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Van der Waals Equation

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

Volume Correction:

Molecules are assumed to be a hard rigid spheres 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 Ïƒ = 2r is the diameter using the molecule as a rigid sphere.
 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) Ï€ r3
When two molecules encounter each other the distance between the centers of the two molecules would be Ïƒ. They can not approach beyond this distance. Thus the sphere of radius Ïƒ will occupy a space unavailable for a pair of molecules.
This excluded volume = 4/3 Ï€ Ïƒ3 for a pair of molecules.
Thus a single molecule,
= 1/2 × 4/3 Ï€ Ïƒ3
The effective volume for Avogadro number of molecules (present in 1 mole gas),
 b = N0 × 2/3 × Ï€ Ïƒ3 or, b = N0 × 2/3 × Ï€ (2r)3 or, b = 4 N0 × 4/3 × Ï€ (r)3 Thus, b = N0 × 2/3 × Ï€ (2r)3 ∴ b = 2/3 × N0 × Ï€ Ïƒ3
Thus b measures and helps to calculate the radius ( or diameter) of the gas molecule. The equation becomes,
Pi (V - b) = RT

Pressure Correction:

Pressure of the gas is developed due to the wall collision of the gas molecules. But due to inter-molecular attraction, the colliding molecules will experience an inward pull and so pressure exerted by the molecules in real gas will be less than that if there had not been inter molecular 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 inter molecular 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,
 Van der Waals equation for n Moles Real Gas
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 ideal gas equation and find the extent of deviation from ideal behavior.
P = 4.904 atm, Pi = 4.92 atm and Deviation=0.325%

Let us consider two hypothetical cases to show the size effect and attraction effect on the pressure of the gas:

(a)For the real gas a=0 (that is no inter molecular attraction )
but b≠0 (size considers):
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.
(b)For the real gas a≠0 (existence of attractive force) but b=0 (no size consider):
It means that the molecular size effect (repulsive interaction) creates higher pressure then 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)ã„‘Pi
Since, Pi = nRT/V only.
Thus, inter molecular attraction effect reduces the pressure of the real gases.

Units of Van der Waals Constant a and b:

From the Van der Waals Equation,
(P+ an2/V2)(V - nb) = nRT
where, Pa = an2/V2
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
Find the Units and Dimensions of Van dar Walls constant 'a' and 'b' from Van der walls Equation.
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

Significance of Van der Waals constants 'a' and 'b':

a’ term originates from the inter molecular 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 inter molecular 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(Ïƒ). 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.

For a Van der Waals gas the expression for Boyle's temperature:

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
Calculate the Boyle temperature of nitrogen gas given a=1.4 atm lit2 mol-2 and b = 0.04 lit mol-1.
TB = 427 K
Comment on the possibility of defining a Boyle Temperature if a=0 and a=b=0.
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 Walls 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.

Explanation of Amagat’s Curves:

 Amagat’s 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 ideal gas equation for 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 uses to explain Amagat curve qualitatively at low pressure and moderate pressure region.
(a) For CO2 ‘a’ is very high since the gas is easily liquefied.
Thus (a/RT) 〉b in the equation and {b - (a/RT)} = - ve .
That is slope of Z vs P curve is ( -) for carbon dioxide at moderate pressure region. That is the value of Z decreases with increase of Pressure and it is also found in the curve of carbon dioxide.
(b) For H2 ‘a’ is very small and It is not easily liquefied.
So, a/RT〈 b and slope of Z vs P curve for H2 becomes (+) ve and the value of Z increases with pressure.
 (i) When T〈 TBor, T〈 a/Rbor, b〈 a/RTand {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/RThence, {b - (a/RT)} = 0
That is Z = 1 the gas shows ideal behavior. The size effect compensates the effect due to inter molecular attraction of the gas. Z runs parallel to P axis up to certain range of pressure at low pressure region.
 (iii) When T 〉TB or, T 〉a/Rbor, b 〉a/RThence {b - (a/RT} = (+)ve.
That is value of Z increases with increase of P when T 〉TB.
The size effect dominates over the effect due to inter molecular attraction. For hydrogen and helium, 0°C is grater 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.

Critical Constants of a Gas

A gas can be liquefied by lowering temperature and increasing of pressure. But influence of temperature is more important. Most of gases are liquefied at ordinary pressure by suitable lowering of the temperature. But a gas can not be liquefied unless its temperature is below of a certain value depending upon the nature of the gas whatever high the pressure may be applied.
This temperature of the gas is called its Critical Temperature (Tc). A gas can only be liquefied when the temperature is kept below Tc of the gas.

Definitions of Critical Constants:

Critical Temperature (Tc):
Critical Temperature (Tc) is the maximum temperature at which a gas can be liquefied, that is the temperature above which a liquid cannot exist.
Critical Pressure (Pc):
Critical Pressure (Pc) is the maximum pressure required to cause liquefaction at the temperature (Tc).
Critical Volume (Vc):
Critical Volume (Vc) is the volume occupied by one mole of a gas at critical temperate (Tc) and Critical Pressure (Pc).

Andrews Isotherms:

In 1869, Thomas Andrews carried out an experiment in which P - V relations of carbon dioxide gas were measured at different temperatures.
 Andrews Isotherms.

Following are observed from this graph:

1. At high temperature, such as T4, the isotherms look like those of an Ideal Gas.
2. At low temperatures, the curves have altogether different appearances. Consider, for examples, a typical curve abcd. As the pressures increases, the volume of the gas decreases (curve a to b).
3. At point b liquefaction commences and the volume decreases rapidly as the gas converted to a liquid with much higher density. This conversion takes place at constant pressure P.
4. At the point C, liquefaction is complete and thus the cd is evidence of the fact that the liquid cannot be easily compressed. Thus, we note that ab represents the gaseous state, bc, liquid and vapour in equilibrium, and cd shows the liquid state only.
5. At still higher temperatures we get a similar type of curve as discussed in the above point, except that the horizontal portion is reduced. The Pressure corresponding to this portion being higher then at lower temperatures.
6. At temperatures Tc the horizontal portion is reduced to a mere point. At temperatures higher then Tc there is no indication of qualification at all.
Thus for every gas, there is a limit of temperature above which it cannot be liquefied, no matter what the pressure is.

Continuity of State:

It appears from the Amagat curve at T there is discontinuity or break during the transformation of gas to liquid. But it is not so.
The continuity of the states from the gas to liquid can be explains from the following isotherm pqrs at T.
 Continuity of State.
The gas at A is heated to B at constant volume (V) along AB. Then the gas is gradually cooled at constant P along BC, the volume is reduced considerably. The gas is again cooled at constant V until the point D is reached.
No where in the process liquid would appear. At D, the system is highly compressed gas. But from the curve, this point is the representation of the liquid state.
Hence there is hardly a distinction between the liquid state and the gaseous state. There are no line of separation between the two phases. This is known as the principle of continuity of state.

Critical Phenomena and Van der Walls Equation:

 (P + a/V2)(V - b) = RTV3 - (b + RT/P)V2 + (a/P)V - ab/P = 0
This equation has three roots in V for given values of a, b, P and T. It found that either all the three roots are real or one is real and the other two are imaginary.
 Critical Phenomena and Van der Walls Equation.

The main characteristics of the above isotherm are as follows:

1. At higher temperature and higher volume region, the isotherms looks much like the isotherms for a Ideal gas.
2. At the temperature lower than Tc the isotherm exhibits a maximum and a minimum. For certain values of pressure, the equation gives three roots of volume as V1, V2 and V3 at pressure P1.
The section AB and ED of the Van der walls curve at T1 can be realized experimentally. ED represents Supersaturated or Super cooled vapour and AB represents super heated liquid. Both these states are meta-stable.
These states are unstable in the seance that slight disturbances are sufficient to cause the system to revert spontaneously into the stable state with the two phases present in equilibrium.
3. The section BCD of the Van der Walls isotherms cannot be realized experimentally. In this region the slope of the P - V curve is positive. Increasing or decreasing the volume of such a system would increase or decrease of pressure. The line BCD represents the meta- stable state.

Expression of Critical Constants in Terms of Van der Waals Constants:

Again with increase of temperature the minimum and maximum points come close to each other and at the critical point (C) both maximum and minimum coalesce. The slope and curvature both becomes zero at this point.
Thus the mathematical condition of critical point is,
(dP/dV)T = 0 and (d2P/dV2)T = 0
Van der Waals equation for 1 mole gas is,
(P + a/V2)(V - b) = RT
or, P = RT/(V - b) - a/V2
Differentiating Van der Waals equation with respect to V at constant T,
We get Slope,
(dP/dV)T = - {RT/(V - b)2} + 2a/V3
And Curvature,
(d2P/dV2)T = {2RT/(V - b)3} - 6a/V4
Hence at the critical point,
- {RTc/(Vc - b)2} + 2a/Vc3 = 0
or, RTc/(Vc - b)2 = 2a/Vc3
and {2RTc/(Vc - b)3} - 6a/Vc4 = 0
or, 2RTc/(Vc -b)3 = 6a/Vc4
Thus, (Vc - b)/2 = Vc/3
 ∴ Vc = 3b
Putting the value of Vc = 3b in RTc/(Vc - b)2 = 2a/Vc3.
We have, RTc/4b2 = 2a/27b3
 ∴ Tc= 8a/27Rb
Again the Van der Walls equation at the critical state,
Pc = RTc/(Vc - b) - a/Vc2
Putting the value of Vc and Tc,
 ∴ Pc = a/27b2
The Critical constants for water are 647 K, 22.09 MPa and 0.0566 dm3mol-1. Calculate the value of Van der Waals constants a and b.
We have Tc = 647 K,
Pc = 22.09 Mpa = 22.09 × 103 kPa
and Vc = 0.0566 dm3 mol-1
Thus, b = Vc/3 = (0.0566 dm3 mol-1)/3
∴ b = 0.0189 dm3 mol-1
a = 3 Pc Vc2
= 3 (22.09 × 103 kPa)(0.0566 dm3 mol-1)2
∴ a = 213.3 kPa mol-2

Compressibility Factor at the Critical State (Zc):

The Critical coefficient is defined as,
 RTc/PcVc
Thus the value of Critical Coefficient,
= {R × (8a/27Rb)}/{(a/27b2) × 3b}
= 8/3
= 2.66
Thus the value of Compressibility factor at the critical state (ZC)
= PCVC/RTC
= 3/8
= 0.375

Van der Waals constants in terms of critical constants:

Van der Waals constants can be determined from critical constants Tc and Pc of the gas. Vc in the expression is avoided due to difficulty in its determination.
We have, b = Vc/3
but, PcVc/RTCc = 3/8
or, Vc = (3/8) × (RTc/Pc)
 ∴ b = (1/8)(RTc/Pc)
Again, a = Pc × 27b2
= 3 × Pc × (3b)2
= 3 PcVc2
= 3Pc × (3RTc/8Pc)2
 ∴ a = (27/64)(R2Tc2/Pc)
Calculate Van der Waals constants for Ethylene. (TC = 280.8 K and PC = 50 atm).
a = 0.057 lit mol-1 and b = 4.47 lit2 atm mol-2
Argon has (TC = - 122°C, PC = 48 atm). What is the radius of the Argon atom?
1.47 × 10-8 cm
 Gas TC (0K) PC (atm.) VC (c.c.) Oxygen 154.28 49.7 74.4 Hydrogen 33.2 12.8 69.7 Nitrogen 125.97 33.5 90.0 Ammonia 405.5 111.3 72.0 Carbondioxide 304.15 72.9 94.2 Helium 5.2 2.25 61.55 Methane 190.3 45.6 98.8

Law of Mass Action

The Low of Mass Action was first formulated by two Norwegian chemists, Guldberg and Waage in 1864. The basis of their formulation is the observation of huge deposit of Sodium Carbonate on Egyptian take shore. Large amount of NaCl in take water and CaCO3 on the take shore made the reverse reaction possible.
 CaCl2(aq) + Na2CO3(aq) ↓ CaCO3(s) + 2NaCl(aq)
The foreword reaction occurs spontaneously in the laboratory.
They States the Low as,
The rate of chemical reaction at a constant temperature, is directly proportional to the active mass of the reactants.
The active mass is thermodynamic quantity. We assumed active mass as,
1. Molar Concentration (moles/lit) When the solution is dilute, that is when the system behaves ideally.
2. Partial pressure in atmosphere unit for gaseous system and when the pressure of the system is very low.
3. For pure solid and pure liquid, active mass is assumed to be unity since their mass does not effect the rate of reaction.

Application of Low of Mass Action to the Chemical Reaction:

Let us Consider a reaction,
 A + B ⇆ C + D
Let the reacting system contains reactants only and CA and CB are their Concentrations in molar units.
According to the mass action low, the rate of the foreword reaction,
 Rf ∝ CA × CB ∴ Rf = Kf × CA × CB
Though the rate equation is an experimental quantity and powers of concentration terms are determined experimentally, here we consider a special type of rate equation where the stoichometric coefficients are raised in the of the conc. term(this is true for the elementary or one step reaction).
As the reaction proceeds in the foreword direction, conc. of A and B decreases and Rf also decreases. When the products are getting accumulated in the system, backwards reaction also starts and the rate of the backward reaction,
 Rb ∝ CC × CD ∴ Rb = Kb × CC × CD
Here Kf and Kb are the rate constants of the foreword and backward reaction and they do not depends on Conc. at a given temperature.
As the reactions proceeds in the foreword reactions Rf is decreasing but Rb is increasing. A state is then attain when they are equal. This state is called the chemical equilibrium. There will be no further change in the composition of the system.
Thus at equilibrium,
Rf = Rb
or, Kf × CA × CB = Kb × CC × CD
 ∴ Kf/Kb = (CC × CD)/(CA × CB)
CA, CB, CC and CD are the equilibrium concentration of A, B, C and D.
Again, Kf/Kb = Kc, called concentration equilibrium constant of the reaction.
At a given temperature for a reaction, Kc is constant does not depend on the concentration of the reacting components.
 Kc = (CC × CD)/(CA × CB)

Equilibrium Constant:

Concentration Equilibrium Constant:
If we write the equation as,
Æ”1A1 + Æ”2A2 Æ”3A3 + Æ”4A4
Where Æ”1, Æ”2, Æ”3 and Æ”4 are stoichiometric coefficient.
 ∴ Kc = (C3Æ”3 × C4Æ”4)/(C1Æ”1 × C2Æ”2)
Where Kc is called concentration equilibrium constant of the reaction and C1C2C3 and C4 are the equilibrium concentration of A, B, C and D.
However the values of equilibrium constant of a chemical reaction depends on the mode of writing its Stoichiometric (balanced) equation.
Thus, for the reaction of formation of NH3 from N2 and H2, we can write the equation as,
N2 + 3 H2 2 NH3
The equilibrium constant can be written as,
 Kc = (CNH3)2/(CN2) (CH2)2
But if the equation written as,
1/2 N2 + 3/2 H2 NH3
 Then, K՛c = (CNH3)/(CN2)3/2(CH2)1/2
It is clear then Kc and c are not Equal in magnitude.
 Thus, Kc = (K՛c)1/2
The general rule is that if the equation is multiplied by n then the relation of Equilibrium concentration n will be raised to the power.
 That is, Kc = (K՛c)n
Pressure Equilibrium Constant:
When all the reactants and products are gases (that is, gas - phase reacting system), the expression of equilibrium constant for the equation,
Æ”1A1 + Æ”2A2 Æ”3A3 + Æ”4A4
Where Æ”1, Æ”2, Æ”3 and Æ”4 are stoichiometric coefficient.
 ∴ Kp = (P3Æ”3 × P4Æ”4)/(P1Æ”1 × P2Æ”2)
Where Kp is called Pressure equilibrium constant of the reaction and P1P2P3 and P4 are the equilibrium partial pressure of reacting components.
For the dissociation N2O4 2 NO2, obtain an expression for the fraction of original N2O4 dissociated at equilibrium in terms of Kp and total pressure.
The reaction is, N2O4 2 NO2
Let a mole of N2O4 is taken initially and x moles of N2O4 is dissociated at equilibrium then mole number of N2O4 and NO2 at equilibrium is (a - x) and 2x.
Total moles number at equilibrium = (a -x + 2x) = (a + x).
Equilibrium partial pressure of the components is {(a - x)/(a + x)}P and {2x/(a + x)}P where P = Total pressure of the reacting system.
The expression of the equilibrium constant,
Kp = (PNO2)2/PN2O4
= {2x/(a + x)}P/{(a - x)/(a + x)}P
Kp = (4x2P)/(a2 - x2)
The fraction of the original N2O4 dissociated at equilibrium É‘ = x/a.
Replacing (x/a) by É‘, we have,
 Kp = (4É‘2P)/(1 - É‘2)
Mole Fraction Equilibrium Constant:
If we write the equation as,
Æ”1A1 + Æ”2A2 Æ”3A3 + Æ”4A4
Where Æ”1, Æ”2, Æ”3 and Æ”4 are stoichiometric coefficient.
 ∴ Kx = (x3Æ”3 × x4Æ”4)/(x1Æ”1 × x2Æ”2)
Where Kx is called Mole fraction Equilibrium Constant of the reaction and x1x2x3 and x4 are the equilibrium mole fraction of reacting components.
 Equilibrium Constant.

Relation Between Kp and Kc:

The interrelations of these Equilibrium constants are as follows,
Kp = (P3Æ”3 × P4Æ”4)/(P1Æ”1 × P2Æ”2)
The ideal gas Equation, PV = nRT, may be written as,
P = (n/V)RT = CRT
Where C is the concentration of gas expressed as amount per unit volume.
= {(C3RT)Æ”3 × (C4RT)Æ”4}/{(C1RT)Æ”1 × (C2RT)Æ”2}
 ∴ Kp = Kc(RT)Î”Æ” Î”Æ” = (Æ”3 + Æ”4) - (Æ”1 + Æ”2)
1. For the reaction in which total number of reactant molecules and of resultant molecules are same:
2. H2 (g) + I2 (g) 2HI
KP = (PHI)2/{(PH2)(PI2)}
= (CHIRT)2/(CH2RT) (CI2RT)
= [(CHI)2/{(CH2) (CI2)}] × [(RT)2/(RT)(RT)]
= Kc
Thus when (Æ”3 + Æ”4) = (Æ”1 + Æ”2),
Kp = Kc
3. For the reactions in which the number of molecules of reactants differ from that of the resultant:
4. 2SO2(g) + O2(g) SO3(g)
Here, Kp = Kc×(RT){1 - (2+1)}
= Kc×(RT)-2
Thus when, (Æ”3 + Æ”4) (Æ”1 + Æ”2),
Kp  Kc
 Reaction (Gaseous System) Relation between Kp and Kc H2 + Cl2 ⇆ 2HCl Kp = Kc CO + H2O ⇆  2H2 + CO2 Kp = Kc CO + NO2 ⇆ NO + CO2 Kp = Kc PCl5 ⇆ PCl3 + Cl2 Kp = Kc RT 2H2 + O2 ⇆ 2H2O Kp = Kc RT -1 2CO + O2 ⇆ 2CO2 Kp = Kc RT -1 N2 + 3H2 ⇆ 2NH3 Kp = Kc RT -2
Calculate the Kc value of the reaction N2 + 3H2 ⇆ 2NH3 at 4000C, Given Kp at the same temperature 1.64 × 10-4.
Kp = 0.5.
For Solution see,
At 1000C the vapour density of N2O4 is 25 at 1 atm. Show that Kp = 9.6.
N2O4 (g) 2NO2 (g)
Let 1 moles of N2O4 is taken initially (t = 0) and x mole of N2O4 has reacted at equilibrium.
So the mole number of each component is (1-x) and 2x and total moles at equilibrium,
(1-x+2x) = (1+x).
So total moles has increased from 1 to (1+x).
Let Volume is increases from V1 to V2
So, (1+x) = V2/V1
As density and hence vapour density is inversely proportional to volume so vapour density will decrease from d1 to d2.
Hence, (1+x) = V2/V1= d1/d2.
Molecular weight of N2O4 is 92 and vapour density,
= 92/2
= 46
Due to dissociation it is = 25.
∴ 1+x = 46/25
or, x = 0.84
Now partial pressure are,
PNO2 = {2x/(1+x)}P
= (2×0.84)/1.84
= 0.913 atm
PN2O4 = {(1-x)/(1+x)}P
= 0.16/1.84
= 0.087
Kp = (PNO2)2/PN2O4
= (0.913)2/0.087
≃ 9.6

Relation Between Kp and Kx:

 ∴ Kp = Kx(P)Î”Æ” Î”Æ” = (Æ”3 + Æ”4) - (Æ”1 + Æ”2)
Show that the equilibrium condition for any chemical reaction is given by Î”G = 0.
Vant Hoff reaction isotherm is Î”G = - RT lnKa + RT lnQa
But when the reaction attain equilibrium, Qa = Ka
Thus, Î”G = 0.

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