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Showing posts with label PHYSICAL CHEMISTRY. Show all posts
Showing posts with label PHYSICAL CHEMISTRY. Show all posts

Jan 1, 2019

Van der Waals Equation

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 σ = 2r is the diameter using the molecule as a rigid sphere.
How can we derived Ven der Waals Equation?
Representation of 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) πr3
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 the effective volume for a single molecule,
= 1/2 × 4/3 π σ3
The effective volume for Avogadro number of molecules (present in 1 mole gas),
How can we derived Ven der Waals Equation?
The effective volume for Avogadro number of molecules
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,
what is the Van der Waals equation?
Van der Waals equation for n Moles Real Gas

Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 27°C. 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/V²ㄑPi 
Since, Pi = nRT/V only.
 Thus, inter molecular attraction effect reduces the pressure of the real gases.

Units of 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 m⁻² m⁶ mol⁻² 
N m⁴ mol⁻²
In CGS system,unit of 'a' 
= dyne cm⁻² cm⁶ mol⁻² 
= dyne cm⁴ mol⁻²
In SI system,unit of 'b' = m³ mol⁻¹
In CGS system,unit of 'b' = cm³mol⁻¹

Significance of 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(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.

Calculation of Boyle Temperature (TB):

Mathematical condition for Boyle’s temperature is,
TB = [d(PV)/dP]T When P0
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, T= (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.

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
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:

We have Amagat Curves,
How can we Explain Amagat Curves By Ven der Walls Equation?
(a) Amagat Curve For Different Gases
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 CO₂a’ is very high since the gas is easily liquefied. 
What is the Ven der Waals Equation?
(b) Amagat Curve For CO₂
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.
For H₂a’ 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.
(b) (i) When T〈 Tв that is 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 = Tв = 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 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, that is T 〉a/Rb or, b 〉a/RT and {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/RTP are negligible and Z=1 that is the gas obeys ideal behavior.

Dec 23, 2018

Van't Hoff Equation

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

Equilibrium Constant KP of a reaction is constant at a given temperature but when temperature is changed, the value of equilibrium constant also changed. The quantitative relation, relation, Known as Van't Hoff equation is derived by using Gibbs - Helmholtz equation.
The Gibbs - Helmholtz equation is 

ΔG0 = ΔH0 + T[d(ΔG0)/dT]P 

Zero superscript is indicating the stranded values.
or, - (ΔH0/T2 ) = -(ΔG0/T2 ) + 1/T[d(ΔG0)/dT]P 
or, - (ΔH0/T2 )  [d/dTG0/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/dTG0/T)]P
Comparing the above two equation we have,
(dlnKP/dT) = (ΔH0/T2)
This is the differential form of Van't Hoff reaction equation.
Greater the value of ΔH0, the faster the equilibrium constant (KP) changes with temperature (T).
Separating the variables and integrating,
dlnKP  = (ΔH0/R) (dT/T2)
(Assuming that ΔH0 is independent of temperature)
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/

For a reaction 2A + B ⇆ 2C, ΔG0(500K) = 2 KJ mol-1 
Find the KP at 500K for the reaction A + ½B  C .

ΔG0(500K) for the reaction A + ½B  C  is 2 KJ mol-1/2 = 1 KJ mol-1/2
The relation is ΔG0 = - RT lnKP 
or, KJ mol-1  = - 8.31 × 10-3 KJ mol-1 K-1 × 500K lnKP 
or, lnKP  = 1/(8.31 × 0.5) = 0.2406

∴ KP = 1.27

Plot of lnKP vs 1/T :

For exothermic reaction, ΔH0 = (-) ve.
Examples are the formation of ammonia from Hand N2.
Van't Hoff Equation - Effect on temperature on Equilibrium Constant
Plot of lnKP vs 1/T
For endothermic reaction, ΔH0 = (+) ve.
Examples are the dissociation of HI into Hand I2.
For the reaction, ΔH0 = 0, lnKP is independent of T. Provided ΔS0 does not change much with T.
Since for idel system, H is not a function of P and ΔH0 = ΔH
and Van't Hoff equation is dlnKP/dT = ΔH/RTand 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 determination of KP1 and KP2 at two temperature helps to calculate the value of ΔH of the reaction.
The above reaction is called Van't Hoff reaction isobar since P remains constant during the change of temperature.

Use Gibbs - Helmholtz equation to derive the Vant Hoff reaction isochore. Under What condition do you expect a linear relationship between logk and 1/T ?

Again KP = KC (RT)ΔƔ
or, lnKP = lnKC + ΔƔ lnR + ΔƔlnT
Differentiating with respect to T,
dlnKP/dT = dlnKC/dT + ΔƔ/T
But dlnKP/dT = ΔH0/RT2
Hence, dlnKC/dT = ΔH0/RT2 - ΔƔ/T = (ΔH0 - ΔƔRT)/RT2
or, dlnKC/dT = ΔU0/RT2
ΔU0 is standerd 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:
(i) The reacting system is assumed to behave ideally and 
(ii) ΔH is taken independent of temperature for small range of temperature change.
Due to assumption involved ΔH and ΔU do not produce precise value of these reactions.
As ΔG is independent of pressure for ideal system a and b also independent of pressure.
Hence ΔG0 = - RT lnKP 
or, [dlnKP/dT]T = 0

Show that the equilibrium condition for any chemical reaction is given by ΔG = 0.

Vant Hoff reaction isotherm is ΔG = - RT lnKa + RT lnQa
But when the reaction attain equilibrium Qa = Ka
Thus, ΔG = 0.

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

Van't Hoff equations gives quantitative expression of Le-Chatelier Principle. From the equation,

lnKP = -(ΔH/R)(1/T) + C

It is evident that for endothermic reaction (ΔH〉0), increase of T increases the value of a of the reaction. But for exothermic reaction (ΔHㄑ 0), with rise 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 is always react 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.