--%>

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 : What are halogen oxoacids? Fluorine

    Fluorine yields only one oxyacid, hypo

  • Q : Describe chemical properties of amines.

    Like ammonia, primary, secondary and tertiary amines have a single pair of electrons on N atom. Hence chemical behavior of amines is similar to ammonia. Amines are basic in nature, and in most of the reactions they act as nucleophiles.      1. Reaction wi

  • Q : Molecular energies and speeds The

    The average translational kinetic energies and speeds of the molecules of a gas can be calculated.The result that the kinetic energy of 1 mol of the molecules of a gas is equal to 3/2 RT can be used to obtain numerical values for the

  • Q : Explain the preparation of phenols. The

    The methods used for the preparation of phenols are given below:    From aryl sulphonic acids

  • Q : How molecule-molecule collisions takes

    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 b

  • Q : Determining concentration in ppm A 500

    A 500 gm tooth paste sample has 0.2g fluoride concentration. Determine the concentration of F in terms of ppm level: (a) 250 (b) 200 (c) 400 (d) 1000Answer: (c) F-ions in ppm = (0.2/500) x 106 = 400

  • Q : Neutralisation of phosphorous acids

    Provide solution of this question. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3 PO3) the volume of 0.1 M aqueous KOH solution required is: (a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL

  • Q : Problem based on molarity Choose the

    Choose the right answer from following. The molarity of a solution of Na2CO3 having 10.6g/500ml of solution is : (a) 0.2M (b)2M (c)20M (d) 0.02M

  • Q : What is electrolysis? Explain with

    Passage of a current through a solution can produce an electrolysis reaction.Much additional information on the properties of the ions in an aqueous solution can be obtained from studies of the passage of a direct current (dc) through a cell containing a s

  • Q : Problem on mole fraction of glucose

    Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998