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Introduction to Equilibrium

Introduction to Equilibrium. AP Chemistry. Equilibrium. Defined as the state when the concentrations of all reactants and products remain constant over time. This occurs when the rate of the forward = rate of the reverse reaction.

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Introduction to Equilibrium

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  1. Introduction to Equilibrium AP Chemistry

  2. Equilibrium • Defined as the state when the concentrations of all reactants and products remain constant over time. • This occurs when the rate of the forward = rate of the reverse reaction. • Equilibrium happens because reactants and products never stop colliding!

  3. Equilibrium Graph • After 60 seconds the reaction has reached equilibrium • The concentrations of the reactants and products are constant, but not necessarily equal!!! • Rate of forward reaction = rate of reverse reaction

  4. Not all reactions will go to completion. The equilibrium expression will tell us how much of each species is present at equilibrium • The Keq is constant at a given temperature and depends on the equilibrium concentrations of products and reactants. • In a general reaction: • aA + bB ⇄ cC + dD • The lower case letters represent the stoichiometric coefficients and the upper case letters represent reactants and products. • The law of Mass Action states that the • Equlibrium constant is: • Keq = [C]c[D]d • A]a[B]b

  5. Example 1 • What is the equilibrium expression (Keq) for the following dissociation? • 2N2O5(g) ⇄ 4NO2(g) + O2(g) [NO2]4[O2] [N2O5]2 Keq =

  6. Equilibrium Constants • When Keq > 1 then products are favored in the reaction • Which means that they are more products than reactants at equlibrium • When Keq < 1 then the reactants are favored in the reaction • Which means there are more reactants then products at equlibrium.

  7. Equilibrium Constants • When a reaction is multiplied by a factor of “n”, then the Keq is raised to the ‘nth’ power. • 2H2O  2H2 + O2Keq = [H2]2[O2] • [H2O]2 • Multiply by 2 4H2O  4H2 + 2O2 (Keq)2 = [H2]4[O2]2 [H2O]4

  8. Equilibrium Constants • The equilibrium constant for the reverse reaction is Keq(rev) = 1/Keq(fwd)

  9. Heterogenous Equilibria • Pure solids and pure liquids, and pure water are not included in equilibrium expressions because there concentrations remain essentially constant over time. • Write the equilibrium expression for the following • PCl5 (s) ↔ PCl3 (l) + Cl2(g) Keq = [Cl2]

  10. Kp versus Kc • In homogeneous gasequilibria, partial pressure may be used in place of the concentrations—the format remains the same. • The term Kp is often used to describe the equilibrium constant in terms of partial pressure. • Kc may be used when the concentrations of the reactants and products are expressed in terms of molarities. • Numerical values can be calculated for any of these. • By convention, units are usually omitted for any equilibrium constant. • Water is excluded in aqueous media because the concentration of water does not change and/or water as a product of a reaction is insignificant in aqueous media.

  11. Example 2 • In the homogeneous gas equilibrium, dinitrogen tetraoxide decomposes as follows: • N2O4 (g) ⇄ 2NO2(g) • Kc = Kp = [NO2]2 [N2O4] (PNO2)2 PN2O4

  12. The Relationship Between Kp and Kc • For the general reaction: • aA(g) ⇄ bB • Using the Ideal Gas Law: • pA V = nRT and pA = nA RT • V • Similarly: pB = nB RT • V

  13. Substituting these terms into the expression for Kp: nB RT V b Kp = nART V a Since n/V is concentration (Molarity):

  14. Since n/V is concentration (Molarity): Kp = [B]b [A]a (RT) Δn Kp = Kc(RT) Δn Δn = Σ (moles products) – Σ(moles reactants) Δn = b - a

  15. Sample Problem • For the reaction: • N2(g) + 3H2(g) ⇄ 2NH3(g) •  The following equilibrium pressures at 472oC were obtained: H2 at 7.38atm, N2 at 2.46 atm and NH3 at 0.166 atm. From these data, calculate the Kp and the Kc at this temperature.

  16. Determining Keq :Quantitative Equilibium What is the equilibrium constant for the following reaction in terms of Kc? • 2NO(g) +O2(g) ⇄ 2NO2(g) • The reaction was run at 230o C and the following concentrations were obtained: • [NO]= 4.065x 10-2 M • [O2] =9.525 x 10-2M • [NO2] = 11.625 M

  17. Sample Problem 2 • The Kp for phosgene gas (COCl2) at 800K is 3.3. At equilibrium, the pressure of CO was found to be 0.455 atm and that of Cl2 was found to be 0.356 atm. What is the partial pressure of the phosgene at equilibrium? The overall reaction is: • COCl2(g) ⇄ CO(g) + Cl2(g)

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