The following two reactions take place in a gas-phase reactor: 

If the system comes to equilibrium at 3000 K and 1 atm, the product gas mole fractions satisfy the relations 

(a) Let n_A0; ... ; n_E0 be the initial number of gram-moles of each species and ξ_e1 and ξ_e2 be the extents of reactions 1 and 2, respectively, at equilibrium (see Equation 4.6-7). Derive expressions for the mole fractions y_A; y_B ; ... ; y_E in terms of n_A0; n_B0; ... ; n_E0, ξ_e1, and ξe2. Then substitute in the equilibrium relations to derive two simultaneous equations for the two extents of reaction. 

(b) One-third of a gram-mole each of CO₂, O₂, and N₂ are charged into a batch reactor and the reactor contents equilibrate at 3000 K and 1 atm. Without doing any calculations, prove that you have enough information to calculate the component mole fractions of the reactor contents at equilibrium. 

(c) Create a spreadsheet and input the sets of values of n_A0, n_B0, n_C0, n_D0, and n_E0 shown below, and two initial guesses for ξ₁ and ξ₂ (make them each 0.1). Then for each set, have the spreadsheet calculate n_A, n_B, n_C, n_D, nE from Eq. (4.6-7); then n_total and y_A, y_B, y_C, y_D, and yE; and then the squares of the left-hand side minus the right-hand side of Equations (1) and (2). Finally, use Solver to find the values of ξ1 and ξ2 that minimize the sum of those two squares. When you call Solver the first time, click on “Options,” and select “Use automatic scaling.” Don’t be surprised if you get negative extents of reaction, which can happen with reversible reactions. 

Equation 4.6-7

Transcribed Image Text:

2CO2=2CO + O2 (A) O2 + N2 (В) (С) 2ΝΟ (С) (D) (E) (C)