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Chapter 7: Chemical Equilibrium

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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 = dn A /v A ( where v A 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, dn A = -dξ On the other hand, dn B = dξ

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Example 1: N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) when the extent of reaction changes from ξ = 0 to ξ = 1.0 mole, what are the changes of each reagent? Solution: identify v j : v(N 2 ) = -1; v(H 2 ) = -3; v(NH 3 ) = 2. since dξ = 1.0 mole, dn(N 2 ) = -1x1.0 mole = -1.0 mole, dn(H 2 ) = -3x1.0 mole = -3.0 moles, dn(NH 3 ) = 2x1.0 mole = 2.0 moles, Example 2: CH 4 ( g ) + Cl 2 ( g ) ↔ CHCl 3 ( l ) + HCl( g ), in which the amount of reactant Cl 2 ( g ) decreases by 2 moles. What is the extent of the reaction? Solution:

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Variation of Gibbs energy during a reaction process

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

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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

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The calculation of reaction Gibbs energy (Δ r G) Consider the reactionA ↔ B initial amount: n A0 n B0 final amount: n Af n Bf G initial = u B n B0 + u A n A0 G final = u B n Bf + u A n Af ΔG = G final - G initial = ( u B n Bf + u A n Af ) – ( u B n B0 + u A n A0 ) = u B (n Bf - n B0 ) + u A (n Af - n A0 ) = u B Δξ + u A (-Δξ) = ( u B - u A )Δξ Δ r G = = u B - u A When u A > u B, the reaction A → B is spontaneous. When u B > u A, the reverse reaction (B → A) is spontaneous. When u B = u A, the reaction is spontaneous in neither direction (equilibrium condition).

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7.2 The description of equilibrium 1. Perfect gas equilibrium: A (g) ↔ B (g) Δ r G = u B – u A

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At equilibrium, Δ r G = 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, Δ r G θ = Δ f G θ (B) - Δ f G θ (A)

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Equilibrium of a general reaction Example: 2A + B ↔ C + 3D The reaction Gibbs energy, Δ r G, 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 α j ν j v j are the corresponding stoichiometric numbers; positive for products and negative for reactants. v A = -2; v B = -1; v C = 1; v D = 3

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Justification of the equation dG = ∑ u j dn j Assuming that the extent of reaction equals dξ, one gets dn j = v j dξ then dG = ∑ u j dn j = ∑ u j v j dξ (dG/dξ) = ∑ u j v j Δ r G = ∑ u j v j because u j = u j θ + RTln(a j ) Δ r G = ∑{ v j ( u j θ + RTln(a j ) )} = ∑( v j u j θ ) + ∑ v j ( RTln(a j ) ) = Δ r G θ + RT ∑ ln(a j ) vj = Δ r G θ + RT ln (∏ (a j ) vj ) = Δ r G θ + RT ln ( Q)

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Again, we use K to denote the reaction quotient at an equilibrium point, K = Q equilibrium = ( ∏ a j vj ) 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: v A = -1, v B = -2, v C = 3, and v D = 4. At the equilibrium condition:

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Examples of calculating equilibrium constants 1.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: Compoundμ θ (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:

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Example 2: Using the data provided in the data section, calculate the standard Gibbs energy and the equilibrium constant at 25 o C for the following reaction CH 4 (g) + Cl 2 (g) ↔ CHCl 3 (g) + HCl(g) Solution: (chalkboard) Δ f G θ (CHCl 3, g) = -73.66 kJ mol -1 Δ f G θ (HCl, g) = - 95.30 kJ mol -1 Δ f G θ (CH 4, g) = - 50.72 kJ mol -1

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