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Chapter 7: Chemical Equilibrium

Chapter 7: Chemical Equilibrium. 7.1 The Gibbs energy minimum. 1. Extent of reaction ( ξ ): The amount of reactants being converted to products. Its unit is mole. In a very general way, the extent of reaction is calculated

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Chapter 7: Chemical Equilibrium

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  1. Chapter 7: Chemical Equilibrium

  2. 7.1 The Gibbs energy minimum 1. Extent of reaction (ξ): The amount of reactants being converted to products. Its unit is mole. In a very general way, the extent of reaction is calculated as dξ = dnA/vA(where vA is the stoichiometric number of the reactant A, which is negative for the reactant!!) • Why do we need a new quantity? Consider a generic reaction: 2A ↔ 3B The amount of A consumed is different from the amount of B produced. Which number should be used in the report? • Consider: A ↔ B Assume an infinitesimal amount dξ of A turns into B, dnA = -dξ On the other hand, dnB = dξ

  3. Example 1: N2(g) + 3H2(g) ↔ 2NH3(g) when the extent of reaction changes from ξ = 0 to ξ = 1.0 mole, what are the changes of each reagent? Solution: identify vj: v(N2) = -1; v(H2) = -3; v(NH3) = 2. since dξ = 1.0 mole, dn(N2) = -1x1.0 mole = -1.0 mole, dn(H2) = -3x1.0 mole = -3.0 moles, dn(NH3) = 2x1.0 mole = 2.0 moles, Example 2: CH4(g) + Cl2(g) ↔ CHCl3(l) + HCl(g), in which the amount of reactant Cl2(g) decreases by 2 moles. What is the extent of the reaction? Solution:

  4. Variation of Gibbs energy during a reaction process

  5. The reaction Gibbs energy: ΔrG the slope of the Gibbs energy plotted against the extent of reaction: ΔrG = here Δr signifies a derivative • A reaction for which ΔrG < 0 is called exergonic. • A reaction for which ΔrG > 0 is called endergonic. • ΔrG < 0, the forward reaction is spontaneous. • ΔrG > 0, the reverse reaction is spontaneous. • ΔrG = 0, the reaction is at equilibrium!!!

  6. Molecular interpretation of the minimum in the reaction Gibbs energy Gibbs energy of the system decreases as the reaction progress Gibbs energy of a system consisting of different portions of reactants and products

  7. The calculation of reaction Gibbs energy (ΔrG) • Consider the reaction A ↔ B initial amount: nA0 nB0 final amount: nAf nBf Ginitial = uBnB0 + uAnA0 Gfinal = uBnBf + uAnAf ΔG = Gfinal - Ginitial = (uBnBf + uAnAf ) – (uBnB0 + uAnA0 ) = uB(nBf- nB0) + uA (nAf - nA0) = uBΔξ + uA(-Δξ) = (uB - uA )Δξ ΔrG = = uB - uA When uA > uB, the reaction A → B is spontaneous. When uB > uA, the reverse reaction (B → A) is spontaneous. When uB = uA, the reaction is spontaneous in neither direction (equilibrium condition).

  8. 7.2 The description of equilibrium 1. Perfect gas equilibrium: A(g)↔ B(g) ΔrG = uB – uA

  9. At equilibrium, ΔrG = 0, therefore Note: The difference in standard molar Gibbs energies of the products and reactants is equal to the difference in their standard Gibbs energies of formation, thus, ΔrGθ = ΔfGθ(B) - ΔfGθ(A)

  10. Equilibrium of a general reaction Example: 2A + B ↔ C + 3D • The reaction Gibbs energy, ΔrG, is defined in the same way as discussed earlier: where the reaction quotient, Q, has the form: Q = activities of products/activities of reactants in a compact expression Q = Π αjνj vj are the corresponding stoichiometric numbers; positive for products and negative for reactants. vA = -2; vB = -1; vC = 1; vD = 3

  11. Justification of the equation dG = ∑ujdnj Assuming that the extent of reaction equals dξ, one gets dnj = vjdξ then dG = ∑ujdnj = ∑ujvjdξ (dG/dξ) = ∑ujvj ΔrG = ∑ujvj because uj = ujθ + RTln(aj) ΔrG = ∑{vj(ujθ + RTln(aj))} = ∑(vjujθ) + ∑vj(RTln(aj)) = ΔrGθ + RT ∑ln(aj)vj = ΔrGθ + RT ln (∏(aj)vj) = ΔrGθ + RT ln (Q)

  12. Again, we use K to denote the reaction quotient at an equilibrium point, K = Qequilibrium = (∏ajvj)equilibrium K is called thermodynamic equilibrium constant. Note that until now K is expressed in terms of activities) Example: calculate the quotient for the reaction: A + 2B ↔ 3C + 4D Solution: first, identify the stoichiometric number of each reactant: vA = -1, vB = -2, vC = 3, and vD = 4. At the equilibrium condition:

  13. Examples of calculating equilibrium constants • Consider a hypothetical equilibrium reaction A(g) + B(g) ↔ C(g) + D(g) While all gases may be considered ideal. The following data are available for this reaction: Compound μθ(kJ mol-1) A(g) -55.00 B(g) -44.00 C(g) -54.00 D(g) -47.00 Calculate the value of the equilibrium constant Kp for the reaction at 298.15K. Solution:

  14. Example 2: Using the data provided in the data section, calculate the standard Gibbs energy and the equilibrium constant at 25oC for the following reaction CH4(g) + Cl2(g) ↔ CHCl3(g) + HCl(g) Solution: (chalkboard) ΔfGθ(CHCl3, g) = -73.66 kJ mol-1 ΔfGθ(HCl, g) = - 95.30 kJ mol-1 ΔfGθ(CH4, g) = - 50.72 kJ mol-1

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