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Diffusion is the movement of gas molecules from high concentration to the low concentration. Diffusion is occurred on the gases because of the random movement of the gas molecules.
Grahams Law of diffusion or effusion
Graham's Law of diffusion or effusion
    The phenomenon of diffusion may be described as the tendency for any substance to spread uniformly throughout the space available to it.
    The passing out of gas through the pinhole in the wall of the container is called effusion.
    The rate of diffusion and effusion of the gas passing out depends on the density, pressure, and temperature of the gases.

Graham's law of diffusion

    At constant temperature and pressure, the rates of diffusion or effusion of different gases vary inversely as the square root of their densities.
    At constant temperature(T) and pressure(P) if the rate of diffusion or effusion = r and density = d.
r ∝ 1/√d
or, r = k/√d where k is a constant.

Relation between rates of diffusion and densities

    If r₁ and r₂ be the rates of diffusion or effusion of two gases having densities d₁ and d₂ at constant temperature(T) and pressure (P).
∴ r₁ ∝ 1/√d₁ and r₂ ∝ 1/√d₂
Thus r₁/r₂ = √d₂/√d₁
Again density(d) ∝ Vapour density (D)
Thus r₁/r₂ = √D₂/√D₁

Relation between rates of diffusion and molecular weight

    If r₁ and r₂ be the rates of diffusion or effusion of two gases having molecular weight M₁ and M₂ at constant temperature(T) and pressure (P).
We know that Molecular weight(M) = 2 × Vapour density (D)
Thus,  r₁/r₂ = √D₂/√D₁ = √M₂/√M₁

Relation between rates of effusion and volume

    Let at constant temperature and pressure V₁ and V₂ be the volume of the two gases passing through the same hole with the time t.
Then V₁/t = r₁ and V₂/t = r₂
∴ r₁/r₂ = V₂/V₁ = √M₂/√M₁

Graham's law from the kinetic gas equation

    The rate of diffusion or effusion can be assumed to be directly proportional to the root mean square speed or any other average speed.
    Thus, r₁/r₂ = √u₁²/√u₂² where u₁ and u₂ are the root mean square speed of the gas molecules.
From the kinetic gas equation PV = ⅓ m N u²
or, u² = 3PV/m N

For 1-mole ideal gas PV = RT and m N = M = molar mass of the gas.

Thus u² = 3RT/M

∴ r₁/r₂ = √u₁²/√u₂² = √(3RT/M₁)/√(3RT/M₂)
or,  r₁/r₂ = √M₂/√M₁
which is Graham's law of diffusion.

Practical application of Graham's law

    Application of Graham's law is in the partial separation of the components in a gas mixture. If the mixture is led out a tube made of porous walls, in a given time the lighter components will diffuse out more than the heavier one.  By repeating the process with each sperate fraction from diffusion, the concentration of one component is considerable increases compared with that of the other. This is called the atmolysis.
    Argon has been concentrated with nitrogen in this way. The isotopes of neon, chlorine, bromine, etc have been partially separated by this method.
    Graham's law also used on for detecting the marsh gas in mines.

Problems solutions

  • Problem
    At constant temperature and pressure 432 ml and 288 ml be the volume of the two gases passing through the same hole with the time 36 min and 48 min respectively. If the molecular weight of one gas is 64 gm mol⁻¹, what is the molecular weight of another gas?
  • Solution
According to Graham's law of diffusion,
r₁/r₂ = √M₂/√M₁
    Here, r₁ = 432 ml/36 min = 12 ml min⁻¹,  r₂ = 288 ml/48 min = 6 ml min⁻¹ and M₂ = 64 gm mol⁻¹.
∴ 12/6 = √64/√M₁
or, M₁ = 64/4 = 16 gm mol⁻¹
  • Problem
    A certain temperature the time required to the complete diffusion of 200 ml of hydrogen gas is 30 min. How many time required for the complete diffusion of 50 ml of oxygen gas at the same temperature?
  • Solution
    Let time required for the complete diffusion of 50 ml of oxygen gas is t min.
According to Graham's law of diffusion,
r₁/r₂ = √M₂/√M₁
    Here, rate of diffusion of hydrogen (r₁) = 200 ml/30 min = 20/3 ml min⁻¹,  rate of diffusion of oxygen (r₂) = 50 ml/t min = 50/t ml min⁻¹,  molecular weight of hydrogen (M₁) = 2 gm mol⁻¹ and molecular weight of oxygen (M₂) = 32 gm mol⁻¹
∴ (200/30)/(50/t) = √(32/2)
or, t₂ = 30 min
  • Problem
    Which of the two gases ammonia and hydrogen chloride diffuse faster and by which factor?
  • Solution
According to Graham's law of diffusion,
rNH₃/rHCl = √MHCl/√MNH₃
    Here molecular weight of ammonia (NH₃) = 17 gm mol⁻¹ and molecular weight of hydrogen chloride (M₂) = 36.5 gm mol⁻¹
∴ rNH₃ = rHCl × (√MHCl/√MNH₃) = rHCl × (√36.5/√17) = 1.46 rHCl
    Thus ammonia will diffuse 1.46 times faseter then hydrogen chloride gas.
  • Problem
    Ladenberg found that a sample of ozonized oxygen containing 86.16% of ozone by weight required 430 seconds to diffuse under conditions where pure oxygen required 367.5 seconds. Determine the vapor density of ozone.
  • Solution
    Let V is the volume diffusing out in each case. Let dm, do and d be the densities of the mixture, pure oxygen, and ozone respectively.
    Here, rate of diffusion of mixure (rm) = V ml/430 seconds = V/430 ml seconds⁻¹,  rate of diffusion of oxygen (ro) = V ml/367.5 seconds = 50/t ml seconds⁻¹
According to Graham's law of diffusion,
rm/ro = √do/√dm = √16/√dm
or, dm = (430/367)² × 16
≈ 21.91

Again for 100 gms, the volume = 100/dm = (86.16/d) + (13.84/do)
or, (100/21.91) = (86.16/d) + (13.84/16)
or, d = 23.3
  • Problem
    Equal moles of hydrogen and oxygen is placed in a container with a pinhole through which escape. What fraction of oxygen escape in the time required for one half of the hydrogen to escape?
  • Solution
    Let the time required for this escape is t, and the fraction of oxygen and hydrogen escape is nO₂ and nH₂ respectively.
Thus according to Graham's law of diffusion,
rO₂/rH₂ = √MH₂√MO₂
or, (nO₂/t)/(nH₂/t) = √2/√32
or, (nO₂/t)/(0.5/t) = 1/4
or, nO₂ = 1/8

The rate at which a radioactive sample disintegrates can be determined by counting the number of particles emitted in a given time.
    Radioactive decay is one important natural phenomenon obeying the first order rate process. Rate of these reactions depends only on the single power of the concentration. of the reactant. The rate can be expressed as,
(dN/dt) = - k N
    Where N is the number of the atoms of the disintegrating radio-element present at any time, dt is the time over which the disintegration is measured and k is the radioactive decay rate constant.
    Again, k = - (dN/dt)/N
    Thus, the rate constant(k) is defined as the fraction decomposing in the unit time interval provided the concentration of the reactant is kept constant by adding from outside during this time interval. The negative sign shows that N decreases with time.
    Let N₀ = number of the atoms present at the time t = 0 and N = number of atom present after the t time interval. Rearranging and integrating over the limits N₀ and N and time, 0 and t.
Rate of radioactive decay half life for radioactive elements
Rate of radioactive decay half-life

Half-life period of radioactive elements

    After a certain period of time the value of (N₀/N ) becomes one half, that is, half of the radioactive elements have undergone disintegration. This period is called half-life of a radioactive element and is a characteristic property of a radioactive element.
Half life of the radioactive elements
Half-life and rate of radioactive decay
    If a radioactive element is 100% radioactive and the half-life period of this element 4 hours. Thus after four hours, it decomposes 50% and the remaining 50%. After 8 hours it decomposes 75% and reaming 25% and the process is continued. The half-life is given by,
    2.303 log(N₀/N ) = kt When t = t½, N = N₀/2 
Putting in the above equation we have,
2.303 log{N₀/(N₀/2)} = k t½
or, t½ = 0.693/k
k = 0.693/t½
    The above relation shows that both the half-life and radioactive decay rate constants are independent of the amount of the radio-element present at a given time.
    ₈₄Po²¹³ has t½ = 4.2 × 10⁻⁶ sec, whereas ₈₃Bi²⁰⁹ is 3 × 10⁷ years.

Calculation of the age of radioactive elements


Age of organic material

    A method of determining the age of organic martial based on the accurate determination of the ratio of carbon-14 and carbon-12. carbon 14 is produced in the atmosphere by the interaction of neutron with ordinary nitrogen.
    ₇N¹⁴ + ₁n⁰ ₆C¹⁴ + ₁H¹ ( Cosmic Reaction)
    The carbon-14 ultimately goes over to carbon-14 dioxide. A steady-state concentration of one ¹⁴C to ¹²C is reached in the atmospheric CO₂. This carbon dioxide is taken in or given out by plants and plant-eating animals or human beings so they all bear this ratio. 
    When a plant or animals died the steady state is disturbed since there is no fresh intake of stratospheric CO₂ the dead matter is out of equilibrium with the atmosphere.
    The ¹⁴C continues to decay so that thereafter a number of years only a fraction of it is left on the died matter. Therefore the ratio of the ¹⁴C/¹²C drops from the steady-state ratio in the living matter.
    ₆C¹⁴ ₇N¹⁴ + ₋₁e⁰(t½ = 5760 years)
    By measuring this ratio and comparing it with the ratio in living plants one can estimate when the plant died.

Determination of the age of rock deposits

    Knowledge of the rate of decay of certain radioactive isotopes helps to determine the age of various rock deposits. Let us consider uranium containing rock formed many years ago.
    The uranium started to decay giving rise to the uranium - 238 to lead -206 series. The half-lives of the intermediate members being small compared to that of uranium -238(4.5 × 10⁹ years), it is reasonable to assume that those uranium atoms that started decaying many many years ago must have been completely converted to the stable lead-206 during this extra long period.
    The uranium-238 remaining and the lead-206 formed must together account for the uranium 238 present at zero time, that is, when the rock solidified. Thus both N₀ and N are known k is known from a knowledge of the half-life of uranium -238. Therefore the age of the rock can be calculated.

Determination of the Avogadro number

If 1 gm of a radioactive element contains N number of atoms.
Then, N = N₀/m
where N₀ = Avogadro number and m = Mass number
Therefore, kN = (k × N₀)/m
where k = 0.693/t½
Thus, kN = (0.693 × N₀)/(t½ × m)

∴ N₀ = (kN × t½ × m)/0.693
    If each atom of the radio-element expels one particle then kN, the rate of decay, is also the rate at which such particles are ejected.
    One gram of radium undergoes 3.7 × 10¹⁰ disintegrations per second, the mass number of Ra-226 is 226 and t½ = 1590 year so that the Avogadro Number can be calculated by the above equation,
    Avogadro Number =(3.7 × 10¹⁰ sec⁻¹ × 1590 × 365 × 24 × 60 sec × 226)/0.693
    = 6.0 × 10²³

Average life period of radioactive elements

    Besides half-life of a radio-element, another aspect of the life of a radio-element must also be known. It is now possible to determine the average life period of a radioactive atom present in aggregate of a large number of atoms.
    The average life period of an atom of the radio-element tells us the average span of time after which the atom will disintegrate.
    The length of time a radio-element atom can live before it disintegrates may have values from zero to infinity. This explains the gradual decay of the radio-element instead of decay of all the atoms at the same time. The average life period may be calculated as follows- 
    tav = Total Life Period/Total Number of Atoms
    Let N₀ atoms of a radioactive element are present in an aggregate of large no of atoms at time zero. Now in the small time interval t to (t+dt), dN atoms are found to disintegrate.
    Since dt is a small time period, we can take dN as the number of atoms disintegrating at the time t. So the total lifetime of all the dN atoms is t dN.
    Again the total number of atoms N₀ is composed of many such small numbers of atoms dN₁, dN₂, dN₃, etc, each with its own life span t₁, t₂, t₃, etc.
Average life of the radioactive elements
Average life and the rate of radioactive decay
    The average life of Radio-element is thus the reciprocal of its radioactive disintegration constant. This result can also be derived in a very simple alternative way.
    Since the radioactive atoms may be regarded as having an average life, then the product of the fractions of atoms disintegrating in unit time (that is k) and average life must be unity. 
    tav k = 1 So that, tav = 1/k
    Thus the relation between an average life and half-life is,
t½ = 0.693/k = 0.693 tav
  • Problem
    A piece of wood was found to have ¹⁴C/¹²C ratio 0.7 times that in the living plant. Calculate the approximate period when the plant died(t₁/₂ = 5760 years).
  • Solution
We know that radioactive decay constant,
k = 0.693/(t½) = 0.693/5760 years
= 1.20 ×10⁻⁴ yr⁻¹

N₀/N = 1.00/0.70
∴ 2.303 log(N₀/N) = kt
or, t = (2.303 log(N₀/N)/k

Putting the value, above equation,
we have, t = (2.303 × 0.155)/(1.20 × 10⁻⁴)
= 2970 years
  • Problem
    A sample of uranium (t₁/t₂ = 4.5 × 10⁹) ore is found to contain 11.9 gm of uranium-238 and 10.3 gm of lead-206. Calculate the age of the ore.
  • Solution
    11.9 gm of uranium-238 = 11.9/238 = 0.05 mole of uranium
    10.3 gm of lead-206 = 10.3/206 = 0.05 mole of lead -206
    Thus mole of uranium -238 present in the ore at zero time = (0.05+0.05) = 0.010 mole.
    And radioactive decay constant = 0.693/(4.5 × 10⁹) = 0.154 × 10⁻⁹ yr⁻¹.
Then 2.303 log(0.10/0.05) = kt
∴ t = (2.303 log2)/(0.154 × 10⁻⁹ yr⁻¹)
= 4.5 × 10⁹ year

Van't Hoff Equation - Effect on temperature on Equilibrium Constant:

Equilibrium Constant Kp of a reaction is constant at a given temperature but when the temperature is changed, the value of equilibrium constant also changed.
    The quantitative relation, relation, Known as Van't Hoff equation is derived by using the Gibbs - Helmholtz equation.
    The Gibbs - Helmholtz equation is,
ΔG0 = ΔH0 + T[d(ΔG0)/dT]P
Zero superscripts are indicating stranded values.
or, - (ΔH0/T2 )= -(ΔG0/T2 )+(1/T)[d(ΔG0)/dT]P
or, - (ΔH0/T2 ) = [d/dT(ΔG0/T)]P
Again Van't Hoff isotherm is,
- RT lnKP = ΔG0
or, - R lnKP = ΔG0/T
Differentiating with respect to T at constant P,
- R [dlnKP/dT]P = [d/dT(ΔG0/T)]P
Comparing the above two equation we have,
dlnKP/dT = ΔH0/T2
    This is the differential form of Van't Hoff reaction equation.
Separating the variables and integrating,
dlnKP = (ΔH0/R)(dT/T2)
Assuming that ΔH0 is independent of temperature.,br />or, lnKP = - (ΔH0/R)(1/T)+C(integrating constant)
The integration constant can be evaluated and identified as ΔS0/R using the relation,
ΔG0 = ΔH0 - TΔS0
Thus the Van't hoff Equation is,
lnKP = - (ΔH0/R)(1/T) + ΔS0/R
  • Problem 1:
    For a reaction 2A + B ⇆ 2C, ΔG0(500 K) = 2 KJ mol-1 Find the KP at 500 K for the reaction A + ½B ⇆ C.
  • Answer:
ΔG⁰(500 K) for the reaction,
A + ½B ⇆ C; (2 KJ mol⁻¹)/2 = 1 KJ mol⁻¹.
The relation is ΔG⁰ = - RT lnKp
or, 1 = - (8.31 × 10-3 ) × (500) lnKP
or, lnKP = 1/(8.31 × 0.5)
or, lnKP = 0.2406
∴ KP = 1.27

Plot of lnKP vs 1/T:

  • For exothermic reaction, ΔH0 = (-) ve.
    Examples are the formation of ammonia from H2 and N2.
N2 + H2 2NH3 ΔH0 = (-) ve
  • For endothermic reaction, ΔH0 = (+) ve.
    Examples are the dissociation of HI into H2 and I2.
HI H2 + I2 ΔH0 = (+) ve
  • For the reaction, ΔH0 = 0.
    lnKP is independent of T. Provided ΔS0 does not change much with T.
Van't Hoff Equation - Effect on temperature on Equilibrium Constant and Expression of Le -Chatelier Principle from Van't Hoff Equation
The plot of dlnKp vs T
    Since for ideal system, H is not a function of P and ΔH⁰ = ΔH and Van't Hoff equation is, dlnKp/dt = ΔH/RT² and the integrated equation is, lnKp = (ΔH/R)(1/T) + ΔS/R
    However, S of an ideal gas depends strongly on P, so ΔS and ΔG per mole of reaction in the mixture differ quite substantially from ΔS⁰ and ΔG⁰.
    If we consider the integrated Van't Hoff equation at two temperature then, it becomes,
ln(KP2/KP1) = (ΔH/R){(T2 - T1)/T1T2)}
    Where Kp₁ and Kp₂ are the equilibrium constants of the reaction at two different temperature T₁ and T₂ respectively.
    Thus the determination of Kp₁ and Kp₂ at two temperature helps to calculate the value of ΔH of the reaction.
    The above relation is called Van't Hoff reaction isobar since P remains constant during the change of temperature.

  • Problem 2:
    Use Gibbs - Helmholtz equation to derive the Vant Hoff reaction isochore. What condition do you expect a linear relationship between logk and 1/T?
  • Answer:
Again Kp = Kc (RT)ΔƔ
or, lnKp = lnKc + ΔƔ lnR + ΔƔ lnT
Differentiating with respect to T,
dlnKp/dT = dlnKc/dT + ΔƔ/T
But dlnKp/dT = ΔH⁰/RT²
Hence, dlnKc/dT = ΔH⁰/RT² - ΔƔ/T
or, dlnKc/dT = (ΔH⁰ - ΔƔRT)/RT²
∴ dlnKc/dT = ΔU⁰/RT²
    ΔU0 is standers heat of reaction at constant volume. This is really the Van't Hoff reaction isochore in differential form.
    The integrated form of the equation at two temperature is,
ln(KC2/KC1) = (ΔU0/R){(T2 - T1)/T1T2)}
    For ideal system ΔU0 = ΔU. Since the reaction occurs in an equilibrium box at constant Volume this is called Van't Hoff reaction isochore.
  • Two important assumptions are:
  1. The reacting system is assumed to behave ideally.
  2. ΔH is taken independent of temperature for a small range of temperature change.
    Due to assumption involved ΔH and ΔU do not produce the precise value of these reactions. As ΔG is independent of pressure for ideal system ΔH and ΔU also independent of pressure.
    Hence, ΔG0 = - RT lnKp or, [dlnKP/dT]T = 0
  • Problem 3:
    Show that the equilibrium condition for any chemical reaction is given by ΔG = 0.
  • Answer:
Vant Hoff reaction isotherm is,
ΔG = - RT lnKa + RT lnQa
But when the reaction attains an equilibrium,
Qa = Ka
∴ ΔG = 0

Expression of Le -Chatelier Principle from Van't Hoff Equation:

Van't Hoff equations give a quantitative expression of Le-Chatelier Principle.
lnKp = -(ΔH/R)(1/T) + C
    It is evident that for endothermic reaction (ΔH〉0), an increase of T increases the value of any of the reaction.
    But for exothermic reaction (ΔHㄑ 0), with rising in temperature, a is decreased. This change of Kp also provides the calculation of quantitative change of equilibrium yield of products.
    This above statement is in accordance with Le - Chatelier Principle which states that whenever stress is placed on any system in a state of equilibrium, the system always reacts in a direction to reduce the applied stress.
    Thus when T is increased, the system in equilibrium will try to move in a direction in which heat is absorbed that is endothermic reaction is favors.

Derivation of van der Waals equation

In 1873, Van der Waals modified the Ideal Gas Equation and formulate Van der Waals equation of state for real gases.
    PiVi = RT
    By incorporating the size effect and intermolecular attraction effect of the real gas. These above two effects are discussing under the Volume Correction and Pressure Correction of the Ideal Gas Equation.

Volume Correction by Van der Waals

    Molecules are assumed to be a hard rigid sphere and in a real gas, the available space for free movement of the molecules becomes less then V. let us take the available space for free movement of 1-mole gas molecules,
Vi = (V-b)
    where V is the molar volume of the gas and b is the volume correction factor. Vi = molar volume of the ideal gas where the gas molecules are regarded as point masses. Let us take, r is the radius of the molecule and σ = 2r is the diameter using the molecule as a rigid sphere.
Encounter of the gas molecules in van der waals equation
An encounter of the gas molecules.


    Then it can be shown that the volume correction term or effective volume of one-mole gas molecules,
    b = 4 × N0 (4/3) π r3
    When two molecules encounter each other the distance between the centers of the two molecules would be σ. They can not approach beyond this distance. Thus the sphere of radius σ will occupy a space unavailable for a pair of molecules.
    This excluded volume = 4/3 π σ3 for a pair of molecules. Thus a single molecule, = 1/2 × 4/3 π σ3 The effective volume for Avogadro number of molecules (present in 1-mole gas),
b = N0 × 2/3 × π σ3
or, b = N0 × 2/3 × π (2r)3
or, b = 4 N0 × 4/3 × π (r)3
Thus, b = N0 × 2/3 × π (2r)3
∴ b = 2/3 × N0 × π σ3
    Thus b measures and helps to calculate the radius ( or diameter) of the gas molecule. The equation becomes,
    Pi (V-b) = RT

Pressure Correction by Van der Waals

    The pressure of the gas is developed due to the wall collision of the gas molecules. But due to intermolecular attraction, the colliding molecules will experience an inward pull and so the pressure exerted by the molecules in real gas will be less than that if there had not been an intermolecular attraction as in ideal gas Pi.
    Thus, Pi 〉P or, Pi = P + Pa
    Where Pa is the pressure correction term originating from attractive forces. Higher the intermolecular attraction in a gas, greater is the magnitude of Pa. Pressure depends on both the frequency of molecular collisions with the walls and the impulse exerted by each collision. Both contributions are diminished by the attractive forces and the strength with which these operate is roughly proportional to the density of the molecules. Therefore the average pressure exerted by the molecules is decreased by Pa is proportional to the square of the density.
    Pa ∝ 1/V2 Since density ∝ 1/V ∴ Pa = a/V2
    Where a = constant for the gas that measures the attractive force between the molecule.
    Thus, Pi = P + (a/V2)
    Using the two corrections we have Van der Waals equation for 1 mole Real Gas is,
    (P + a/V2 )(V-b) = RT
    To convert the equation for n moles volume has to change as it is the only extensive property in the equation. Let V be the volume of n moles gas hence, V = v/n. Putting this term in the Van der Waals equation we have,
what is the Van der Waals equation?
Van der Waals equation for n Moles Real Gas



For the real gas a=0 but b≠0

    We have Van der Waals Equation, (P+ an2/V2)(V - nb) = nRT but a=0 Hence, P = nRT/(V - nb) 〉Pi Since, Pi = nRT/V only.

For the real gas a≠0 but b=0

    It means that the molecular size effect (repulsive interaction) creates higher pressure than that observed by the ideal gas where molecules have no volume.
    We have Van der Waals Equation, (P+ an2/V2)(V - nb) = nRT but b=0(no size). Hence, P = (nRT/V - an2/V2)ㄑPi Since, Pi = nRT/V only. Thus, intermolecular attraction effect reduces the pressure of the real gases.

Units of Van der Waals Constant

From the Van der Waals Equation, (P+ an2/V2)(V - nb) = nRT.
Where Pa is the pressure correction term, called often the internal pressure of the gas.
Then, a = Pa × (V2/n2)
    Thus the unit of a = atm lit2 mol-2
    Again, nb = Unit of volume (say liter)
    Hence, the unit of b = lit mol-1

Significance of Van der Waals constants

    ‘a’ term originates from the intermolecular attraction and Pa = an2/V2. Thus, ‘a’ is a measure of internal pressure of the gas and so the attractive force between the molecules. Higher the value of ‘a’ grater is the intermolecular attraction and more easily the gas could be liquefied. Thus, a of CO2 = 3.95 atm lit2 mol-2 while, a of H2 = 0.22 atm lit2 mol-2. Another constant, ‘b’ measures the molecular size and also a measure of repulsive forces. The value of b can also be utilized to calculate the molecular diameter(σ). The grater, the value of ‘b’ larger the size of the gas molecule. Thus, b of CO2 = 0.04 lit mol-1 while, b of H2 = 0.02 lit mol-1.

Boyle's temperature from Van der Waals equation

Mathematical condition for Boyle’s temperature is,
TB = [d(PV)/dP]T When P→0
From the Van der Waals Equation for 1 mole gas,
(P + a/V2)(V - b) = RT
or, P = RT/(V - b) - a/V2
Hence, PV = {RTV/(V - b)} - a/V
Thus, TB = [d(PV)/dP]T
= [RT/(V-b)-{RTV/(V-b)2}+a/V2][(dV/dP)]T
= [{RT(V - b) - RTV
}/(V - b)2 + a/V2] [(dV/dP)]T
=[{- RTb/(V - b)2} + a/V2] [(dV/dP)]T
But when T = TB, [d(PV)/dP]T = 0 and [(dV/dP)] ≠ 0
Thus, We have, RTBb/(V - b)2 = a/V2
or, TB = (a/Rb) {(V - b)/V}2 Since P → 0, V is large.
Thus, (V - b)/V ≃ 1
Hence,TB = a/Rb

Amagat curve from Van der Waals Equation

Explanation of Amagat Curves by van der waals equation
Amagat Curves
Van der Waals equation for 1-mole real gas,
(P + a/V2)(V - b) = RT
or, PV - Pb + (a/V) + (a/V2) = RT
Neglecting the small term (a/V2), we get,
PV = RT + Pb - (a/V)
Using the ideal gas equation for the small term,
a/V = aP/RT and taking Z = (PV/RT),
We have, Z = 1 + (1/RT){b - (a/RT} P
This Shows that Z = (T, P)
    This equation can be used to explain Amagat curve qualitatively at low pressure and moderate pressure region.
For CO2  Van der Waals constant ‘a’ is very high
    Thus (a/RT) 〉b in the equation and {b - (a/RT)} = - ve . That is the slope of Z vs P curve is ( -) for carbon dioxide at moderate pressure region. That is the value of Z decreases with the increase of Pressure and it is also found in the curve of carbon dioxide.
For H2  Van der Waals constant ‘a’ is very small
    So, a/RT〈 b, slope of Z vs P curve for H2 becomes (+) ve and the value of Z increases with pressure.
(i) When T〈 TB
or, T〈 a/Rb
or, b〈 a/RT
and {b−(a/RT)} = (-)ve.
    That is the value of Z decreases with increases with P at the moderate pressure region of the curve of CO2, Z〈 1 and more compressible.
(ii) When T = TB = a/Rb
or, b = a/RT
hence, {b - (a/RT)} = 0
    That is Z = 1 the gas shows ideal behavior. The size effect compensates the effect due to the intermolecular attraction of the gas. Z runs parallel to P axis up to a certain range of pressure at the low-pressure region.
(iii) When T 〉TB
or, T 〉a/Rb
or, b 〉a/RT
hence {b - (a/RT} = (+)ve.
    That is the value of Z increases with the increase of P when T 〉TB. The size effect dominates over the effect due to the intermolecular attraction. For hydrogen and helium, 0°C is greater then their TB values and Z vs P slope becomes (+)ve. At very low-pressure P→0 and high temperature, the volume is very large and both the size effect and attraction effect becomes negligible that is Pb and (a/RT) P are negligible and Z=1 that is the gas obeys ideal behavior.

Questions and Answers of Van der Waals equation

  • Question
    Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 270C. Given, a = 1.4 atm lit2 mol-2 and b=0.04 lit mol-1 using Van der Waals equation. Also, calculate the pressure of the gas using the ideal gas equation and find the extent of deviation from ideal behavior.
  • Answer
    P = 4.904 atm, Pi = 4.92 atm and Deviation=0.325%
  • Question
    Find the units and dimensions of Van der Waals constant 'a' and 'b' from Van der Waals Equation.
  • Answer
    In SI system,unit of 'a' = N m4 mol-2 In CGS system,unit of 'a' =dyne cm4 mol-2 In SI system,unit of 'b' = m3 mol-1 In CGS system,unit of 'b' = cm3 mol-1
  • Question
    Calculate the Boyle temperature of nitrogen gas given a=1.4 atm lit2 mol-2 and b = 0.04 lit mol-1.
    • Answer
      TB = 427 K
    • Question
      Comment on the possibility of defining a Boyle Temperature if a=0 and a=b=0.
    • Answer
    1. When, a = 0, Van der Waals Equation, P (V - b) = RT or, PV = RT + Pb Differentiating with respect of pressure at constant temperature, [d(PV)/dP]T = b, but b ≠ 0 hence [d(PV)/dP]T ≠ 0. The gas has no Boyle temperature since at any temperature since at any temperature, [d(PV)/dP]T ≠ 0.
    2. Again when a = b = 0, Van der Waals equation becomes PV = RT; or, [d(PV)/dP]T = 0.
      That is at any temperature this becomes zero and so all the temperature are Boyle Temperatures.

    Ionization energy is supplied to an electron of a particular atom, it can be promoted to successively higher energy levels. If the supplied energy is large enough it may possible to remove the electron completely from the pull of the positive nucleus.
      The amount of energy required to remove the most loosely bound electron, that is outermost electron from an isolated gaseous atom of an element in its lowest energy state, that is ground state to produce cation is known as Ionization Potential or Ionization Energy of that element.
      M(g) + Ionization Energy M⁺(g) + e
      It is generally represented as I or IP and it is measured in electron volt (eV) or kilocalories (kcal) per gram atom. One electron volt (eV) being the energy gained by an electron falling through a potential difference of one volt.
    ∴ 1 eV = (Charge of an electron) × (1 volt)
    = (1.6 × 10⁻¹⁹ Coulomb) × (1 Volt)
    = 1.6 × 10⁻¹⁹ Joule
    = 1.6 × 10⁻¹² erg

    Successive Ionization Potentials

    1. The electrons are removed in stages one by one from an atom. The energy required to remove the first electron from a gaseous atom is called its first Ionization potential.
      M (g) + IP₁  M⁺ (g) + e
    2. The energy required to remove the second electron from the cation is called second Ionization potential.
      M⁺ (g) + IP₂ M⁺² (g) + e
    3. Similarly, we have third, fourth Ionization potentials.
      M⁺² (g) + IP₃ M⁺³ (g) + e
      M⁺ (g) + IP₄ M⁺⁴ (g) + e

    Periodic trends ionization energy

      The following factors influence the magnitude of the Ionization Potentials:

    Atomic radius ionization potential

      The greater the distance of an element from the positive charge nucleus of an element, the weaker will be the attraction and hence the lower the ionization potential of that element.
      If an atom is raised to an excited state by promoting one electron to a higher energy level, then that electron is further from the nucleus and it is observed that this excited electron is more easily detached than when the electron is in its ground state.
      Thus on moving from top to bottom in a group the ionization potential of the elements decreases with the increase of their nucleus electron separation.
      The successive increase in their values is due to the fact that it is relatively difficult to remove an electron from a cation having higher positive charge than from a cation having a lower positive charge or from a neutral atom.
    Ionization Potential
    The magnitude of Ionization Potential

    Nuclear charge and ionization potential

      The higher the charge on the nucleus the more difficult it is to remove an electron and hence the higher the value of ionization potential.
      Thus the value of ionization potential generally increases in moving from left to right in a period, since the nuclear charge of the elements (atomic number) also increases in the same direction.
    Ionization potential vs nuclear charge
      With increasing, atomic number electrons are added to orbitals of the same principal quantum number. These electrons add little to the size of the atom, the increasing nuclear charge brings about a contraction in size. In effect, therefore, ionization potential steadily increases along a period.

    Filled and half-filled orbitals and ionization potential

      According to the Hund's rule atoms having half-filled or completely filled orbitals are comparatively more stable and hence more energy is needed to remove an electron from such atom.
      The ionization of such atoms is therefore relatively difficult than expected normally from their position in the periodic table. A few regulations that are seen in the increasing value of ionization potential along a period can be explained on the basis of the concept of the half-filled and completely filled orbitals.
      Be and N in the second period and Mg and P in the third period have a slightly higher value of ionization potentials than those normally expected.
      This is explained on the basis of extra stability of the completely - filled 2S - orbital in Be (2S²) and 3S - orbital in Mg (3S²) and of half-filled 2P - orbital in N (2S² 2P³) and 3P - orbital in P (3S² 3P³).

    Screening Effect and ionization potential

      Electrons provide a screening effect on the nucleus. The outermost electrons are shielded from the nucleus by the inner electrons.
      The radial distribution functions of the S, P, d orbitals shows that for the same principal quantum number the S - orbital is the most penetrating., next is P - orbital and least penetrating is the d - orbital. Screening efficiency falls off in the order, S〉P〉d.
      As we move down a group, the number of inner - shells increases and hence the ionization potential tends to decreases.
    Elements of II A Group: Be〉Mg〉Ca〉Sr〉Ba

    Overall Charge on the Ionizing Species

      An increase in the overall charge on the ionizing species (M⁺, M⁺², M⁺³, etc) will enormously influence the ionization potential since electron withdrawal from a positively charged species is more difficult than from a neutral atom.
      The first ionization of the elements varies with their positions in the periodic table. In each of the table, the noble gas has the highest value and the alkali metals the lowest value for the ionization potential.

    Ionization potential questions answers




    • Question
      Calculate the ionization potential of the Hydrogen atom in eV.
    • Answer
      We know that the energy corresponding to the transition from n = ∞ to n = 1 gives the ionization potential of the hydrogen atom.
    Thus the IP of the Hydrogen atom,
    EH = (2π²me⁴/h²)[(1/n₁²) - (1/n₂²)]
    ∴ EH = 2.179 × 10⁻¹¹ erg
    = 2.179 × 10⁻¹⁸ Joule
    = (2.179 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV
    = 13.6 eV
    • Question
      Calculate the second ionization potential of Helium (given IP of Hydrogen = 13.6 eV)
    • Answer
      The ground state electronic configuration of helium is 1S². The second ionization potential means the removal of the second electron from the 1S orbital against the nuclear charge of +2.
    So we have: IPHe = (2π²mZ²e⁴/h²)[(1/n₁²) - (1/n₂²)]
    = Z² × IPH
    ∴ Second Ionization Potential of Helium, = 22 × 13.6
    = 54.4 eV
    • Question
      In first transition series electron filling up processes begins in the 3d level below a filled 4S level. During ionization process will a 4S electron or a 3d electron be lost first? Explain with reference to Chromium.
    • Answer
      During ionization, the 4S electron lost first.
      For details follow the link
    • Question
      Arrange the following with the increasing order of Ionization Potentials Li, Be, B, C, N, O, F, Ne.
    • Answer
      LiㄑBㄑBeㄑCㄑOㄑN〈FㄑNe

    Zero-order kinetics, the rate of these reactions does not depend on the concentration of the reactants.

    Mathematical derivation of zero-order kinetics

      Let us take a reaction represented as
        Product
      Let the initial concentration of the reactant a and product is zero. After the time interval t, the concentration of the reactant is (a-x) and concentration of the product is x. Thus x is decreases of concentration in zero-order reaction.
    • Mathematical derivation of zero-order kinetics in terms of product.
      Thus the mathematical equation of zero-order kinetics in terms of product,
    dx/dt = k₀
    Where k₀ is the rate constant of the zero-order reaction.
    or, dx = k₀dt

    Integrating the above reaction,
    dx = k₀ dt
    or, x = k₀t + c
    where c is the integration constant of the reaction.

    When t = o, x is also zero thus, C = o Thus the above equation is,
    x = k₀ t
      This is the relationship between decreases of concentration of the reactant(x) within time(t).
    • Mathematical derivation of zero-order kinetics in terms of reactant.
    Rate equation in terms of reactant,
    -d[A]/dt = k₀ [A]⁰ = k₀
    Where [A] is the concentration of the reactant at the time t.
    or, - d[A] = k₀dt
    Integrating the above equation,
    We have - d[A] = k₀ ∫ dt
    or, - [A] = k₀t + c
    where c is the integration constant of the reaction.
    If initial at the time t = 0 concentration of the reactant [A]₀ Then from the above equation,
    - [A]₀ = 0 + c
    or, c = -[A]₀
      Putting the value on the above equation,
    - [A] = kt - [A]₀
      This is another form of the rate equation in zero-order kinetics.

    The half-life of zero-order kinetics

      The time required for half of the reaction to be completed is known as the half-life of the zero-order reaction. It means 50% of reactants disappear in that time interval.

    Half-life in zero-order kinetics

      If in a chemical reaction initial concentration is [A]₀ and after t time interval the concentration of the reactant is [A].
      Then, [A]₀ - [A] = kt
      Thus when t = t½, that is the half-life of the reaction, the concentration of the reactant [A] = [A]₀/2. Putting the value on the above equation,
    We have [A]₀ - [A]₀/2 = k t½
    or, k t½ = [A]₀/2
    t½ = [A]₀/2k
      Thus for the zero-order kinetics the half-life of the reaction proportional to its initial concentration.

    Examples of the zero-order kinetics

      The only heterogeneous catalyzed reactions may have zero-order kinetics.
    Zero order kinetics integrated equation with examples
    Examples of zero-order kinetics

    Characteristics of zero-order kinetics

    1. The rate of the reaction is independent of concentration.
    2. Half-life is proportional to the initial concentration of the reactant.
    3. The rate of the reaction is always equal to the rate constant of the reaction at all concentration.

    Unit of the rate constant in zero-order kinetics

      The rate equation in terms of product for the nth-order reaction is,
    d[A]/dt = k [A]n
    or, k = (d[A]/dt) × (1/[A]n)
      Thus the unit of rate constant(k) = (unit of concentration)/{unit of time × (unit of concentration)n}
      = (unit of concentration)1-n/unit of time
      Thus if zero-order kinetics the concentration is expressed in lit mole⁻¹ and time in sec
    Then the rate constant = (lit mol⁻¹)/sec
    = mol lit⁻¹sec⁻¹

    Questions and Answers of zero-order kinetics


    • Question
      The rate constant of a chemical reaction is 5 × 10⁻⁸ mol lit⁻¹sec⁻¹. What is the order of this reaction? How long does it take to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹?
    • Answer
      The reaction is a zero-order reaction and 3.92 × 10⁵ Sec take to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹.
    • Questions
      The half-life of a zero-order reaction is x and the reaction is completed on t₁ time. What is the relation between x and t₁?
    • Answer
      2x = t₁
    • Question
      If the rate of the reaction is equal to the rate constant. What is the order of the reaction?
    •  Answer
      Zero-order reaction.
    • Question
      For a reaction, N₂ + 3 H₂ → 2NH₃, if d[NH₃]/dt = 2 × 10⁻⁴mol lit⁻¹sec⁻¹, What is the order and the value of - d[H₂]/dt of this reaction?
    • Answer
      Zero-order reaction and the value of d[H₂]/dt = 3 × 10⁻⁴ mol lit⁻¹sec⁻¹.
    • Question
      For the reaction N₂O₅ → 2NO₂ + ½ O₂, the rate of disappearance of N₂O₅ is 6.25 × 10⁻³ mol lit⁻¹sec⁻¹, what is the rate of formation of NO₂ and O₂ respectively?
    • Answer
      1.25 × 10⁻² and 3.125 × 10⁻³ mol lit⁻¹sec⁻¹
    • Question
      For the reaction H₂ + Cl₂ → 2HCl on sunlight and taking place on the water. What is the order of the reaction?
    • Answer
      This is a zero-order reaction.

    Inorganic Chemistry

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