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Group 14

Group 14. CHAPTER EIGHT REACTIONS INVOLVING GASES. 8.1 Introduction. As result of no interactive forces among the particles of perfect gases, no heat effects occur when two (or more) of such gases are mixed together.

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Group 14

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  1. Group 14

  2. CHAPTER EIGHT REACTIONS INVOLVING GASES

  3. 8.1 Introduction • As result of no interactive forces among the particles of perfect gases, no heat effects occur when two (or more) of such gases are mixed together. • On the other hand, when H2 and O2 gases are mixed in presence of a suitable catalyst, the heat released is of considerable magnitude. • The thermodynamics of such a system can be treated in either of two ways: either the mixture can be regarded as being a highly non-ideal gas mixture of H2 and O2, the thermodynamic equilibrium state of which at given T and P can be described in terms of the fugacities of H2 and O2; or the H2 and O2 have reacted with one another to some extent to give use to the physical appearance of H2O. The latter, is by far the more convenient one.

  4. 8.2 Reaction Equilibrium in Gas Mixture to form Reaction's Products • Consider the process: • A(g) + B(g) = 2C(g) , • and by starting with 1 mole of A and 1 mole of B; and as 1 atom of A reacts with 1 atom of B to produce 2 atoms of C, stociometric conditions, and assuming that • the extent of the process at time t is y then at time t: • n(A) = n(B) = 1-y • where n(A) and n(B) are the number of moles of A and B, respectively, at time t.

  5. Consider that the process occurs at constant T and P, thus at any moment during the process, the integral free energy of the system is given by: G' = n(A) G(A) + n(B) G(B) + n(C) G (C) • Let us consider the following setting: • A B C total n at t = 0 1 1 0 2 • n at t = t n n 2(1-n) 2 • mole fraction (x) n/2 n/2 (1-n) -- • extent of the process (1-n) (1-n) -- -- • since: G(i) = Go(i)+ RTlnx(i)P

  6. thus, if the pressure=1 atm, the free energy of the system can be given by: G' = n(Go(A)+ RT (A)) + n (Go(B)+RT (B)) • +2(1-n) (Go(B (C)+RT • =nGo(A) + nGo(B) + 2(1-n) Go(C) +2n RT ln (n/2)+2(1-n)RT ln (1-n) • = n Go A + n Go(B) + 2Go(C) – 2n Go(C) + 2n RT ln (n/2)+2(1-n)RT ln (1-n) therefore: G' – (Go(A) + Go(B)) =(1-n)(2Go(C) - Go(A) - Go(A)) + 2RT(n ln(n/2) + (1-n) ln (1-n) or: ΔG = (1-n) ΔGo(R) + 2RT(n ln (n/2) + (1-n) ln (1-n))

  7. The right-hand side of the previous equation has two terms, the first, (1-n) ΔG0(R), represents the standard free energy change due to the chemical reaction, and the second term 2RT [ n ln n/2 + (1-n) ln (1-n) ] represents the free energy change due to mixing of the gases. • Figure (8.1) shows the variation of the contribution to the free energy change due to the chemical reaction (line II), the contribution to the free energy change due to gas mixing (curve III) and the sum of these two contribution ( curve I ) with the extent of the reaction: • A(g) + B(g) = 2C(g) • for which ΔG0(R) = -1000 cal at 500K

  8. Figure (8.1): the variation of the contribution to the free energy change for the reaction A+B=2C for which ΔG0(R)= -1000 cal at 500K.

  9. 8.3 The Equilibrium Constant • The equilibrium point of the previous reaction occurs at the minimum point of curve 1, point S, which can be obtained at the condition of (∂ΔG/∂n)T,P= 0 ; this means that: -ΔGo(R)+ 2RT(1+ln(n/2)-1-ln(1-n))=0 i.e. ΔGo(R)= -RT ln (1-n)2/(n/2)2 or ΔGo(R) = -RT ln [X(C) /x (A) x (B)] if P is constant, thus: ΔGo(R) = -RT ln [P(C)2 /P(A) P(B)]

  10. in general as: G' = n(A) G(A) + n(B) G(B) + 2(1-n(A)) G (C) and by differentiation and setting (∂G'/∂n)T,P= 0 ; we have: • at equilibrium, G(A)+G(B)-2G(C)=0 i.e. the reaction equilibrium occurs at Go(A) + RT ln a(A) + Go(B) +RT ln a(B)= 2 Go(C) + 2RT ln a(C) where: a(i) =P(i) / P0(i) thus; ΔGo(R) = -RT ln (a2(C) /a(A) a(B))eq = -RT ln (P2(C)/P(A) P(B))eq = -RT ln (P2(C)/P(A) P(B)) and hence: ΔGo(R) = -RT lnKp thus; Kp = ( P(C)2/P(A) P(B) ) where Kp is known as the equilibrium constant.

  11. 8.4 The Effect of Temperature on theEquilibrium Constant (van't Hoff equation) • The Gibbs – Helmholtz equation states that: As ΔGo = -RT lnKP then (∂ln K/∂(1/T))P= -ΔHo/R • Thus, for endothermic reaction, the equilibrium constant increases with increasing temperature, i.e, the equilibrium shift in that direction which involves absorption of heat, i.e. in the forward direction.

  12. Conversely, for exothermic reaction, the equilibrium constant decreases with increasing temperature, i.e. the equilibrium shift in that direction which involves the absorption of heat, i.e. in the backward direction. It is clear now that the variation of K­pwith temperature obeys the le-Chatelier's principle, i.e if heat is added to a system, the reaction equilibrium is displaced in that direction which involves absorption of heat van't Hoff equation shows also that if ΔHo is independent of temperature, then In Kp varies linearly with (1/T) which agrees with the experimentally obtained Arrhenius equation

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