Question 14.11: Determine the number of degrees of freedom F for each of the...

(a) The system consists of two nonreacting species in two phases. If there were no azeotrope, Eq. (14.36) would apply:

F = 2 − π + N − r        (14.36)

F = 2 − π + N − r = 2 − 2 + 2 − 0 = 2

This is the usual result for binary VLE. However, a special constraint is imposed on the system; it is an azeotrope. This provides an equation, x_1 = y_1 , not considered in the development of Eq. (14.36). Thus, Eq. (14.37) with s = 1 yields F = 1. If the system is an azeotrope, then just one phase-rule variable—T, P, or x_1 (= y_1)—can be arbitrarily specified.

(b) Here, a single chemical reaction occurs:

F = 2 − π + N − r − s              (14.37)

\mathrm{CaCO}_3(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_2(g)

and r = 1. Three chemical species are present, and three phases: solid CaCO_3,  solid  CaO,  and  gaseous  CO_2. One might think a special constraint has been imposed by the requirement that the system be prepared in a special way—by decomposing CaCO_3. This is not the case, because no equation connecting the phase-rule variables can be written as a result of this requirement. Therefore,

F = 2 − π + N − r − s = 2 − 3 + 3 − 1 − 0 = 1

and there is a single degree of freedom. Thus, CaCO_3 exerts a fixed decomposition pressure at fixed T.

(c) The chemical reaction here is:

\mathrm{NH}_4 \mathrm{Cl}(s) \rightarrow \mathrm{NH}_3(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})

Three species, but only two phases, are present in this case: solid NH_4Cl and a gas mixture of NH_3 and HCl. In addition, a special constraint is imposed by the requirement that the system be formed by the decomposition of NH_4Cl. This means that the gas phase is equimolar in NH_3 and HCl. Thus a special equation, y_{\mathrm{NH}_3}=y_{\mathrm{HCl}} connecting the phase-rule variables can be written. Application of Eq. (14.37) gives:

F = 2 − π + N − r − s = 2 − 2 + 3 − 1 − 1 = 1

and the system has but one degree of freedom. This result is the same as that for part (b), and it is a matter of experience that NH_4Cl has a given decomposition pressure at a given temperature. Nonetheless, this conclusion is reached differently in the two cases.

(d) This system contains five species, all in a single gas phase. There are no special constraints. Only r remains to be determined. The formation reactions for the compounds present are:

Systematic elimination of C and O_2 , the elements not present in the system, leads to two equations. One such pair of equations is obtained in the following way. Eliminate C from the set of equations by combining Eq. (B), first with Eq. (A) and then with Eq. (D). The two resulting reactions are:

\text { From }(B) \text { and }(A): \quad \mathrm{CO}+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{CO}_2         (E)

\text { From }(B) \text { and }(D) \text { : } \quad \mathrm{CH}_4+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2+\mathrm{CD}_2         (F)

Equations (C ), (E), and (F ) are the new set, and we now eliminate O_2 by combining Eq. (C ), first with Eq. (E) and then with Eq. (F ). This gives

From (C) and (E ): \mathrm{CO}_2+\mathrm{H}_2 \rightarrow \mathrm{CO}+\mathrm{H}_2 \mathrm{O}         (G)

From (C) and (F ): \mathrm{CH}_4+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CO}_2+4 \mathrm{H}_2         (H)

Equations (G) and (H) are an independent set and indicate that r = 2. The use of different elimination procedures produces other pairs of equations, but always just two equations.

F = 2 − π + N − r − s = 2 − 1 + 5 − 2 − 0 = 4

This result means that one is free to specify four phase-rule variables, for example, T, P, and two mole fractions, in an equilibrium mixture of these five chemical species, provided that nothing else is arbitrarily set. In other words, there can be no special constraints, such as the specification that the system be prepared from given amounts of CH_4 and  H_2O. This imposes special constraints through material balances that reduce the degrees of freedom to two. (Duhem’s theorem; see the following paragraphs.)