Evaluation of Properties During an Irreversible Chemical Reaction

4. Evaluation of Properties During an Irreversible Chemical Reaction

The change in entropy between two equilibrium states for an open system is repre- sented by the relation (cf. Chapter 3 and Chapter 8)

dS = dU/T + P dV/T – Σµ k dN k /T. (9a) Similarly, dS = dU/T + P dV/T – Σµ k dN k /T. (9a) Similarly,

Thus, dU = TdS- PdV, and the change in internal energy ≈ heat added - work performed. How- ever, if mass crosses the system boundary and the system is no longer closed (e.g., pumping of air into a tire), chemical work is performed for the species crossing the boundary (= Σµ k dN k ). Likewise,

dH = T dS + V dP + Σµ k dN k , (9c) dA = –S dT – P dV + Σµ k dN k ,

(9d) dG = –S dT + V dP + Σµ k dN k

(9e) It is apparent from Eqs. (9a) to (9e) that -T ( ∂S/∂N k ) U,V =-T ( ∂S/∂N k ) H,P = ( ∂U/∂N k ) S,V =( ∂A/∂N k ) T,V =( ∂G/∂N k ) T,P = ˆg k = µ k , and (9f)

-TdS U,V,,m = - TdS H,,P,m = dU S,V,m = dH S,P,m = dA T,V,,m = dG T,P,,m = Σµ k dN k . (9g)

a. Nonreacting Closed System In a closed nonreacting system in which no mass crosses the system boundary dN k =

0. Therefore for a change in state along a reversible path, dS = dU/T + P dV/T,

(10) dU = T dS – P dV, dH = T dS + V dP dA = –S dT – P dV, and dG = – S dT + V dP.

It is apparent that for a closed system Σµ k dN k = 0.

b. Reacting Closed System Assume that 5 kmole CO, 3 kmole of O 2 and 4 kmole of CO 2 (with a total mass equal to 5 ×28+3×32+4×44= 412 kg) are introduced into two identical piston–cylinder–weight as-

semblies A and B. We will assume that system A contains anti-catalysts or inhibitors which suppress any reaction while system B can engage in chemical reactions which result in the

final presence of 4.998 kmole of CO, 2.999 kmole of O 2 and 4.002 kmole of CO 2 . The species changes in system B are dN CO = -0.002, dN O2 = -0.001 and dN CO2 = 0.002 kmole, respectively. The Gibbs energy change dG of system B is now determined by hypothetically injecting 0.002

kmole of CO 2 into system A and withdrawing 0.002 kmole of CO and 0.001 kmole of O 2 from it (so that total mass is still 412 kg) so as to simulate the final conditions in system B. The Gibbs energy G A = G + dG T,P . System A is open even though its mass has been fixed. The change dG T,P , during this process is provided by Eq. (9e). Thus,

(12) Since the final states are identical in both systems A and B, the Gibbs energy change dG T,P;B

dG T,P;A = (–0.002) µ CO + (–0.001) µ O 2 + (+0.002) µ CO 2 .

during this process must then equal dG T,P;A . Therefore, during this process must then equal dG T,P;A . Therefore,

(13) The sum of the changes in the Gibbs energy associated with the three species CO, CO 2 ,O 2 are

dG T,P;B =–T δσ = (µ CO dN CO + µ O2 dN O2 + µ CO2 dN CO2 )< 0, i.e.,

(14a) dG T,P,B =(–0.002) µ CO + (–0.001) µ O 2 + (+0.002) µ CO 2 =- Τδσ < 0

(14b) Note that system B is a chemically reacting closed system of fixed mass.

Recall that for irreversible processes involving adiabatic rigid closed systems dS U,V,m > 0 and from Eq. (9g), Σµ k dN k < 0. This inequality involves constant (T,V) processes with A

being minimized or constant (T,P) processes with G being minimized. If (S,V) are maintained constant for a reacting system (e.g., by removing heat as a reaction occurs), then dU S,V,m = Σµ k dN k < 0 In this case S is maximized at fixed values of U, V and m, while U is minimized

at specified values of S, V, and m. The inequality represented by Eq. (13) is a powerful tool for determining the reaction direction for any process.

c. Reacting Open System If in one second, a mixture of 5 kmole of CO, 3 kmole of O 2 , and 4 kmole of CO 2

flows into a chemical reactor and undergoes chemical reactions that oxidize CO to CO 2 , the same criteria that are listed in Eqs. (4) through (8) can be used as long we follow a fixed mass. As reaction proceeds inside the fixed mass, the value of G should decrease at specified (T, P) so that dG T,P ≤ 0.