as follows: The fraction of particles having a kinetic energy in the range of E to E + de (or ) is given as f(e)de (or f(e) ). These expression have particular use with regards to chemical reactions. As two particles collide, there must be a minimum energy of collision in order to overcome close-range repulsion forces (electron-electron cloud repulsions). This minimum energy is molecule-specific and has been introduced to general chemistry students as the "Activation Energy" It is thus, useful to use the Maxwell-Boltzmann distribution in order to calculate the fraction of molecules possessing the minimum threshold or activation energy required in order to overcome repulsions and "react". Without derivation, which requires the Calculus, the result is given here as: Fraction of Energies greater than E a. f E a T E a e πrt E a RT erfc E a RT 4

15 What on EARTH is an "erfc"??!!! Okay, there is a new math value given here. Panic not, for reasons we will see in a moment. Most everyone at this point is aware of a "Normal Curve", or sometimes called a "Bell Curve" due to it's shape. It is an extremely important curve used an enormous number of applications. The most common application you are familiar with is it's use in laboratory error. When you calculate standard deviations, for example, you are using a normal distribution. Well, the erfc, called the "error function compliment", is a mathematical function of a particular type related to this normal distribution curve. But..but...do I have to worry about it??? Good news here for most systems. It turns out that as the argument of the erfc, larger (in this case as E a >> T), the value of the erfc becomes negligible. E a RT, gets However!! I would be lying to completely dismiss it. If E a gets close to T, (e.g. high temperatures and low E a 's) then including this is necessary. But, these cases are few and far between and only necessary for high precision work. Example erfc values The Examining this function for values of E a, gives... x erfc(x) erfc value E a_by_t Ratio of Activation Energy to Temperature 5

16 Collisions between Particles In our previous discussion on collisions, we were concerned with collisions against a wall, giving rise to pressure. Here, we are interested in collisions between particles. Why? Well, think again about chemical reactions. In order for a reaction to occur, the particles must collide into each other. So, if we can determine the collision rate, this will give us wonderful information about the reaction rate. And, as we see in later chapters, this information gives insight into reaction mechanisms. So this is a very valuable application of kinetic theory! Consider a single particle of effective diameter, A. As the particle moves through space with average velocity, v avg it will sweep out a volume dependent on the diameter. If there are B molecules within this volume having an effective diameter, B then they will collide if they come within a distance, AB defined as: σ AB π r A r B It is not quite this simple, as we have assumed for the moment that all the B particles are stationary while the A particle sweeps out the volume of the cylinder. If we consider that all particle are in motion, then we need to use the "relative velocity" between particles. We also need the distribution of velocities (good news..we have the Boltzmann Distribution!). If we consider all these elements, we arrive at a paarticle collision rate. I won't go through the derivation (whew again!) but present the results here. 6

17 Z AB σ 8k bt N A N B for two differing particles A and B. πμ If the particles are identical, then Z A σ 8k b T πm N A Note that we are talking about ideal gases in this analysis. As such we can write N A P so an alternative expression is k b T Z A 8k b T P σ σp πm k b T 8 πm k b T The rate of collisions with depend on the relative velocity between the two is given as v rel 8k b T where k boltzmans's constant and μ πμ reduced mass m A m B m A m B If we denote the relative velocity as defined above and let N A n a P σ we get Z A RT RT v rel P The mean free path.. If the molecule travels at a given velocity, then we can readily calculate the distance it travels in the time it takes to collide with another particle as: λ v rel Z 7

18 Example: Calculate the average velocity and collision frequency of Ar gas at 300 K and 0 bar. The cross section for Argon is nm kg mol M Ar M Ar kg M Ar 8k b T μ P 0bar T 98.5K v rel πμ σ Ar 0.36nm v rel 56.6 ms v rel 56.6 m s σ Ar v rel P Z Z s RT Calculatibg the distance it travels in the time it takes to collide with another particle, or the mean free path... v rel λ λ m Z 8

19 Calculation of a simple rate coefficient Although simplistic in it's treatment, it is reasonable to use what has been presented so far in order to calculate a nominal rate constant from kinetics. Example: Consider HI gas at 500 K at a concentration of.5 mol/l. The reaction threshold is 5000 J/mol and the reaction cross section for HI is 5 nm. Calculate a rate constant for such a system: M mol Molarity HI 0.5M σ 5nm MW HI 7.9 gm L mol As discussed, we can consider a simple bimolecular reaction to occur when two particles collind with the minimum energy required to overcome repulsions of the particles. The collision rate for two identical is given above as: Z A σ 8k T πm N From the Maxwell-Boltzmann distribution, we know that the fraction of particles with the minimum energy required to react is given as: f E a T E a E a RT e if E a is sufficiently larger than RT. πrt Combining will give us a way to estimate the rate of the reaction as: σ Rate E a T E a 8k T N E a RT e km HI πm πrt 9

20 Now, let us set upon carrying out the calculations We need a value for N in order to use the collision frequency expression. N is the number of particles. Thus, converting from Moles per Liter to particles in a Liter.. L N HI Molarity HI N HI (Note the lack of units. We understand this to mean particle MW number.) HI Mass HI Now.. σ Rate E a T 8k b T N HI πmass HI E a e πrt E a RT J calculating... Rate K mol s This is the number of reactions in reactions per second per m 3.! We want the reactions in moles. in order to do this we recognize that there are avogadros number of reactions per mole of reaction. Thus, we need to convert or Molarity HI Rate Molar Rate5000 J 500K Rate Molar M mol s Finally, recognizing that from phenomenological kinetics, a uni-molecular reaction has the form Rate khi the rate constant can be found by dividing by the molarities, giving k HI Rate Molar k HI s Molarity HI Note: Experimental rate measurements of this system gives a value of 8.8 x 0 0 s -. 0