1. Sorption Properties of Food
Water Activity and
Sorption Properties of Food

2. CRITERIA OF EQUILIBRIUM
The criterion for thermal equilibrium is equality of temperatures while the criterion for mechanical equilibrium is equality of pressures.
Physicochemical equilibrium is characterized by equality of chemical potential (μ) of each component.
The chemical potential determines whether a substance will undergo a chemical reaction or diffuse from one part of the system to another.
The chemical potential of a component in the liquid (L) phase is equal to that in the vapor (V) phase if vapor and liquid are in equilibrium:
Chemical potential is the partial molar free energy and can be expressed as:

3. where G is Gibbs free energy and ni is the moles of component i
where G is Gibbs free energy and ni is the moles of component i . The definition shows that the chemical potential of a component of a homogenous mixture is equal to the ratio of the increase in Gibbs free energy on the addition of an infinitesimal amount of the substance.
Gibbs free energy is defined as a combination of enthalpy (H), temperature (T ), and entropy (S):
G = H − TS
Enthalpy can be expressed in terms of internal energy (U), pressure (P), and volume (V):
H = U + PV
For a reversible process in a closed system of constant composition, the first and second laws of thermodynamics may be combined to yield:
dU = TdS − PdV

4. IDEAL SOLUTION—RAOULT’S LAW
A solution can be defined as ideal if the cohesive forces inside a solution are uniform.
This means that in the presence of two components A and B, the forces between A and B, A and A, and B and B are all the same.
Equation can be written in terms of partial molar quantities and since the partial molar free energy is the chemical potential, for compound A in a solution: -
where VA is the molar volume of component A in solution which is the volume divided by the number of moles of A.

5. Using ideal gas law and , μA can be related to the partial vapor pressure by:
If μ0A is the value of chemical potential when the pressure is 1 atm,

6. Partial vapor pressure of a component, which is a measure of tendency of the given component to escape from solution into the vapor phase, is an important property for solutions. For a solution in equilibrium with its vapor:
Thus, chemical potential of component A in solution is related to the partial vapor pressure of A above the solution. Equation above is true only when vapor behaves as an ideal gas.
A solution is ideal if the escaping tendency of each component is proportional to the mole fraction of the component in the solution.

7. The escaping tendency of component A from an ideal solution, as measured by its partial vapor pressure, is proportional to the vapor pressure of pure liquid A and mole fraction of A molecules in the solution. This can be expressed by Raoult’s law as:
where PA is the partial vapor pressure of A, XA is its mole fraction, and P0A is the vapor pressure of pure liquid A at the same temperature.
If component B is added to pure A, vapor pressure is decreased as:

8. HENRY’S LAW
Consider a solution containing solute B in solvent A. If the solution is very dilute, a condition is reached in which each molecule B is completely surrounded by component A.
Solute B is then in a uniform environment irrespective of the fact that A and B may form solutions that are not ideal at higher concentrations.
In such a case, the escaping tendency of B from its environment is proportional to its mole fraction, which can be expressed by Henry’s law as:
PB = kXB
where k is the Henry’s law constant. Henry’s law is only valid for fairly and extremely dilute solutions.

9. COLLIGATIVE PROPERTIES
Colligative properties are properties of solutions that depend upon the ratio of the number of solute particles to the number of solvent molecules in a solution.
Vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure are the colligative properties.
These properties are used to determine the molecular weights and to measure water activity. The properties include:
Boiling Point Elevation
Freezing Point Depression
Osmotic Pressure

10. A) Boiling Point Elevation
If a small amount of nonvolatile solute is dissolved in a volatile solvent and the solution is very dilute to behave ideally, the lowered vapor pressure can be calculated from equation.
Boiling point elevation (T – T0) can be expressed as ∆TB. When boiling point elevation is not too large, the product of T times T0 can be replaced by T02 .
Considering the mole fraction of solute as XB, the mole fraction of solvent is XA = 1 − XB
ln (1 – XB) can be expanded in a power series as:
For dilute solutions where XB is a very small fraction, the equation can be written as:
Where 
In which wB and wA are the masses of solute and solvent, respectively and MB and MA are the molecular weights of the solute and solvent, respectively.

11. Therefore,
Where ∆TB is boiling point elevation (T-T0), lv is the latent heat of vaporization per unit mass
The term (wB/wAMB) is expressed in terms of molality as (m/1000). Then:
where KB is the molal boiling point elevation constant.

12. This equation is valid for all non-electrolytes in which equal numbers of moles are involved before and after the solution is formed. For electrolytes, the equation can be expressed as:
where the i factor is the ratio of number of total moles after the solution to the number of moles before the solution. For non-electrolytes i = 1.
To predict boiling point rise due to the solutes in the solution, an empirical rule known as Dühring’s rule can be used. In this rule, a straight line is obtained if the boiling point of solution is plotted against the boiling point of pure water at the same pressure for a given concentration. For each concentration and pressure a different line is obtained.

13. Exercise 1
(a) Determine the boiling temperature of 10% (w/w) NaCl solution under atmospheric pressure.
Check the result also from the Dühring plot.
(b) By using the Dühring plot, estimate the boiling point of the same solution under a pressure of kPa. What is the boiling point elevation at this pressure if at kPa water boils at 80◦C?
Data: Molecular weight of NaCl: 58.4 g/g-mole
Enthalpy of saturated vapor: kJ/kg at 100◦C
Enthalpy of saturated liquid: kJ/kg at 100◦C
R, gas constant: kJ/kg-mole K.

14. Dühring chart for aqueous solutions of NaCl

15. Solution
(a) For electrolytes such as NaCl, KB can be calculated using
Molality is the number of moles of solute in 1000 g of solvent. Molality of 10% sodium chloride solution can be calculated by using its molecular weight:

16. The i factor for NaCl is 2. Then,
△TB = i KBm = (0.513)(1.9)(2) = 1.95◦C
TB = = ◦C
From Dühring chart, the boiling point of the solution is read as 103◦C
(b) At kPa water boils at 80◦C. From the Dühring plot, the boiling point of 10% NaCl solution for the pressure yielding boiling point of water at 80◦C can be read as 83◦C. Therefore, the
boiling point elevation is:
∆TB = 3◦C

17. (B) Freezing Point Depression
When equilibrium exists between solid A and its solution, the chemical potentials of A must be the same in both phases:
μSA = μLA
For small freezing point depression:
where ∆T f is the freezing point depression.
Then, the equation can be written as:
where

18. In freezing point depression, the term (wB/wAMB) is expressed in terms of molality as (m/1000).
Then equation becomes:
where Kf is the molal freezing point depression constant.
Freezing point depression is useful for determining the molar mass of the solutes. These relations apply to ideal solutions. For electrolytes,
where i is the ratio of number of total moles after the solution to the number of moles before the solution. For non-electrolytes i = 1.

19. Exercise 2
In a city, the administrative board of the municipality is discussing using salt (NaCl) or glycerol (C3H8O3) for road treatment to eliminate ice. Assuming that you are one of the engineers
taking part in the discussion, help them to decide which is the more effective type of antifreeze.
Compare the performances of both types of solutions at the same concentration such as 10% (w/w).
Data: Latent heat of fusion for ice at 0◦C: kJ/kg-mole
Molecular weight of glycerol: 92 kg/kg-mole
Molecular weight of water: 18 kg/kg-mole
The gas constant, R: kJ/kg-mole·K

20. Solution
Freezing point depression is calculated as
For NaCl solution, molality of 10% sodium chloride solution is 1.9 m,
The i factor for NaCl solution is 2. Then, freezing point depression is:
∆Tf,NaCl = (1.85)(1.9)(2) = 7.03◦C

21. For a 10% glycerol solution molality is calculated as:
The i factor for glycerol is 1.
∆Tf,gly = (1.85)(1.21)(1) = 2.24◦C
Since ∆Tf,NaCl is greater than ∆Tf,gly for the same concentration, NaCl is a more effective antifreeze.

22. C) Osmotic Pressure
Osmotic pressure is a colligative property that is closely related to the vapor pressure, freezing point, and boiling point.
The activity of the solvent affects mostly osmotic pressure.
An osmotic pressure arises when two solutions of different concentrations are separated by a semipermeable membrane.
Figure below shows the osmometer showing conditions at equilibrium.

23. The solution and solvent are separated by a semipermeable membrane that allows only passage of solvent. Osmotic flow continues until the chemical potential of the diffusing component is the same on both sides of the barrier.
The equilibrium occurs in an osmometer in which the extra pressure π necessary for equilibrium is produced by the hydrostatic pressure of a water column on the solution side.
At equilibrium there will be no net flow of solvent across the membrane so the chemical potential of pure solvent at pressure P must be equal to the chemical potential of solvent in solution at pressure P + π.
μA(solution, P + π) = μA(solvent, P)
At constant temperature and at 1 atm pressure:
μA(solution, 1 + π) = μ0A(solvent)

24. The chemical potential of A at 1 atm is decreased owing to the presence of solute in solution and it can be calculated as:
μosmosis = μA(solution, 1) − μ0A(solvent)
After derivation from related equations:
where π is osmotic pressure (Pa), V is the volume of solution (m3), and nB is the number of moles of solute in solution.
In dilute solutions nA VA0 is equal to the volume of solution since the volume of the solute is negligible.
Therefore:
πV = nBRT
⇒ π = cRT
where R is the gas constant ( m3· Pa/kg-mole·K) and c is the concentration of solute (kg-mole/m3).