--%>

How molecule-molecule collisions takes place?

An extension of the kinetic molecular theory of gases recognizes that molecules have an appreciable size and deals with molecule-molecule collisions.


We begin studies of elementary reactions by investigating the collisions between the molecules of a gas. We are led to expression for the average distance that a molecule of a gas travels between collisions with other molecules and to two quantities that express the number of molecule-molecule collisions which occur in a unit time travel.

Consider a particular molecule A with diameter d, moving in the direction indicated. If the speed of molecule A is v, m remain stationary, molecule A will collide in 1 s with all the molecules that have remain centered within the cylinder. The volume of the cylinder whose radius is equal to the molecular diameter d is ∏d2-vN*, is the diameter of molecules per unit volume. The mean free path, i.e. the distance traveled between collisions, is the free path length.

L = -v/∏d2-vN* = 1/∏d2N*

A more detailed calculation shows that this result is not exactly correct. The assumption that only molecule A moves implies a relative speed of the colliding molecules of v. in fact if the molecules are all moving with speed v-, all types of collisions will occur, ranging from glancing collisions, where the relative angles to each other and the relative speed is √2v-. a correct result can be obtained in place of these recognitions that although molecule A moves a distance v- in 1 s, it collides with other molecules with a relative speed of √2v-. The mean path is then written as:

L = 1/ √2∏d2N*

How far a molecule travels between collisions has now been shown to depend on the number of molecules per unit volume and so on, the molecular diameter d.

The second matter to be investigated is the number of collisions per second that a molecule makes. This collision frequency is denoted by Z1. In relation to the other molecules, the molecule A travels with an effective speed equals to the number of molecules in a cylinder of radius d and of length √2v. We therefore have:

Z1 = 9√2u-) (∏d2)N* = √2∏d2vN*

The last matter to be investigated is the number of collisions occurring in a unit volume per unit time. As can be imagined, this quantity is of considerable importance in understanding the rates of chemical reactions. The number of collisions per second per unit volume is called the collision rate, denoted Z11.

The collision rate Z11 is closely related to the collision frequency Zt. Since there are N*molecules per unit volume and each of these molecules collided and not contacted twice. We therefore obtain 

Z11 = ½ √2∏d2v- (N*)2 = 1/√2 ∏d2v- (N*)

The mean free path, the collision frequency, and the collision have now been expressed in equations that involves the molecular diameter d. since the molecular speeds and the number of molecules per cubic meter of a particular gas can be determined, only molecular diameters need be known in order to evaluate l, Z1 and Z11. Many methods are available for determining the size of molecules.

Instance: use the collision diameter value of d = 374 pm to calculate the collision properties L, Z1 and Z11 for N2 at 1 bar and 25 degree C.

Answer: the number of molecules in 1 m3 is:

N* = 6.022 Χ 1023/ 0.0248 m3 = 2.43 Χ 1025 m-3

The mass of mole of N2 molecules is:

M = 0.02802 kg

The average molecular speed form v- = [8kT/(∏m)]½ = [8RT/∏M]½ here we have;

v- = [8(8.314 JK-1 mol-1) (298 K)/ ∏ (0.02802 kg mol-1)] = 475 ms-1

   Related Questions in Chemistry

  • Q : Question relatede to calculate molarity

    Select the right answer of the question. What is molarity of a solution of HCl that contains 49% by weight of solute and whose specific gravity is 1.41 : (a) 15.25 (b) 16.75 (c) 18.92 (d) 20.08

  • Q : Sedimentation and Velocity The first

    The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment. A

  • Q : Problem on preparing of a solution Give

    Give me answer of this question. How many grams of CH3OH should be added to water to prepare 150 solution of@M CH3 OH: (a) 9.6 (b) 2.4 (c) 9.6x 103 (d) 2.4 x103

  • Q : Lab question Explain how dissolving the

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately 5.

  • Q : Molal elevation constant of water The

    The boiling point of 0.1 molal aqueous solution of urea is 100.18oC  at 1 atm. The molal elevation constant of water is: (a) 1.8    (b) 0.18   (c) 18    (d) 18.6Answer: (a) Kb

  • Q : Entropy on molecular basis. The

    The equation S = k in W relates entropy to W, a measure of the number of different molecular level arrangements of the system.In the preceding developments it was unnecessary to attempt to reach any "explana

  • Q : Crystals of covalent compounds Crystals

    Crystals of the covalent compounds always contain:(i) Atoms as their structural units  (ii) Molecules as structural units  (iii) Ions held altogether by electrostatic forces (iv) High melting pointsAnswer: (i)

  • Q : Molar and Volumetric flow rate problem

    Cyclohexane (C6H12) is produced by mixing Benzene and hydrogen. A process including a reactor, separator, and recycle stream is used to produce Cyclohexane. The fresh feed contains 260L/min C6H6 with 950 L/min of H2

  • Q : Difference among hcl gas and hcl acid

    What is the basic difference among hcl gas and hcl acid? Briefly state the difference?

  • Q : Problem on colligative properties

    Choose the right answer from following. The magnitude of colligative properties in all colloidal dispersions is : (a) Lowerthan solution (b)Higher than solution(c) Both (d) None