Notes On Gaseous State This page provides you all details about the gaseous state of BSc standard. Concerned problems will be solved at the end of each concept. Continue till the end to find the easiest explanation of every concept you need in this regard . Postulates of Kinetic Theory of Gases: 1. All gas consist of a very large number of minute particles, called molecules. 2. The gas molecules are extremely small in size and are separated by large distance. The actual volume of the gas molecules is thus negligible as compared to the total volume occupied by the gas. 3. The pressure exerted by the gas is due to the bombardment of the molecules on the walls of the vessel. 4. The gas molecules collide with one another and also with the walls of the vessels. These collisions are perfectly elastic and there is no loss of energy during these collisions. 5. The distance between the gas molecules are very large. Thus, there is no effective force of attraction or r
Degree of dissociation, mole fraction and associated equilibrium:
The
number of molecules dissociated out of total number of molecules originally
taken is called the degree of dissociation.
Mathematically,
Degree of dissociation = α = (Number of molecules dissociated
/ total number of molecules) X 100
It is expressed in percentage or as
fraction or decimals. If 40 molecules are dissociated out of 100 molecules than
the degree of dissociation is expressed as 40% or 0.4.
The ratio
of number of moles of a component to the total number of moles of the entire
component in a given system is called the mole fraction. If nA and nB are
the numbers of moles of two components A and B of a system then the total
number of moles = nA + nB. Thus the mole fraction of component A = xA or XA
(read as chi A) = nA / (nA + nB). Similiarly, the molefraction of B = xB or XB
= nB / (nA + nB).
The mole fraction of a component multiplied
by the total pressure inside a container due to all components gives the
partial pressure of that component. Using the partial pressures of various components
the equilibrium constant can be calculated.
pA =
partial pressure of the component A = xA (total pressure) and
pB = partial
pressure of the component B = xB (total pressure)
Example: At 300k and 1 atm pressure N2O4 is 20% dissociated. What will be the degree of dissociation or percentage of dissociation at 300k and 0.1 atm pressure.
Solution:
Total moles: 1 — α + 2 α = 1 + α
Mole fraction of N2O4 at eq. =
xN2O4 = number of moles N2O4 /
total moles = 1 — α / 1 + α
Mole fraction of NO2 at equilibrium = xNO2 =
number of moles NO2 / total moles = 2 α / 1 + α
If the total pressure is = P
The partial pressure of N2O4 = pN2O4
= P. xN2O4 = P X (1 — α / 1 + α)
The partial pressure of NO2 = pNO2 = P.
xNO2 = P X (2 α / 1 + α)
Now Kp = (pNO2)2 / pN2O4 = { P X (2 α / 1 + α)}2 / { P X (1 — α / 1 + α)} = {4α2 / 1 — α2} P
At the first equilibrium, α = 20% 0r 0.2 and the total pressure, P = 1 atm
Thus, Kp = {4 (0.2)2 / 1 — (0.2)2} P = 0.166, this is constant and remains constant as long as the temperature remains
constant. Thus when a new equilibrium is established due to change in pressure,
the value of equilibrium constant remains the same though the degree of
dissociation changes.
Let the new degree of dissociation is = β, when
pressure (P) changes to 0.1 atm.
Thus,
Kp = 0.166 = {4β 2 / 1 — β 2} 0.01, solving which we get β = 0.54 or 54%
Got it sir. Thank you so much.
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