Thermodynamics of Solutions Partial Molar Properties

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1 MME3: Lecture 6 Thermodynamics of Solutions Partial Molar Properties A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka omposition of solutions Partial molar properties

2 Introduction Materials of practical applications are either: mixtures of metals (alloys), or mixture of compounds (slags, mattes, glasses, ceramics) When mixed, materials can exist either as separate mixtures of several components, or as solutions. During mixing, changes in chemical composition can occur in either of the following two ways:. Transfer of atoms or molecules across the boundary (during the formation of multi-component, homogeneous non-reacting systems, also known as the solutions). Occurrence of chemical reactions within the boundary (in the multi-component, reactive systems) A solution is a homogeneous phase composed of two or more chemical substances, whose concentration may be varied without the precipitation of a new phase. Formed as a result of physical mixing involving: intermolecular interactions in such a way that the molecules maintain their individual identity. Nearly all substances of metallurgical interest are solutions. The thermodynamic properties of a component in solution are significantly different from the properties of that component when it is pure. Some basic concepts in thermodynamics of solution include: concept of activity concept of free energy of mixing

3 Number of moles of component k (n k ) and mole fraction (X k ) X k = n k n T Weight percentage of component k (wt.% k) wt.% k = w k w T x 00 Molality of solute k in solution, m k n k m k = x 000 = w n k M 0 n 0 n k = gram moles of solute k w 0 = gram of solvent n 0 = gram moles of solvent M 0 = molecular weight of solvent Molarity of solute k in solution, c k c k = n k V V volume of solution in litres Use of Nomogram Example 6. The solubility of naphthalene ( 0 H 8 ) in benzene ( 6 H 6 ) at 0 is 34.3 wt.%. Find its mole fraction. Answer:. onnect the point corresponds to the molecular weight of naphthalene, M =8 (the first scale on the left) to the point corresponds to the molecular weight of benzene, M =78 (the third scale from the left).. The line gives the ratio M /M on the second scale. 3. Now connecting this point to the point 34.3 wt.% on the fifth scale (i.e., the rightmost scale) will give the desired mole fraction 4 mole % (or, X =0.4) on the fourth scale. 3

4 How can we express the molar properties of a component in solution? For single component systems, volume occupies by two moles of oxygen at 98 K and atm is exactly twice that occupies by one mole under the same conditions, i.e., V = V / n. What is the molar volume of each of the components present in solution? We can measure this by adding a very small quantity of a component in solution without significantly changing the overall composition of the solution and then converting the measured change in volume to one mole of the component being added. For a multi-component system, V is a function not only of T and P (as in one-component system) but also of the number of moles of each component in the system (n, n,.., n ). V = V (T, P, n, n,., n ) For an arbitrary infinitesimal change in state V V dv = dt + dp T P, n k P T, n k + V n dn + V dn +.. P, T, n, n 3,., n n P, T, n, n 3,., n V + dn n P, T, n, n,., n - 4

5 V V dv = dt + dp T P, n k P T, n k V + dn n k k P, T, nj n k The coefficient of each of the changes in number of moles V V k (k =,, 3,. ) n k P, T, nj n k These quantities are defined to be the partial molar volumes for each component in the system. It represents the rate of change of volume with respect to addition of substance k and is equal to the increase in volume resulting the addition of one mole of k to an infinite amount of all the components in the system without significantly altering the overall composition of the system. V V dv = dt + dp T P, n k P T, n k V + dn n k k P, T, nj n k For any extensive variable Z, = MdT + NdP + Z k dn k where the partial molar property of Z of component k Z Z k n (k =,, 3,. ) k P, T, nj n k Self-Assessment Question 6. Write an expression for the partial molar Gibbs free energy of component A in the A-B binary solution. 5

6 General Properties of Partial Molar Properties Z = n Z + n Z n Z = n k Z k Dividing both sides with the total number of moles, n T Z = X Z + X Z X Z = X k Z k Z = Z / n T X k = n k / n T Differentiating completely = (X k k + Z k dx k ) = MdT + NdP + Z k dn k At constant T and P = Z k dn k and = Z k dx k Since = (X k k + Z k dx k ) X k k = 0 (At constant T and P) Gibbs-Duhem Equation 6

7 Each of the four categories of relationships developed earlier for the total properties of system between the various state functions (the laws, definitions, coefficient relations, and Maxwell relations) are applicable also to the partial molar properties of the component of a system. H k = U k + PV k G k = H k TS k Gk P T, n k V k dg k = S k dt + V k dp Gk T P,n k - S k Evaluation of Partial Molar Properties Two methods:. Measurements of the partial molar properties from the total properties of the solution. Z Z k. Measurements of the partial molar properties of one component knowing the partial molar properties of another. Z Z 7

8 Partial molar properties from the total molar properties Z = Z k X k = Z X + Z X Z = Z X + X Z + Z = Z X + X + X Z = Z + X dx Z = Z X dx dx dx = Z + ( X ) = Z dx + Z dx X + X = dx = - dx = ( Z Z ) dx Z = Z + dx dx Z = Z X dx = Z + ( X ) dx Z and Z can be found from knowledge of Z either analytically or graphically Analytical Solution Example 6.3 Derive expressions for the partial molar volumes of each components as functions of composition if the volume change in a binary solution obeys the relation V =.7 X X cc/mol. dv/dx =.7 ( X + X. X. dx /dx ) =.7 (X X X ) dv/dx = dv/dx =.7 (X X X ) V = V X (dv/dx ) =.7 X X X.7 (X X X ) V =.7 ( X 3 X X ) V = V X (dv/dx ) =.7 X X X.7 (X X X ) V = 5.4 X X 8

9 Graphical Solution Molar Property, Z Z 0 E Z = Z X (/dx ) Z = Z + ( X ) (/dx ) Z P B Z D Z G X 0 A Pure Pure Mole Fraction, X Schematic representation of molar property, Z, versus composition, X, diagram of a binary solution at constant temperature and pressure F Z 0 Slope of curve at point P / dx = B / PB Z = Z + ( X ) (/dx ) = AB + PB (B/PB) = A Z = GD To determine partial molar properties for a solution of a given composition, construct a tangent to the Z curve at that composition and read the intercepts on the two sides of the graph. Partial molar properties of one component from partial molar properties of another Using Gibbs-Duhem equation for a binary solution X + X = 0 X = X Integrating this equation from X = to X =X Z Z X = X X = X = X X = X 9

10 Z X = X Z X = X = X X = X For pure component, partial molar property does not exist. So at X =, Z = 0 X = X X Z = X Thus, knowing the value of Z, the partial molar properties of one component, and a reference state, the partial molar property of the other component can easily be calculated. X = X X X = If a relationship between Z and X is given, the above relation can be modified as = dx dx Z = X = X X X dx dx If, on the other hand, the input relationship is between Z and X, then Z = X = X X dx X dx 0

11 Example 6.4 The partial molar enthalpy of component in a binary solution is given by the equation H = ax. ompute the partial molar enthalpy for component and the molar enthalpy for the solution. H H X = X X = d(h ) d(h ) dx X dx = ax dx X = X X X = X =. ax X. dx = a X dx X = X X = 0 H = + a X dx = ax H = X H + X H = X. ax + X. ax H = ax X (X + X ) = ax X Next lass Lecture 7 Thermodynamics of Solutions The Ideal and Nonideal Solutions

Thermodynamic Variables and Relations

MME 231: Lecture 10 Thermodynamic Variables and Relations A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka Today s Topics Thermodynamic relations derived from the Laws of Thermodynamics Definitions