# Chapter 6 Phase Equilibria¶

## Example 6_1 pgno:385¶

In :
MW=249.6;#molecular weight of CuSO4.5H2O in grams#
w=0.3120;#weight of CuSO4.5H2O in grams#
V=0.25;#volume of the solution in litres#
print'From Equation (a) 2 mol of CuSO4.5H2O liberates 1 mol of I2,i.e. 2 equivalents.\nHence the equlivalent weight of CuSO4.5H2O=mol.wt/1.'
print'\nFrom equation (b) the equivalent weight of CuSO4.5H2O is mol.wt/2 since 1mol of CuSO4.5H2O reacts with 2 mol of OH-,i.e 2 equivalents.'
W=w/V;#weight of CuSO4.5H2O in one litre solution in grams#
print'\nWeight of CuSO4.5H2O in a litre of the solution=W=grams.',W
Na=W/MW;#Normality of the solution for (a)#
print'\nNormality of the solution for (a)=Na=',Na
Nb=W*2/MW;#Normality of the solution for (b)#
print'\nNormality of the solution for (b)=Nb=',Nb
print'\nIn the first case 1ml of the solution contains 5*10^-3equivalents or 5 equivalents of CuSO4.5H2O,\nand in the second case 1ml of the solution will contain 10m eq of CuSO4.5H2O.'

From Equation (a) 2 mol of CuSO4.5H2O liberates 1 mol of I2,i.e. 2 equivalents.
Hence the equlivalent weight of CuSO4.5H2O=mol.wt/1.

From equation (b) the equivalent weight of CuSO4.5H2O is mol.wt/2 since 1mol of CuSO4.5H2O reacts with 2 mol of OH-,i.e 2 equivalents.

Weight of CuSO4.5H2O in a litre of the solution=W=grams. 1.248

Normality of the solution for (a)=Na= 0.005

Normality of the solution for (b)=Nb= 0.01

In the first case 1ml of the solution contains 5*10^-3equivalents or 5 equivalents of CuSO4.5H2O,
and in the second case 1ml of the solution will contain 10m eq of CuSO4.5H2O.


## Example 6_2 pgno:386¶

In :
v=180.;#volume of conc. H2SO4 in ml#
n=6.61;#Normality of the solution#
N=1000*n/v;
print'The Noramality or Strength of the Conc. acid=N=N',N
print'\n1 eq.per litre=0.5mol per litre in the case of H2SO4 since the eq.wt=0.5 the mol.wt.'
print'\n 6.6N soln=6.61 eq per litre=3.305mol per litre.\n Strength of the diluted solution=3.305M'
SG=1.84;#super gravity of Conc. H2SO4#
w=SG*v;#weight of 180ml of conc. H2SO4 in grams#
print'\nWt of 180ml of conc.H2SO4=w=grams.',w
print'\nThis actually contains 6.61*49grams of H2SO4.\n percentage of H2SO4 by weight=97.8'
sg=1.198;#specific gravity of the diluted solution#
V=1000;#volume of the diluted solution in ml#
W=sg*V;#weight of one litre of the diluted solution in grams#
print'\nWt of 1 litre of the diluted solution=W=grams ',W
WH2O=w+W;#weight of water in grams#
print'\ntherefore Weight of water=WH2O=grams.',WH2O
print'\nIf the percent of H2SO4 by wt in the diluted solution is y.\nWt of H2SO4 in 1litre of the diluted solution=49*6.61grams.so y value comes as 27.04percent'
M=3.305*1000/WH2O;#molality of the solution#
print'\nMolality of the solution=M=',M
mf=0.064;#mole fraction of H2SO4#
mfH2O=1-mf;#mole fraction of water#
print'\nMol of sulphuric acid is 329.9/98=3.305.\nMol of water=874.1/18=48.561.\nMol fraction of H2SO4=0.064.'
print'\nMole fraction of water=mfH2O=',mfH2O

The Noramality or Strength of the Conc. acid=N=N 36.7222222222

1 eq.per litre=0.5mol per litre in the case of H2SO4 since the eq.wt=0.5 the mol.wt.

6.6N soln=6.61 eq per litre=3.305mol per litre.
Strength of the diluted solution=3.305M

Wt of 180ml of conc.H2SO4=w=grams. 331.2

This actually contains 6.61*49grams of H2SO4.
percentage of H2SO4 by weight=97.8

Wt of 1 litre of the diluted solution=W=grams  1198.0

therefore Weight of water=WH2O=grams. 1529.2

If the percent of H2SO4 by wt in the diluted solution is y.
Wt of H2SO4 in 1litre of the diluted solution=49*6.61grams.so y value comes as 27.04percent

Molality of the solution=M= 2.16126078996

Mol of sulphuric acid is 329.9/98=3.305.
Mol of water=874.1/18=48.561.
Mol fraction of H2SO4=0.064.

Mole fraction of water=mfH2O= 0.936


## Example 6_3 pgno:388¶

In :
N2=79.2;#percentage of Nitrogen in air#
O2=20.8;#percentage of Oxygen in air#
b=76.93;#Weight percent of N2 in air#
print'Weight percent of N2 in air=b=',b
a=100-b;#weight percent of O2 in air#
print'\nWeight percent of O2 in air=a=',a

Weight percent of N2 in air=b= 76.93

Weight percent of O2 in air=a= 23.07


## Example 6_4 pgno:390¶

In :
N2=0.79;#partial pressure of Nitrogen in air#
O2=0.21;#partial pressure of Oxygen in air#
AN2=0.015;#Absorption coefficient of N2#
AO2=0.028;#Absorption coefficient of O2#
l=22.4;
print'Absorption coefficient being the solubility of the gas at partial pressure of 1atm of the gas,\nThe solubilities in mol per litre of the two gases are'
SN2=N2*AN2/l;#solubility of N2#
SO2=O2*AO2/l;#solubility of O2#
print'\nSolubility of N2=SN2==5.29*10^-4mol per litre.',SN2
print'\nSolubility of O2=SO2==2.625*10^-4mol per litre',SO2
VO2=(SO2*100)/(SN2+SO2);
print'\nThe mole or Volume percent of O2=',round(VO2,2)
VN2=100-VO2;
print'\nThe mole or volume percent of N2=',round(VN2,2)

Absorption coefficient being the solubility of the gas at partial pressure of 1atm of the gas,
The solubilities in mol per litre of the two gases are

Solubility of N2=SN2==5.29*10^-4mol per litre. 0.000529017857143

Solubility of O2=SO2==2.625*10^-4mol per litre 0.0002625

The mole or Volume percent of O2= 33.16

The mole or volume percent of N2= 66.84


## Example 6_5 pgno:393¶

In :
print'Upon solving the equations PA=0.9atm,PB=0.3atm'
PA=0.9;#vapour pressure of A#
PB=0.3;#Vapour pressure of B#
xA=0.33;
xB=0.66;
yA=(xA*PA)/(xA*PA+xB*PB);
print'\nComposition of Vapour A in the mixture=yA=',yA
yB=1-yA;
print'\nComposition of Vapour B in the mixture=yB=',yB
VP=yA*PA+yB*PB;#total vapour pressure of the mixture#
print'\nTotal vapour pressure of the mixture=VP=',VP

Upon solving the equations PA=0.9atm,PB=0.3atm

Composition of Vapour A in the mixture=yA= 0.6

Composition of Vapour B in the mixture=yB= 0.4

Total vapour pressure of the mixture=VP= 0.66


## Example 6_6 pgno:393¶

In :
yA=0.60;
xA1=0.40;
xA=0.5*yA+0.5*xA1;
print'Let PA and PB represent the vapour pressures of pure A and pure B respectively.'
print'\nFrom 1 mol of solution after distillation,we get 0.5mol of distillate and 0.5mol of residue.'
print'\nVapour pressure of substance A=PA=900.00000mm of Hg'
print'\nVapour pressure of substance B=PB=400.00000mm of Hg'

Let PA and PB represent the vapour pressures of pure A and pure B respectively.

From 1 mol of solution after distillation,we get 0.5mol of distillate and 0.5mol of residue.

Vapour pressure of substance A=PA=900.00000mm of Hg

Vapour pressure of substance B=PB=400.00000mm of Hg


## Example 6_7 pgno:402¶

In :
wA=162.;
wB=100.;
VPB=641.;#vapour pressure of water#
VPA=119.;#vapour pressure of oraganic substance#
MB=18.;#Molecular weight of H2O#
print'Even though the boiling part of A might be higher,it distills out at a low temperature 95.3degrees.'
print'\nIf A were to distill at 95.3degrees,the distillation will have to be carried out at a reduced pressure of about 119mm of mercury'
MA=(wA*MB*VPB)/(wB*VPA);
print'\nMolecular weight of A=MA=grams',round(MA)

Even though the boiling part of A might be higher,it distills out at a low temperature 95.3degrees.

If A were to distill at 95.3degrees,the distillation will have to be carried out at a reduced pressure of about 119mm of mercury

Molecular weight of A=MA=grams 157.071932773


## Example 6_8 pgno:404¶

In :
w=50.;#weight of acid A in grams#
x=1.;
y=0.2;
K=5.;
n=5.;
wn=w*(x/(x+K*y))**n;
print'wn=grams',round(wn,3)
y1=1;
w0=w*(x/(x+K*y1));
print'\nw0=grams',round(w0,3)
print'\nIt is seen that the first process leaves only 1.563grams of A with the aq. layer,\nwhereas the secondone using all available solvent in a single lot leaves 8.333grams in aqueous layer.\nIn the process (a)96.88percent of A is extracted,whereas in (b) only 83.67percent A is extracted.'

wn=grams 1.563

w0=grams 8.333

It is seen that the first process leaves only 1.563grams of A with the aq. layer,
whereas the secondone using all available solvent in a single lot leaves 8.333grams in aqueous layer.
In the process (a)96.88percent of A is extracted,whereas in (b) only 83.67percent A is extracted.


## Example 6_9 pgno:405¶

In :
AN=0.096;#normality of H2SO4 in aqua layer#
ON=0.014;#normality of H2SO4 in org. layer#
AV=13.3;#amount of H2SO4 required in aq. layer for neutralization#
OV=7.15;#amount of H2SO4 requred in org. layer for neutralization#
AS=AN*AV/10;#strength of NH3 in aq. layer#
print'Strength of NH3 in aq. layer=AS=N.',AS
OS=ON*OV/20;#strength of NH3 in org. layer#
print'\nStrength of NH3 in org. layer=OS=N.',OS
K=AS/OS;#equilibrium constant#
print'\nEquilibrium constant=K=',K
AV1=20.0;#amount of H2SO4 required in aq. layer at equilibrium#
OV1=8.0;#amount of H2SO4 required in org. layer at equilibrium#
AN1=AV1*AN/5;#Normality of NH3 in aq. layer#
print'\nNormality of NH3 in aq. layer=AN1=N.',AN1
ON1=OV1*ON/10;#Normality of NH3 in org. layer#
print'\nNormality of NH3 in org. layer=ON1=N.',ON1
print'\nIn the aq.layer NH3 includes the free ammonia(uncombined).\nNH3t and that which has combined with Cu2+ to form the complex ion NH3.'
print'\nNH3aq=NH3t+NH3combined.\nThe value of NH3t can be obtained from the value of K.\nK=25.49=NH3t/NH3combined.'
NH3t=K*ON1;
print'\nSince nernsts law holds good for the same species present in both phases.\nNH3t==0.2855N.',NH3t
NH3c=AN1-NH3t
print'\nNH3c=0.0985N',NH3c
print'\n0.025mol per litre of Cu2+ combines with 0.0985mol per litre of NH3.\n1 mol per litre of Cu2+ combines with 0.0985/0.025=3.936mol per litre of NH3.'
print'\nor 1mol of Cu2+ combines with 4mol of NH3,i.e the value of x is 4.\nThe formula of the complex ion is thus (Cu(NH3)4)2+'

Strength of NH3 in aq. layer=AS=N. 0.12768

Strength of NH3 in org. layer=OS=N. 0.005005

Equilibrium constant=K= 25.5104895105

Normality of NH3 in aq. layer=AN1=N. 0.384

Normality of NH3 in org. layer=ON1=N. 0.0112

In the aq.layer NH3 includes the free ammonia(uncombined).
NH3t and that which has combined with Cu2+ to form the complex ion NH3.

NH3aq=NH3t+NH3combined.
The value of NH3t can be obtained from the value of K.
K=25.49=NH3t/NH3combined.

Since nernsts law holds good for the same species present in both phases.
NH3t==0.2855N. 0.285717482517

NH3c=0.0985N 0.0982825174825

0.025mol per litre of Cu2+ combines with 0.0985mol per litre of NH3.
1 mol per litre of Cu2+ combines with 0.0985/0.025=3.936mol per litre of NH3.

or 1mol of Cu2+ combines with 4mol of NH3,i.e the value of x is 4.
The formula of the complex ion is thus (Cu(NH3)4)2+