simple mixtures thermodynamic description of mixtures aryo abyoga a (080358395) gerald mayo l...
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SIMPLE MIXTURES
THERMODYNAMIC DESCRIPTION OF MIXTURES
ARYO ABYOGA A (080358395)
GERALD MAYO L (0806472212)
LEONARD AGUSTINUS J (0806472225)
Simple Mixtures Often in chemistry, we encounter mixtures
of substances that can react together. Chapter 7 deals with reactions, but let’s
first deal with properties of mixtures that don’t react.
We shall mainly consider binary mixtures – mixtures of two components.
xA xB 1
Dalton’s Law
The total pressure is the sum of all the partial pressure.
We already used mole fraction to descrice the partial pressure of mixtures of gases which refers to a total pressure
p j x j p
pA pB (xA xB )pp
The partial molar volume is the contribution that one component in amixture makes to the total volume of a sample
H2O EtOHAdd 1.0 mol H2O Add 1.0 mol H2O
Volume increasesby 18 cm3 mol-1Volume increasesby 14 cm3 mol-1
Molar volume of H2O:18 cm3 mol-1
Partial molar volume ofH2O in EtOH: 14 cm3 mol-1
The different increase in total volume in the H2O/EtOH example dependson the identity of the molecules that surround the H2O. The EtOHmolecules pack around the water molecules, increasing the volume byonly 14 cm3 mol-1Partial molar volume of substance A in a mixture is the change in volumeper mole of A added to the large volume of the mixture
partial molar volume
Partial Molar Volumes Imagine a huge volume of pure water at
25 °C. If we add 1 mol H2O, the volume increases 18 cm3 (or 18 mL).
So, 18 cm3 mol-1 is the molar volume of pure water.
Partial Molar Volumes Now imagine a huge volume of pure
ethanol and add 1 mol of pure H2O it. How much does the total volume increase by?
Partial Molar Volumes When 1 mol H2O is added to a large
volume of pure ethanol, the total volume only increases by ~ 14 cm3.
The packing of water in pure water ethanol (i.e. the result of H-bonding interactions), results in only an increase of 14 cm3.
Partial Molar Volumes The quantity 14 cm3 mol-1 is the partial
molar volume of water in pure ethanol. The partial molar volumes of the
components of a mixture varies with composition as the molecular interactions varies as the composition changes from pure A to pure B.
VJ VnJ
p,T ,n'
The partial molar volume of components of a mixture vary as themixture goes from pure A to pure B - that is because the molecularenvironments of each molecule change (i.e., packing, solvation, etc.)Partial molar volumes of a water-ethanolbinary mixture are shownat 25 oC across all possibleCompositions. The Partial molar volume, Vj, of a substance j define as :
The partial molar volume is the slope of a plot of total volume as theamount of J in the sample is changed (volume vs. composition)Partial molar volumes vary
with composition (different slopes atcompositions a and b) - partial molar volume at b is negative (i.e., theoverall sample volume decreases as A is added)
dV VnA
p,T ,nB
dnA VnB
p,T ,nA
dnB
When a mixture is changed by When a mixture is changed by dndnAA of of A and A and dndnBB of B, then the total volume of B, then the total volume changes by: changes by:
Partial Molar Volumes
If partial molar volumes are known for the twocomponents, then at some temperature T, thetotal volume V (state function, always positive)of the mixture is
dV VnA
p,T ,nB
dnA VnB
p,T ,nA
dnB
dV VAdnA VBdnB
V VAdnA0
nA VBdnB0
nB
Partial Molar Volumes
V VAdnA0
nA VBdnB0
nBV VA dnA0
nA VB dnB0
nBV VAnA VBnB
Partial Molar Volumes
V (NaCl, aq) = volume of sodium chloride solution
V_
(NaCl, aq) = partial molar volume of sodium chloride
in water
Partial Molar Volumes
Partial Molar Volumes How to measure partial molar volumes? Measure dependence of the volume on
composition. Fit a function to data and determine the
slope by differentiation.
Partial Molar Volumes Ethanol is added to 1.000 kg of water. The total volume, as measured by
experiment, fits the following equation:
V 1002.93 54.6664x 0.36394x 2 0.028256x 3
x nE
Partial Molar Volumes
VE VnE
p,T ,nw
Vx
p,T ,nw
dV
dx54.6664 (2)0.36394x (3)0.028256x 2
Partial Molar Volumes Molar volumes are always positive, but
partial molar quantities need not be. The limiting partial molar volume of MgSO4 in water is -1.4 cm3mol-1, which means that the addition of 1 mol of MgSO4 to a large volume of water results in a decrease in volume of 1.4 cm3.
Partial Molar Gibbs energies
The concept of partial molar quantities can be extended to any extensive state function.
For a substance in a mixture, the chemical potential is defined as the partial molar Gibbs energy.
J GnJ
p,T ,n'
Partial Molar Gibbs energies
For a pure substance:
GnJGJ ,m
J GnJ
p,T ,n'
nJGJ ,m
nJ
p,T ,n '
GJ ,m
Partial Molar Gibbs energies Using the same arguments for the derivation of partial molar volumes,
Assumption: Constant pressure and temperature
GnAA nBB
Partial Molar Gibbs energies
Fundamental equation of chemical thermodynamics
dGVdp SdT AdnA BdnB
dGAdnA BdnB (at constant p and T)
dwadd,max AdnA BdnB (at constant p and T)
Chemical Potential
GH TS U pV TSU pV TS GdU pdV Vdp SdT TdS dGdU pdV Vdp SdT TdS (Vdp SdT AdnA BdnB )
dU pdV TdS (AdnA BdnB )
dU AdnA BdnB (at constant S and V)
J UnJ
S,V ,n '
Chemical Potential
J HnJ
S,p,n'
J AnJ
V ,T ,n'
Gibbs-Duhem equation
GnAA nBBdGnAdA nBdB AdnA BdnB
dGAdnA BdnB (at constant p and T)
nAdA nBdB 0
nJdJ 0J
Gibbs-Duhem equation
nAdA nBdB 0
dB nAnBdA
Molarity and Molality
Molarity, c, is the amount of solute divided by the volume of solution. Units of mol dm-3 or mol L-1.
Molality, b, is the amount of solute divided by the mass of solvent. Units of mol kg-1.
Using Gibbs-Duhem
The experimental values of partial molar volume of K2SO4(aq) at 298 K are found to fit the expression:
B 32.280 18.216x1 2
B VK 2SO4
x molality of K2SO4
Using Gibbs-Duhem
nAdVA nBdVB 0
dA nBnAdB
dA A*
A nBnAdB
A A* nB
nAdB
A A*
nBnAdB
Using Gibbs-Duhem
A A*
nBnAdB
B 32.280 18.216x1 2
dB
dx9.108x 1 2
dB 9.108x 1 2dx
Using Gibbs-Duhem
A A*
nBnAdB
dB 9.108x 1 2dx
A A* nB
nA9.108x 1 2dx A
* 9.108nBnAx 1 2dx
Using Gibbs-Duhem
A A*
nBnA
9.108x 1 2dx A* 9.108
nBnAx 1 2dx
nBnA
nB1 kg MA
nBMA
1 kgxMA
A A* 9.108 xMA x
1 2dx A* 9.108MA x1 2dx
A A* 9.108MA x1 2dx
0
bBA A
* 2 3 9.108MAbB3 2
Using Gibbs-Duhem
A A* 2 3 9.108MAbB
3 2
A* 18.079 cm3 mol 1
MA 0.01802 kg mol-1
A 18.079 0.1094bB3 2
Thermodynamics of mixing
So we’ve seen how Gibbs energy of a mixture depends on composition.
We know at constant temperature and pressure systems tend towards lower Gibbs energy.
When we combine two ideal gases they mix spontaneously, so it must correspond to a decrease in G.
Thermodynamics of mixing
Thermodynamics of mixing
Gm Gm RT ln
p
p
RT lnp
p
RT ln p
Thermodynamics of mixing
RT ln p
Gi nA A RT ln p nB B
RT ln p G f nA A
RT ln pA nB B RT ln pB
mixGnART lnpAp
nBRT lnpBp
Thermodynamics of mixing
mixGnART lnpAp
nBRT lnpBp
pAp
xApAp
xB
mixGnART ln xA nBRT ln xBxAn nA xBn nBmixGnRT(xA ln xA xB ln xB )
mixG 0
Thermodynamics of mixing
Gibbs energy of mixing
A container is divided into two equal compartments. One contains 3.0 mol H2(g) at 25 °C; the other contains 1.0 mol N2(g) at 25 °C. Calculate the Gibbs energy of mixing when the partition is removed.
Gibbs energy of mixing
Two processes: 1) Mixing2) Changing pressures of the gases.
Gi 3.0 H2
RT ln 3p 1.0 N2
RT ln p G f 3.0 H2
RT ln 3 2 p 1.0 N2
RT ln1 2 p mixG3.0(RT ln
32 p
3p) 1.0(RT ln
12 p
p)
mixG3.0(RT ln 12) 1.0(RT ln 1
2)
mixG 6.9 kJ
Gibbs energy of mixing
mixGnRT(xA ln xA xB ln xB )
mixG3.0(RT ln34
) 1.0(RT ln14
)
mixG 2.14 kJ 3.43 kJ
mixG 5.6 kJ
p
p
Other mixing functions
GT
p,n
S
mixS mixG
T
T
nRT(xA ln xA xB ln xB ) nR(xA ln xA xB ln xB )
mixS nR(xA ln xA xB ln xB )
Other mixing functions
mixGnRT(xA ln xA xB ln xB )
mixS nR(xA ln xA xB ln xB )
GH TSnRT(xA ln xA xB ln xB ) H T( nR(xA ln xA xB ln xB )
nRT(xA ln xA xB ln xB ) H nRT(xA ln xA xB ln xB )
H 0 (constant p and T)