3.00 Thermodynamics Fall 2002 Problem Set 6 Solutions Solution 6.1 CH(g) = C(graphite) + 2H2(g) + AG 69120-22.25TlogT+65.357 2CO(g)=2C(graphite) + O2(g)+AG= 223400+175.37 CO2(g) C(graphite) + O2(g) → AG 394100+0.87 1 H2(g) + O2(g) = H2O(9) AG -246400+54.87 The number of elements in the reaction 3 (C, H, O) Number of gaseous com- ponents in the reactions 6 (CO₂,CO,CH₂H₃O₂H₂O, excluding graphite) The above reactions are the reactions at equilibrium and these can be arranged in any order. There are two ways to go about solving this problem. The first is easier. We examine the mole balance in the system. The system contains the species CO, CO,CHA, H₂,O₂, H₂O (excluding graphite) and thus ne, the number of moles of C, NCH, +1800+neo (1) no, the number of moles of O, 2RCO₂+CO+0 (2) ng, the number of moles of H, 2он +20.0 + ACH (3) Initially, 1 mole of CH, was mixed with 1 mole of CO₂, and thus, in a closed system, ne = 1 2 70 (4) (5) Using the above relations, (6) or, as n o p at constant P, (7) and also, (8) PCHPCO₂+PH₂O (9) In addition, P=PCH + Poo+PCO+PH+PHO (10) It is now possible to solve for pH, PH₂O, and poo in terms of P, and pow. and poo, using Eqs. ??. 1 (11) PCOP-PCO (12) I PHP-2рон + PCO (13) The reaction (i) below has an associated change in Gibbs free energy of AG and equilibrium reaction constant K at T-1750K. CH+CO2H₂+2CO (14) AG=239820-22.257 log 7-109.157 Joules (15) Kp.exp-AG (1750K, latm)/1750R exp[241953/1750-8.314] 16679009 (16) Kp is related to the partial pressures of the gases by POOPH POOPCH (17) And the following reaction (ii) has an associated change in Gibbs free energy of AG and equilibrium reaction constant Kp, at T-1750K. H₂CO₂H₂O+CO (18) AG36000-32.057.Joules (19) Kai exp[-AG(1750K, 1atm)/1750R = exp(20088/1750-8.314]3711114 (20) Kai is related to the partial pressures of the gases by KPCOPHO POOPHY (21) The two equilibrium reaction constants, K and Kps are related to the two independent variables per, and poo,, with P= latm as Kp. 16679009- (P-poo) (P-2рон, +роо) (22) PCO PCH Кран = 3711114 = (Р-реros) (рен, - Poo) (23) POO (P-2PCH + PCO₂) The simultaneous solution of these equations with P-latm gives the solution to the problem. Another way is presented below. (There are several ways to go about solving this problem, depending on the reactions that are chosen.) Combining the first three reaction equations, CH+CO₂=2H+2CO (24) The standard free energy for the above equation is given by AG, AGI+AG AG AGAGAGAG77468.29.J (25) 4dnch, +2dnn 0 (26) 2dnco,+ dnco = 0 (27) ducn, duco,+dnco = 0 (28) (29) Since there are three equations with four variables, one independent variable can be identified, which we identify as = docu Reaction CHA CO₂→2H₂ 200 Initial 1 0 0 1 Equilibrium change Combining reaction equations 2 and 3, 1-x 1-x → 2x 2x The change in standard free energy is given by. AG2AG-AG 435825.J The constraints in the number of moles of components are, (30) (31) 2dnco+2dn.co+dano 0 (32) 2dn+2dn200 (33) Reaction 2002 200 20 Initial 1-x 2x 0 Equilibrium change From the last reaction equation 1-x-2y 2x+2y y H+1/2O H2O (34) The change in the standard free energy (for T-1750K) is given in the problem statement as AG-246400+54.81750-150500/ (35) The constraint equations are, Reaction Initial H₂ 1/2 02 →H₂O 2dno,+ dano = 0 (36) (37) 2x → Equilibrium change 2x-z y-z/2 Thus at equilibrium, taking into account all the above reactions, neu, 1-2co, 1--2, п.22-2, поо 2х+2 noy-/2 and not2+22+y-2/2.