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Partial Molar Volume of That Component Chapter 6 Chemical Potential in Mixtures Partial molar quantities Thermodynamics of mixing The chemical potentials of liquids Chapter 7 : Slide 1 Duesberg, Chemical Thermodynamics Concentration Units • There are three major concentration units in use in thermodynamic descriptions of solutions. • These are – molarity – molality – mole fraction • Letting J stand for one component in a solution (the solute), these are represented by • [J] = nJ/V (V typically in liters) • bJ = nJ/msolvent (msolvent typically in kg) • xJ = nJ/n (n = total number of moles of all species present in sample) PARTIAL MOLAR QUANTITIES Recall our use of partial pressures: p = xA p + xB p + … pA = xA p is the partial pressure For binary mixtures (A+B): xA + xB= 1 We can define other partial quantities… Chapter 7 : Slide 3 Duesberg, Chemical Thermodynamics PARTIAL MOLAR QUANTITIES: Volume In a system that contains at least two substances, the total value of any extensive property of the system is the sum of the contribution of each substance to that property. The contribution of one mole of a substance to the volume of a mixture is called the partial molar volume of that component. V f p,T,nA,nB... V VA At constant T and p nA p,T ,n A V V dV dnA dnB ... nA nB Chapter 7 : Slide 4 Duesberg, Chemical Thermodynamics Partial molar volume Add nA of A to mixture Very Large Mixture of A and B When you add nA of A to a large mixture of A and B, the composition remains essentially unchanged. In this case: VA can be considered constant and the volume change of the mixture is n V . A A V VA const nA p,T ,n A Likewise for addition of B. The total change in volume is nAVA + nBVB . (Composition is essentially unchanged). Scoop out of the reservoir a sample containing nA of A and nB of B its volume is nAVA + nBVB . Because V is a state function: V VAnA VBChapternB 7 :... Slide 5 Duesberg, Chemical Thermodynamics Partial molar volume Example: What is the change in volume of adding 1 mol of water to a large volume of water? The change in V 3 VH O 18cm 3 2 n volume is 18cm H2O p,T A different answer is obtained if we add 1 mol of water to a large volume of ethanol. The change in V 3 VH O 14cm 3 2 n volume is 14cm H2O p,T ,n(CH 3CH 2OH) Chapter 7 : Slide 6 Duesberg, Chemical Thermodynamics Introduction to mixtures Homogeneous mixtures of a solvent (major component) and solute (minor component). Introduce partial molar property: contribution that a substance makes to overall property. V = nAVA + nBVB VA is not generally a constant: It is a function of composition, because the environment and therefore their interaction of the The partial molar volumes of water molecules changes with the and ethanol at 25C. Note the different composition . scales (water on the left, ethanol on : the right). Exercise Calculate the density of a mixture of 20 g of water and 100g of ethanol. MM: EtOH: 46.07g/mol d=0.789 g/ml • First calculate the mole fractions. – 20 g H2O = 1.11 mol; – 100 g EtOH = 2.17 mol – xH2O = 0.34; xEtOH = 0.66 • Then interpolate from the mixing curve 3 -1 – VH2O = 17.1 cm mol ; 3 -1 – VEtOH = 57.4 cm mol • Then plug the moles and partial molar volume – (1.11 mol)(17.1 cm3/mol) + (2.17 mol)(57.4 cm3/mol) =18.1 cm3 + 125 cm3 = 143.1 cm3 • Finally, the total mass is divided by the total volume: 120 g/143 cm3 = 0.84 g/ cm3 Task: You mix 300 ml alcohol with a bottle of juice (water, 0.7 l) - Will you really end up with for your x-mas party supply (1l)? -Of course not… - How much do you have to mix to get 1 l of the same strength? PARTIAL MOLAR QUANTITIES VA is not generally a constant: It is a function of composition, because the environment and therefore their interaction of the molecules changes with the composition : : The partial molar volume of a substance is the slope of the variation of the total volume of the sample plotted against the composition. In general, partial molar quantities vary with the composition, as shown by the different slopes at the compositions a and b. Note that the partial molar volume at b is negative: the overall volume of the sample decreasesChapter 7 : Slide as a9 is added. Duesberg, Chemical Thermodynamics PARTIAL MOLAR QUANTITIES • We can extend the concept of partial molar properties to state functions, such as Gibbs energy, G. Or Chemical Pot m G = nAGA + nBGB or G = nAmA + nBmB Consider this single substance system: m1 m2 dG = -m1 dn dG = +m2 dn Total dG = 0 only if m1 = m2 10 The Partial Molar Gibbs Energy The partial molar Gibbs energy is called the “chemical potential” G f p,T,nA,nB... At constant T and p G G G dG dn dn ... μA A B n nA nB A p,T,n A dG m dn m dn ... A A B B G = nAµA + nBµB (At equilibrium dG = 0) The implication of G = f ( p,T, ni ) is that we now write: dG SdT Vdp mAdnA mBdnB ... Fundamental dG SdT Vdp midni Equation of Chemical Thermodynamics.Chapter 7 : Slide 11 Duesberg, Chemical Thermodynamics Chemical Potential = Partial Molar Gibbs Energy In case of constant temperature and pressure reduces to: dG midni The chemical potential of a substance is the slope of the total Gibbs energy of a mixture with respect to the amount of substance of interest. In general, the chemicaldG potentialSdT variesVdp withm composition,idni as shown for the two values at a and b. In this case, both chemical potentials areChapter positive. 7 : Slide 12 Duesberg, Chemical Thermodynamics The Wider Significance of the Chemical Potential G G dG SdT Vdp dni ni T , p,n j ni T , p,n j H dH TdS Vdp dni ni S, p,n j H A m ni i ni S , p,n j T ,V ,n j A dA SdT pdV dni ni T ,V ,n j U ni U S ,V ,n j dU TdS pdV dni ni S,V ,n j All extensive thermodynamic properties change with composition!Chapter 7 : Slide 13 Duesberg, Chemical Thermodynamics Chemical Potential In case of constant temperature dw dG m dn and pressure also: add,max i i dw m dn "chemical work" Example: E-chem cell: chem jj j dU TdS PdV m dn jj UH j mi nnii S,,,, V nj i S P n j i dH TdS VdP m jj dn j AG dA SdT PdV m jj dn nnii j T,,,, V nj i T P n j i dG SdT VdP m jj dn j The Gibbs-Duhem Equation dG m1dn1 m2dn2 ... G m1n1 m2n2 (For binary systems) by differentiating: dG m1dn1 m2dn2 n1dm1 n2dm2 At equilibrium: dG = 0 recall in case of constant temperature m m and pressure dG midni => 1dn1 + 2dn2 = 0 0 n dm n dm 1 1 2 2 The Gibbs-Duhem Equation. Mixtures can not be change independentely: If or 0 n dm i i one is changed the other changes as n2 well… or dm1 dm2 A similar expression may be deduced n1 for all partial molar quantitiesChapter 7 : Slide 15 Duesberg, Chemical Thermodynamics The Gibbs-Duhem Equation SdT VdP njj dm 0 j Chapter 7 : Slide 16 Duesberg, Chemical Thermodynamics THE THERMODYNAMICS OF MIXING Imagine a system of two perfect gases in amounts nA and nB at equal T and p are separated by a barrier. The initial total Gibbs energy of the system Gi is given by Gi = nAµA + nBµB p G n m RT ln i A A p p n m RT ln B B p The initial and final states considered in the calculation of the Gibbs energy of mixing of gases at different initial pressures.Chapter 7 : Slide 17 Duesberg, Chemical Thermodynamics The Thermodynamics of mixing After mixing each gas exerts a partial pressure pJ, where pA + pB = p. The final G is given by p G n m RT ln A f A A p N , p, T A NB, p, T p n m RT ln B B B p pA+ pB = p,T mixG Gf Gi Duesberg, Chemical Thermodynamics The Thermodynamics of mixing p p G G G n m RT ln A n m RT ln B mix f i A A B B p p p p n m RT ln n m RT ln A A B B p p p p n RT ln A n RT ln B A p B p N , p, T A NB, p, T Chapter 7 : Slide 19 pA+ pB = p,T Duesberg, Chemical Thermodynamics The Thermodynamics of Mixing p p G n RT ln A n RT ln B mix A p B p Writing xA for the mole fraction of n n component A A A nx n nA = xA n therefore pA /p = xA, (same for any component), so The Gibbs energy of mixing of two perfect gases and (or of two liquids that form an ideal solution). The mixG = nRT (xA ln xA + xB ln xB) Gibbs energy of mixing is negative for all compositions and mixG is always negative, since x<1 temperatures, so perfect gases mix spontaneously inChapter all proportions. 7 : Slide 20 Duesberg, Chemical Thermodynamics Gas mixtures What About the Enthalpy of Mixing? G = H – TS Compare Gmix = nRT {xAln xA+ xB ln xB} Smix = − nR {xAln xA+ xB ln xB} Therefore H mix= 0 All due to the effect of entropy 21 The Entropy of Mixing Δ G Δ S mix mix T p,nA ,nB nR(xAln xA xBln xB ) which is positive as xA and xB are <1.
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