In the Classroom edited by
Resources for Student Assessment
John Alexander University of Cincinnati Cincinnati, OH 45221
Probability of Collisions Igor Novak Department of Chemistry, National University of Singapore, Singapore 119260
Most standard physical chemistry courses discuss kinetic theory of (ideal) gases as a necessary prerequisite for diverse topics such as effusion, barometric formula, heat capacity, and chemical kinetics. The appropriate mathematical equations are derived and, when used, produce very large (or small) numbers that are difficult to relate to everyday experience. For instance, calculation of the average number of collisions in m᎑3 s᎑1 conveys little to the student, since the numbers are too large to visualize and do not have any immediate “feel”. Another important topic related to kinetic (collision) theory concerns three-body collisions. Textbooks usually state only that the probability of such collisions is negligibly small, which induces chemical reactions to proceed via several elementary (unimolecular or bimolecular) steps. We present two problems whose aim is to illustrate, in probabilistic terms, collision processes. The problems can be used for assessment of students taking physical chemistry courses.
Substituting necessary values into eqs 1–3 gives pNN = .655, pOO = .036, and pNO = .309. It is interesting to estimate the collision probabilities on the basis of probability theory alone (i.e., without using the equation of kinetic theory). Using the standard expression for a number of combinations of n objects taken k at a time Ckn, the ratio of N2–N2, O2–O2, and N2–O2 collision probabilities becomes
Question 1 Assuming that the composition of air at STP (1 atm, 298 K) is: 80% (v/v) N 2 and 20% (v/v) O 2, deduce probabilities of N2–N2, O2–O2, and N2–O2 collisions. The molecular diameters of N2 and O2 are d N = 0.37 nm and dO = 0.36 nm. Estimate the probabilities with and without using collision theory!
Question 2
Acceptable Solution The standard expressions for number of collisions between like and unlike molecules (1) are:
pNN = .64 ;
pNO = .32
Comparison with kinetic theory values shows that the values are close, the reason being of course that N 2 and O2 have similar molecular masses.
The number of three-body collisions for an ideal gas is given as (1): ZABC = 4πdAB2d BC2 δ[(8πkT/ µAB)1/2+ (8πkT/µBC)1/2 ]N ANBN C (4)
where δ = 0.1 nm is the distance of the closest approach of 3 molecules and all other symbols having their usual meaning. Estimate the probabilities of 2-body and 3-body collisions in N2 gas at STP if dN has the same value as in Question 1.
Z NO = NN NO dNO 2 (8 π kT/ µNO) 1/2
(2)
Acceptable Solution Equation 4 can be adapted for the case of collision of three identical molecules by recognizing that
(3)
NA = NB = NC = NN; dAB = dBC = dN; µ AB = µ BC = mN/2
2
1/2
where NN and NO are numbers of N2 and O2 molecules per unit volume; d N, dO, dNO = (dN + dO)/2 molecular diameters; mN , mO are molecular masses; µ is reduced mass; k is the Boltzmann constant; and T is temperature. One must next calculate the total number of collisions in unit time and volume Z T : Z T = Z OO + Z NO + Z NN The relevant N 2–N 2, O 2 –O2 , and N 2 –O2 probabilities are then: p NN = Z NN/Z T ; p OO = Z OO /Z T ; p NO = Z NO /Z T We give an expression for pOO as an example: 2
᎑1/2
N O d O m O᎑1/ 2 + N N d N m N᎑1/ 2 + N N N O d NO(2/µNO) 2
2
2
Equation 4 then becomes ZAAA = 32(1/3!)πdN4 δ (πkT/mN)1/2 NN3
(5)
where the 1/3! factor takes into consideration the permutations of A in the same collision. Entering the appropriate values into eqs 1 and 5 gives the ratio of collision frequencies as ZNN/Z NNN = 354.4. The probability of a 3-body N2 collision is then p NNN = Z NNN/(Z NN + ZNNN ) = .003 This value is very low (0.3%), which indicates clearly why we are justified in neglecting 3-body contributions! Literature Cited
N O d O m O᎑1/ 2 2
852
pOO = .04 ;
(1)
Z OO = 2 NO dO ( π kT/mO)
2
Since the sum of all probabilities must equal unity we can deduce that
Z NN = 2 NN 2 dN2 ( πkT/mN) 1/2 2
p OO =
pNN : pOO : pNO = C280 : C220 : [C2100 – C280 – C220 ] = 3160 :190 :1600
2
1. Wilkinson, F. Chemical Kinetics and Reaction Mechanisms; Van Nostrand Reinhold: New York, 1980; p 112.
Journal of Chemical Education • Vol. 75 No. 7 July 1998 • JChemEd.chem.wisc.edu
