--%>

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 on Mole fraction Mole fraction

    Mole fraction of any solution is equavalent to: (a) No. of moles of solute/ volume of solution in litter (b) no. of gram equivalent of solute/volume of solution in litters (c) no. of  moles of solute/ Mass of solvent in kg (d) no. of moles of any

  • Q : Maximum vapour pressure Provide

    Provide solution of this question. Which solution will show the maximum vapour pressure at 300 K: (a)1MC12H22O11 (b)1M CH3 COOH (c) 1MNacl2 (d)1MNACl

  • Q : Polyhalogen compounds introduction for

    introduction for polyhalogen compound

  • 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 : Problem on molecular weight of solid

    The vapor pressure of pure benzene at a certain temperature is 200 mm Hg. At the same temperature the vapor pressure of a solution containing 2g of non-volatile non-electrolyte solid in 78g of benzene is 195 mm Hg. What is the molecular weight of solid:

  • Q : Latent heat of vaporization Normal

    Normal butane (C4H10) is stored as a compressed liquid at 90°C and 1400 kPa. In order to use the butane in a low-pressure gas-phase process, it is throttled to 150 kPa and passed through a vaporizer. The butane emerges from the vaporizer as a

  • Q : Difference in Mendeleevs table and

    Briefly describe the difference in the Mendeleev’s table and modern periodic table?

  • Q : Explosions produce carbon dioxide

    Illustrate all the explosions produce carbon dioxide?

  • Q : Molar concentration Choose the right

    Choose the right answer from following. Molar concentration (M) of any solution : a) No. of moles of solute/Volume of solution in litre (b) No. of gram equivalent of solute / volume of solution in litre (c) No. of moles os solute/ Mass of solvent in kg  (

  • Q : The Liver Is Responsible For Much Of

    The Liver Is Responsible For Much Of The Pentose Phosphate Activity Explain