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∂n k /∂t + ∇.n k r V = ˙ , ′′′N k gen . (35) 1. Steady State System Under steady state conditions dN k /dt = 0, dN/dt = 0,dm k /dt = 0, dm/dt = 0 and from Eq. (28), ˙ m i = ˙ m e . Likewise, from Eq.(33), ˙ N i + ˙ N gen = ˙ N e . n. Example 13 efficiency of 90%, write the mass conservation and mole balance equations. Solution The reaction equation is written as follows: 2 CO + 3 O 2 → N CO,e CO + N CO2,e CO 2 + N O2,e O 2 . (A) Since mass does not accumulate within the reactor, the atom balance for C and O at- oms yields 2 = N CO,e + N CO2,e , and (B) 2 + 6 = N CO,e + 2 N CO2,e + 2 N O2,e , i.e., (C) 0.4 = (2 - N CO,e )/2. (D) With the three equations Eqs. (B) through (D), we can solve for the three unknowns N CO,e , N CO2,e and N O2,e , i.e., N CO,e = 1.6, N CO2,e = 0.4, and N O2,e = 2.8. The mass conservation implies dm CO /dt = 2×28+ ˙ m CO,gen - 1.6×28 = 11.2 + ˙ m CO,gen . Similarly, dN CO /dt = 2 + ˙ N CO, gen - 1.6 = 0.4 + ˙ N CO gen Normally ˙ N CO, gen is a negative quantity since CO is consumed. The term dN CO /dt represents the accumulation (destruction, since negative) rate of CO in the reactor. Similarly for CO 2 , dN CO2 /dt = 0 + ˙ N CO2 gen - 0.8, where ˙ N CO2, gen > 0, since CO 2 is a product that is generated. In the initial periods when the combustor is being fired, the CO 2 concentration gradually increases due to the term dN CO2 /dt. Similarly for 90% efficiency, N CO,e = 0.2 N CO2,e = 1.8, and N O2,e = 2.1 dm CO /dt = 2×28+ ˙ m CO,gen - 0.2×28×28 = 50.4 + ˙ m CO,gen . The combustor is operating steadily so that dm cO /dt = 0, ˙ m CO,gen = - 50.4 kg/s Similarly , dN CO /dt = 2 - 0.2 + ˙ N CO, gen = 1.8+ ˙ N CO, gen . Since dN CO /dt = 0, ˙ N CO, gen = - 1.8 k mole s -1 , and A combustor is fired with 2 kmole of CO and 3 k mole of O 2 . When the combustor is just started, very little CO burns. As it warms up, more and more CO are burnt. As- sume that mass does not accumulate within the reactor. At the point when the com- bustor achieves 40% efficiency, write the mass conservation, mole balance and en- ergy conservation equations. If the combustor reaches a steady state with combustion dN CO2 /dt = 0 + ˙ N CO 2,gen - 1.8, If the combustor is operating steadily then dN CO2 /dt = 0, and ˙ N CO 2,gen = 1.8 k mole s -1 , ˙ N O 2,gen = -0.9 kmole s -1 . H. SUMMARY Chemical reactions occur when species rearrange their atoms and different com- pounds with different bond energies are produced. Dry and wet gas analyses are presented in this chapter, which are an analytical tool to measure species transformations. Examples are presented for determining (A:F) from dry gas analyses. The enthalpy of formation or chemical enthalpy, thermal enthalpy and the total enthalpy are defined. Energy conservation (First law) and entropy balance (Second law) of reacting systems are introduced and illustrative examples are provided. Finally mass conservation and mole balance equations for reacting systems are presented. Chapter 12 12. REACTION DIRECTION AND CHEMICAL EQUILIBRIUM A. INTRODUCTION Nature is inherently heterogeneous and, consequently, natural processes occur in such a direction so as to create homogeneity and equilibrium (which is a restatement of the Second Law). In the previous sections, we assumed that hydrocarbon fuels react with oxygen to pro- duce CO 2 , H 2 O, and other products. We now ask the question whether these products, e.g., CO 2 , and H 2 O, can react among themselves to produce the fuel and molecular oxygen. If not, then why not? What governs the direction of reaction? Now we will characterize the parame- ters that govern the predominant direction of a chemical reaction. We will also discuss the composition of reaction products under equilibrium conditions. B. REACTION DIRECTION AND CHEMICAL EQUILIBRIUM 1. Direction of Heat Transfer Prior to discussing the direction of a chemical reaction, we will consider the direction of heat transfer. Heat transfer occurs due to a thermal potential, from a higher to a lower tem- perature. Thermal equilibrium is reached when the temperatures of the two systems (one that is transferring and the other that is receiving heat) become equal. Due to the irreversible heat transfer from a warmer to a cooler system, δσ > 0. 2. Direction of Reaction The direction of heat transfer is governed by a thermal potential. For any infinitesimal irreversible process δσ > 0. (For heat transfer to take place from a lower to a higher tempera- ture δσ < 0, which is impossible.) Likewise, the direction of a chemical reaction under speci- fied conditions is also irreversible and occurs in such a direction such that δσ > 0.The direction of a chemical reaction within a fixed mass is also such that δσ > 0 due to chemical irreversi- bility. For instance, consider the combustion of gaseous CO at low temperatures and high pressures, i.e., CO + 1/2 O 2 → CO 2 (1a) At high temperatures and at relatively low pressures CO 2 → CO + 1/2 O 2 . (1b) Reaction (1a) is called an oxidation or combustion reaction, and Reaction (1b) is termed a dissociation reaction. The direction in which a reaction proceeds varies depending upon the temperature and pressure. We will show that the chemical force potentials F R (= ) g CO +(1/2) ) g O2 ) for the reactants and F P (= ) g CO2 ) for the products govern the direction of chemical reaction at specified values of T and P. If F R >F P , Reaction 1a dominates and vice versa if F R <F P , just as the thermal potential T governs the direction of heat transfer. Consider a premixed gaseous mixture that contains 5 kmole of CO, 3 kmole of O 2 and 4 kmole of CO 2 , placed in a piston–cylinder–weight assembly (PCW) at a specified constant temperature and pressure. It is possible that the oxidation of CO within the cylinder releases heat, in which case heat must be transferred from the system to an ambient thermal reservoir. An observer will notice that after some time the oxidation reaction (Reaction (1a)) ceases when chemical equilibrium is reached (at the specified temperature and pressure). Assume that the observer keeps an experimental log that is reproduced in Table 1. In the context of Reaction 1a, if 0.002 kmole of CO (dN CO ) are consumed, then 1/2×(0.002) = 0.001 kmole of O 2 (dN O 2 ) are also consumed and 0.002 kmole of CO 2 (dN CO2 ) are produced. Assigning a negative sign to the species that are consumed and using the associ- ated stoichiometric coefficients, Table 1: Experimental log regarding the oxidation of CO at specified conditions. Time, sec CO, kmole CO 2 , kmole O 2 , kmole 05 3 4 t A 4.998 2.999 4.002 t B 4.5 2.75 4. 5 t C 4.25 2.625 4.75 t D 3.75 2.375 5.25 t E 3.75 2.375 5.25 dN dN dN CO O CO − = − − =+ − =+ + =+ 1 0 002 1 0 002 12 0 002 1 0 002 22 . .; / .; . , yielding a constant number of 0.002. One can now define the extent of the progress of reaction ξ by the relation d ξ = dN CO 2 / ν CO 2 , (2) where dN CO 2 denotes the increase in the number of moles of CO 2 as a result of the reaction and ν CO 2 the stoichiometric coefficient of CO 2 in the reaction equation. During times 0 < t < t A , dξ = (4.002–4.0)/1 = 0.002. Generalizing, we obtain the relation dξ = dN k /ξ k . The production of CO 2 ceases at a certain mixture composition when chemical equilibrium is attained (time t E in Table 1). 3. Mathematical Criteria for a Closed System i. Specified Values of U, V, and m For a closed, fixed mass system (operating at specified values of U, V and m) under- going an irreversible process (cf. Chapter 3) (dS – δQ/T b = δσ) > 0. We wish to determine irreversibility due to reaction alone and eliminate other irreversibilities due to temperature and pressure gradients within the system, we set T b =T. Thus (dS – δQ/T = δσ) ≥ 0 with “>0” for irreversible, “=0” for reversible process (3) For an adiabatic reactor, (dS U,V = δσ) ≥ 0 (4) Thus for adiabatic reactions within a rigid vessel, the entropy S reaches a maximum. ii. Specified Values of S, V, and m Recall from Chapter 3 that dU = TdS – P dV – T δσ. (5a) Note that Eq. (5a) is valid for a process where irreversible process (δσ > 0) or reversible proc- ess (δσ =0) occurs. For a system operating at specified values of S, V, and m, (dU S,V = – T δσ) ≤ 0, with “<0” for irreversible, “=0” for reversible process (5b) iii. Specified Values of S, P, and m Likewise, since dH = T dS + V dP – Tδσ, for specified values of S, P, and m (dH = – T δσ) ≤ 0. (6a) iv. Specified values of H, P, and m (dS H,P = δσ) ≥ 0. (6b) which is similar to Eq. (4). Note that adiabatic reactions in a constant pressure closed system involves constant enthalpy and the entropy reaches a maximum value. v. Specified Values of T, V, and m Recall that dA = –S dT – P dV – T δσ, i.e., (dA T,V = –T δσ) ≤ 0. (7) For isothermal reactions within a rigid vessel, the Helmholtz function A reaches a minimum. vi. Specified Values of T, P, and m Similarly, dG = – S dT + V dP – T δσ, i.e., (8a) (dG T,P = –T δσ) ≤0. (8b) For isothermal and isobaric reactions within a fixed mass, the Gibbs’ function G reaches a minimum. Note that Eq. (8a) is valid for a process where irreversible process (δσ >0) or re- versible process (δσ =0) occurs. However if state change occurs reversibly between two equi- librium states G and G+dG, then dG = – S dT + V dP. In order to validate equation (8b) for an irreversible process of a fixed mass, we must deter- mine the value of either δσ or dG during an irreversible process (cf. Example 8). As was shown in Chapter 3 when we considered an irreversible mixing process and in Chapter 7 when we considered an irreversible evaporation process at specified (T,P), we must determine the value of G for the reacting system as the reaction proceeds. 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, dU = T dS – P dV + Σµ k dN k . (9b) Equation (9b) can be briefly explained as follows. For a fixed mass closed system, dN k =0. 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 = ˆ g 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. (11) 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, dG T,P;A = (–0.002)µ CO + (–0.001) µ O 2 + (+0.002) µ CO 2 . (12) Since the final states are identical in both systems A and B, the Gibbs energy change dG T,P;B during this process must then equal dG T,P;A . Therefore, dG T,P;B = – Tδσ = dG T,P;A , i.e., (Σµ k dN k ) A < 0. (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 = – Tδσ = (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. 5. Criteria in Terms of Chemical Force Potential The reaction CO+ 1/2O 2 → CO 2 , must satisfy the criterion provided by Eq. (14b) to pro- ceed. In the context of the above discussion, dividing Eq. (15) by 0.002 or the degree of reac- tion, dG T,P /0.002 = (–1) µ CO + (–1/2) µ O 2 + µ CO 2 < 0. (14c) Recall that µ κ = ˆ g k , i.e., dG T,P = Σ ν k ˆ g k = Σ ν k µ k ≤ 0, (14d) where k and ν k represent the reacting species and its stoichiometric coefficient for a reaction. In case of the reaction CO+ 1/2O 2 → CO 2 , ν CO = -1, ν O2 = -1/2, ν CO2 = 1. Equation (14c) can be alternately expressed in the form µ CO + ν O2 µ O 2 > µ CO 2 . Defining the chemical force for the reactants and products as F R = µ CO + (1/2) µ O 2 , and F P = µ CO 2 for reaction CO+ 1/2O2 →CO2 (15) The criterion dG T,P < 0 leads to the relation F R > F P . This criterion is equally valid for an adia- batic closed rigid system (U, V, m specified), and adiabatic and isobaric system (H, P, m speci- fied), isentropic rigid closed system (S, V, m specified), isentropic and isobaric systems (S, P, m specified), and, finally, isothermal and isovolume systems (T, V, m specified). From Eq. (2), dξ = dN k /ν k = dN CO /(-1) = dN CO2 /(+1) = 0.002. Replacing 0.002 in Eq. (14c) by dξ, the stoichiometric coefficients by ν k , and generalizing for any reaction (∂G/∂ξ) T,P = Σµ k ν k < 0. (16) The Gibbs energy decreases as the reaction progresses and eventually reaches a minimum value at equilibrium. Defining the chemical affinity as F = –(∂G/∂ξ) T,P , (17) Equation (16) assumes the form (-(∂G/∂ξ) T,P = F = -Σµ k ν k ) > 0. (18a) Similarly, following the relations for dA, dU and dS, we can show that (-(∂A/∂ξ) T,V = F = -Σµ k ν k ) > 0, (18b) (-(∂U/∂ξ) S,V = F = - Σµ k ν k ) > 0, (19a) (-(∂H/∂ξ) S,P = F = - Σµ k ν k ) > 0, and (19b) (T(∂S/∂ξ) U,V = T(∂S/∂ξ) H,P =F = -Σµ k ν k ) > 0. (20) The last expression shows that the entropy increases in an isolated system as chemical reaction proceeds. For a reaction to proceed under any of these constraints, the affinity F > 0. In the CO oxidation example, the values of F for the reactants and products are F R = µ CO + (1/2) µ O 2 , and F P = µ CO 2 . (21) Since (F R – F P ) > 0 for oxidation to proceed, F R > F P , (22) which is similar to the inequality T hot > T cold that allows heat transfer to occur from a hotter to a colder body. In a manner similar to the temperature (thermal potential), F R and F P are analo- gous intensive properties called chemical force potentials. The chemical potential µ k is the same as partial molal Gibb’s function ) g k , (= ) h k - T ) s k ), which is a species property. Each spe- cies has a unique way of distributing its energy and, thus, fixing the entropy. A species distrib- uting energy to a larger number of states has a low chemical potential and is relatively more stable. During chemical reactions, the reacting species proceed in a direction to form more stable products (i.e., towards lower chemical potentials). The physical meaning of the reaction potential is as follows: For a specified temperature, if the population of the reacting species (e.g., CO and O 2 ) is higher (i.e., higher value of F R ) than the product molecules (i.e., CO 2 at lower F P ), then there is a high probability of collisions amongst CO and O 2 resulting in a reac- tion that produces CO 2 . On the other hand, if the population of the product molecules (e.g., CO 2 ) is higher (larger F P value) as compared to the reactant molecules CO and O 2 (i.,e., lower F R ), there is a higher probability of collisions amongst CO 2 molecules which will break into CO and O 2 . If the temperature is lowered, the molecular velocities are reduced and the transla- tional energy may be insufficient to overcome bond energy among the atoms in the molecules that is required to the potential F(T, P X i ). a. Example 1 Five kmole of CO, three of O 2 , and four of CO 2 are instantaneously mixed at 3000 K and 101 kPa at the entrance to a reactor. Determine the reaction direction and the val- tor? Solution We assume that if the following reaction occurs in the reactor: CO+ 1/2 O 2 → CO 2 , then (A) F R > F P (B) so that the criterion dG T,P < 0 is satisfied. The reaction potential for this reaction is F R = (1) µ CO + (1/2) µ O 2 , and (C) F P = (1) µ CO 2 . (D) For ideal gas mixtures, µ CO = ˆ g CO = g CO (T,P) + R T ln X CO = g CO (T,p CO ). (E) The larger the CO mole fraction, the higher the value of µ CO and, hence, F. g CO (T,P) = h CO (T,P) – T s CO (T,P) = ( h f,CO 0 + ( h t,3000K – h t,298K ) CO )– 3000×( s CO o (3000) – 8.314(ln×P/1)) = (–110530+93541) –3000×273.508–8.314×ln 1) g CO = –837513 kJ per kmole of CO. (F) Similarly, at 3000K and 1 bar, g O 2 = –755099 kJ kmole –1 , and g CO 2 = –1242910 kJ kmole –1 . (G) The species mole fractions X CO = 5÷(5+3+4) = 0.417, X O 2 = 3÷(5+4+3) = 0.25, and X CO 2 = 0.333. (H) Further, µ CO = ˆ g CO (3000K, 1 bar, X CO = 0.417) = g CO (3000K, 1 bar) + 8.314×3000×ln(0.417) = –837513 + 8.314 × 3000 × ln 0.467 = –856504 kJ kmole –1 of CO in the mixture. (I) Similarly, µ O 2 = (3000K, 1 bar, X O 2 =0.25) = –789675 kJ per kmole of O 2 . (J) µ CO 2 = (3000K, 1 bar, X CO 2 =0.333) = –1270312 kJ per kmole of CO 2 . (K) Therefore, based on the oxidation of 1 kmole of CO, F R = –856504 + 1/2(–789675) = –1254190 kJ, and (L) F P =–1270312 kJ, i.e., (M) ues of F R , F P , and G. What is the equilibrium composition of the gas leaving the re- actor? How is the process altered if seven kmole of inert N 2 is injected into the reac- F R > F P , (N) which implies that assumed direction is correct and hence CO will oxidize to CO 2 . The oxidation of CO occurs gradually. As more and more moles of CO 2 are produced, its molecular population increases, increasing the potential F P . Simultaneously, the CO and O 2 populations decrease, thereby decreasing the reaction potential F R until the reaction ceases when chemical equilibrium is attained. Thus chemical equilibrium is achieved when F R = F P , i.e., dG T,P =0 . This is illustrated in Figure 1. The correspond- ing species concentrations are N CO 2 = 5.25 kmole, N CO = 3.75 kmole, and N O 2 = 2.375 kmole. (Recall the evaporation example discussed in Chapter 7 where A reaches a min imum value at specified values of T, V and G. From a thermodynamic perspective, this problem is similar to placing a cup of cold water in bone dry air. Evaporation will oc- cur when dG T,P < 0, but after a finite amount of water is transformed into the vapor, evaporation will cease at which g H2O( l ) = g H2O(g) and dG T,P = 0.) The Gibbs energy at any section G = Σµ k N k = µ CO N CO + µ O 2 N O 2 + µ CO 2 N CO 2 , i.e., G = -856504×5 -789675×3-1270312×4 = -11,732,793 kJ. Figure 2 plots values of G vs N CO2 . The plot in Figure 2 shows that G reaches a mini- mum value when F R =F P . Nitrogen does not participate in the reaction. Therefore, dN N 2 = 0 and, so, the expres- sions for F R and F P are unaffected. However, the mole fractions of the reactants change so that the values of F R and F P are different, as is the equilibrium composition. The G expression for this case is G = Σµ k N k = µ CO N CO + µ O 2 N O 2 + µ CO 2 N CO 2 + µ CO 2 N CO 2 + µ N2 N N2 -1280 -1270 -1260 -1250 -1240 4 4.5 5 5.5 6 N CO2 , kmole F, MJ F R F P Equilibrium BDCE Figure 1: The reaction potential with respect to the number of moles of CO 2 produced. [...]... 49.999) ≈ 0.00002, and Consequently, s O2 = 205.03 – 8. 314 × ln (0.999) ≈ 205.03 kJ K–1 kmole–1, s CO2 = 213.74 – 8. 314 × ln (0.0002) ≈ 303.70 kJ K–1 kmole–1, g O2 = –61099 kJ kmole–1, and g CO2 = -393546- 2 98 × 303.70 = – 484 0 48 kJ kmole–1 For this reaction FR = –1710+(–61099) = –6 281 0 kJ, and FP = = – 484 0 48 kJ kmole–1, i.e., FR – FP = –6 281 0 + 484 0 48 = 4212 38 kJ (G) -3.00E+07 -3.01E+07 -3.01E+07 -3.02E+07... 0 (T) - T sk 0 (T) Using Table A -8, o gCO = –110530 + 49517 – 180 0 × 254 .8 = –519650 kJ kmole–1 Similarly, using Table A-9, o gO2 = 51660 - 180 0 x 264.701 = -42 480 0 kJ kmole–1, and from Table A-19 o gCO2 = – 393546 +79399 – 180 0 × 302 .89 2 = -85 9355 kJ kmole–1, and ∆Go = (1)×(–519650)+(1/2)×(–42 480 0)+(–1)×( 85 9355)= 127305 kJ (If one uses the “g’ values in Tables A -8, A-9, and A-19, then ∆Go = (1)×(–269164)+(1/2)×(0)+(–1)×(–396425)=... l ) )313 K, 1 bar H H 2 2 2 = – 285 830+4. 184 ×(40–25)× 18. 02–313×(69.95+4. 184 × 18. 02×ln (313÷2 98) ) = –307752 kJ kmole–1 = g o O ( l ) (313 K, 10 bar), H 2 (M) which almost equals the previous answer (Eq (H)) ˆ g id O ( l ) (T,P) = g H O ( l ) (T,P) + R T ln X H2O , i.e., H 2 2 ˆ g id O ( l ) (313 K, 10 bar) = g H O ( l ) (313 K, 10 bar) + H 2 2 8. 314 × 313 × ln 0 .85 = –3 081 59 kJ kmole–1 Remarks We have... H 2O = – 285 830 – 2 98 69.95 = –306675 kJ kmole–1 of H2O(l) , o gCO = –110530 – 2 98 197.56 = –169403 kJ kmole–1 of CO, o gCO2 = –393520 – 2 98 213.7 = –457203 kJ kmole–1 of CO2, and o gH2 = 0 – 2 98 130.57 = – 389 10 kJ kmole–1 of H2 Therefore, ∆Go = – 389 10 – 457203 – (–169403 – 306675) (A) = –20035 kJ kmole–1 of CO, and K (2 98 K) = exp (20,035÷ (8. 314×2 98) ) = 3250.4 o Remarks Since Ko(2 98 K) is extremely... is a pure component and hence the activity a C(s) = 1 Further, hC = hfo,C + ∫ T 2 98 K c p,C dT where hfo,C = 0 kJ kmole–1, and o s C= sC (298K) + ∫ (B) T 2 98 K (c p,C / T)dT Now, o sC (298K) = 5.74 kJ kmole–1 K–1 (C) o Hence, using Eqs (B) and (C), gC = g C(298K, 1 bar) = h 298K – 2 98 s C (298K), i.e., o gC = 0 – 2 98 5.74 = –1711 kJ kmole–1 (D) o For solids and liquids, gk (T, P) ≈ gk (T) Assume... respectively ∆G o (2 98 K) = g CO – ( g C + 1/2 g O2 ), I (J) ∆G o (2 98 K) = g CO2 – ( g C + g O2 ),and II (K) ∆G o (2 98 K) = g CO2 – ( g CO + 1/2 g O2 ), IV (L) o where the g k’s are evaluated at 2 98 K, i.e., g k= gk Equation (L) assumes the form ∆G o (2 98 K) = g CO2 –( g C + g O2 )–{ g CO–( g C+1/2 g O2 )}= ∆G o – ∆G o , i.e., (M) IV II I ∆G o (2 98 K) =∆Go II- ∆Go I = g CO2 (2 98 K) – g CO(2 98 K) IV = –394390... 0.07 384 bar and H2O(g)(T, fH2O( l ) (313 K, 1 bar) = 0.074 bar Psat) (E) Using Eqs (A) and (E), g o O ( l ) (313 K,10 bar) ≈ g o O ( g ) (313 K, 1 bar) + 8. 314×313×ln (0.07 384 /1) H H 2 2 (F) Now, o g o O ( g ) (313 K, 1 bar) = ( h o O ( g ) – T s H O ( g ) )313 K, 1 bar H H 2 2 2 = –241321 – 313 × 190.33 = –30 089 4 kJ kmole–1 (G) Using Eqs (A), (E), and (F), g o O ( l ) (313 K,10 bar) = –30 089 4 + 8. 314... 2 98 K, ∆Ho = 336500 kJ kmole–1 and ∆So = 455 .8 kJ kmole–1 K Determine the transition temperature (i.e at which K0(T) =1) Solution Recall from Eq (65) that Ko(T) = Ko(To) exp{(–∆Ho/ R ) (1/T – 1/To)}, (A) where K(T0) = exp (–∆Go/ R T0) and ∆Go = ∆Ho – T∆So = 336500 – 2 98 455 .8 = 200672 kJ kmole–1, i.e., Ko(To) = exp (–∆Go/ R To) = 6.67×10–36, or (B) Ko(T) = 6.67×10–36 exp {– (336500 /8. 314)(1/T – 1/2 98) }... 9 .86 8 and B = 33742.4 in the appropriate units Assume that CO is in trace amounts Solution Assume that the overall combustion reaction is represented by the equation (B) CH4 + a (O2 + 3.76 N2) → CO2 + 2H2O + b N2 + d O2 It is by now straightforward to determine that a = 2.2 982 , b = 8. 641, and d = 0.2 982 The carbon dioxide concentration in the exhaust on a wet basis is 100 × (1 ÷ (1 + 2 + 3.76 × 2.2 982 ... Therefore, s O2 = 205.03 – 8. 314 × ln 0.9999 = 205.03 kJ K–1 kmole–1, i.e., -3.00E+06 -3.05E+06 GI -3.10E+06 -3.15E+06 N G, kJ -3.20E+06 -3.25E+06 G II -3.30E+06 -3.35E+06 -3.40E+06 M -3.45E+06 -3.50E+06 0 0.2 0.4 0.6 NC, consumed 0 .8 1 Figure 4: Variation in GI and GII with respect to the number of moles of carbon consumed for reactions I and II at 2 98 K g O2 (298K, 1 bar) = 0 – 2 98 × 205.03 = –61099 kJ . kmole –1 . For this reaction F R = –1710+(–61099) = –6 281 0 kJ, and F P = = – 484 0 48 kJ kmole –1 , i.e., (G) F R – F P = –6 281 0 + 484 0 48 = 4212 38 kJ. -3.50E+06 -3.45E+06 -3.40E+06 -3.35E+06 -3.30E+06 -3.25E+06 -3.20E+06 -3.15E+06 -3.10E+06 -3.05E+06 -3.00E+06 0. (l) g HO o 2 ()l (313 K, 1 bar) = ( h HO o 2 ()l – T s HO o 2 ()l ) 313 K, 1 bar = – 285 830+4. 184 ×(40–25)× 18. 02–313×(69.95+4. 184 × 18. 02×ln (313÷2 98) ) = –307752 kJ kmole –1 = g HO o 2 ()l (313 K, 10 bar), (M) which. kmole –1 K –1 . (C) Hence, using Eqs. (B) and (C), g C o = g C (298K, 1 bar) = h 298K – 2 98 s C (298K), i.e., g C o = 0 – 2 98 5.74 = –1711 kJ kmole –1 . (D) For solids and liquids, gTP k (,) ≈