Mass Balances for Chain Association

The thermodynamics and phase behavior are sufficiently described by Eqns. 19.75 and 19.77, but you may be curious about the true mole fractions of the species. Furthermore, it is interesting to see how this “fraction of acceptor sites not bonded” is closely related to the fraction monomer, x_M. This turns out to be a bit subtle, and it should not distract you from the primary issue of phase behavior. If you are interested, we can use material balances to obtain two simple relations between the true number of moles in the solution, n_T, and the apparent number of moles that we would expect if there was no association,¹⁹ n_o. Note that n_o is the number of moles one would compute based on dividing the mass of solution by the molecular weight of a monomer as taught in introductory chemistry. For example, in 100 cm³ of water one would estimate

n_o = 100 cm³ · 1.0 g/cm³/(18 g/mole) = 5.556 moles

But how many moles of H₂O monomer do you think truly exist in that beaker of water? We will return to this question shortly. Note that each i-mer contains “i” monomers, such that the contribution to the apparent number of moles is i·n_i. Note also that the true mole fractions, x_i, are given by n_i/n_T, but it may not look so simple at first.

We begin by noting that K_i = K_i–1 implies that x_i = x_M(x_i–1Δ). This leads to a recursive relation originating with the monomer, so x_i = x_M(x_MΔ)ⁱ.

Substituting Eqn. 19.41–19.44,

n₀ = n_T Σ i·x_M(x_MΔ)ⁱ = x_M n_T [1 + 2(x_MΔ) + 3(x_MΔ)² + 4(x_MΔ)³ + …]

This series may not appear to be familiar but it is a common converging series. Referring to series formulas in a math handbook, we find that

Since the mole fractions must sum to unity, we can write a second balance, for x_i,

and again recognizing the series,

Substituting x_M for (1 – x_MΔ) in Eqn. 19.79 results in,

This equation makes clear that the properties of the mixture are closely related to the properties of the monomer.

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Example 19.5. Molecules of H₂O in a 100 ml beaker

Assuming Δ is about 100 at room temperature and ρ = 1 g/cm³, estimate the moles of H₂O monomer in a 100 ml beaker of liquid water.

Solution

Note that the problem statement requests moles of H₂O, not (H₂O)₂ or (H₂O)₃, and so on, so we are interested in the true number of H₂O monomer moles. We know n₀ = 5.556 by applying the monomer molecular weight, but the number of monomer moles n_M = x_M·n_T will be significantly less. Proceeding, using Eqn. 19.73,

Therefore, the true number of moles is 100 times less than the apparent number of moles.