Chapter 3 notes colligative properties PDF

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CHEM 239 Chapter 3 Notes: Thermodynamics of solutions and colligative propertiesSection numbers refer to the “big Atkins”!8 Entropy and Gibbs energy of mixing; liquid mixturesIn Unit 4, we found for the entropy of mixing of nA moles A + nB moles B in the gas phase:Smixing = R(nAlnXA + nBlnXB) 4.If...

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CHEM 239 Chapter 3 Notes:

Thermodynamics of solutions and colligative properties

Section numbers refer to the “big Atkins”! 8.7

Entropy and Gibbs energy of mixing; liquid mixtures

In Unit 4, we found for the entropy of mixing of nA moles A + nB moles B in the gas phase: Smixing = R(nAlnXA + nBlnXB)

4.21

If A and B are ideal gases, Hmixing = 0, and with G = H  TS we can write: Gmixing = +RT(nAlnXA + nBlnXB)

8.17

We note that the Gibbs energy of mixing in the gas phase is always negative (and the entropy of mixing is always positive), in other words mixing is a spontaneous process. How do we apply this equation to the case of liquid mixtures? The ultimate answer will be that for ideal liquid mixtures the expression for the Gibbs energy of mixing is the same as eq. 8.17, i.e. eq. 8.17 applies to both gas and liquid mixtures, under conditions of ideality for both, but we have to go through a few steps before we can see why this is the case. Consider once again the L-G equilibrium for a pure substance A. We will work with the molar Gibbs energy of components A and B in a binary mixture, thus we can replace the Gibbs energies of A and B with the molar Gibbs energies or chemical potentials, A and B (refer to sections 5.4 and 6.7). If at a given temperature the vapour pressure of the pure liquid is pAo we can write, with the equilibrium condition A(l) = A(g), and the fact that A(l) is the pure liquid, A(l) = Ao(l): Ao(l) = Ao(g)+ RTlnpAo

8.18

If on the other hand the vapour pressure of A above a mixture A + B is p A, and its Gibbs energy in the liquid mixture A(l), for the G-L equilibrium of A between this liquid mixture and its vapour we can write: A(l) = A(g) = Ao(g) + RTlnpA Subtracting eq. 8.18 from 8.19: A(l)  Ao(l) = RT(lnpA  lnpAo) = RTln(pA/pAo) or: A(l) = Ao(l) + RTln(pA/pAo)

8.19 8.20 8.21

Raoult’s law states that the vapour pressure of a component in a mixture is proportional to its mol fraction in the mixture:

Physical Chemistry 1, Lecture Notes Unit 8 Part II

122

p A = X A pA o

8.22

Raoult’s law defines the properties of an ideal liquid mixture, as we will see from the reasoning below. Combining eq. 8.21 and 8.22: A(l) = Ao(l) + RTlnXA

8.23a

We can equally apply the same to component B in the mixture (even when component B is non-volatile, in principle there will still be a vapour pressure, however small). B(l) = Bo(l) + RTlnXB

8.23b

For the total Gibbs energy of a mixture of nA mol A + nB mol B we have, from eq. 5.19 with 7.23: Gtotal = nAA + nBB

8.19

Gtotal = nAAo + nBBo +RT{nAlnXA + nBlnXB}

8.24

In eq. 7.84 we recognize the first two terms on the right hand side (n AAo + nBBo) as the Gibbs free energy of the two separate components, thus with Gmixing = Gtotal  (nAAo + nBBo) we arrive back at our starting point, eq. 8.17 which we can now apply also to ideal liquid mixtures: Gtotal(ideal liquid mixture) = nAAo(l) + nBBo(l) + RT(nAlnXA + nBlnXB) Gmixing(ideal liquid mixture) = +RT(nAlnXA + nBlnXB)

8.25

In eq. 8.25, note that because X A and XB are...