Ch. 14: Chemical Equilibrium

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Ch. 14: Chemical Equilibrium. Dr. Namphol Sinkaset Chem 201: General Chemistry II. I. Chapter Outline. Introduction The Equilibrium Constant (K) Values of Equilibrium Constants The Reaction Quotient (Q) Equilibrium Problems Le Châtelier’s Principle. I. Introduction. - PowerPoint PPT Presentation

Transcript of Ch. 14: Chemical Equilibrium

Ch. 14: Chemical EquilibriumCh. 14: Chemical Equilibrium

Dr. Namphol SinkasetChem 201: General Chemistry II

I. Chapter OutlineI. Chapter Outline

I. IntroductionII. The Equilibrium Constant (K)III. Values of Equilibrium ConstantsIV. The Reaction Quotient (Q)V. Equilibrium ProblemsVI. Le Châtelier’s Principle

I. IntroductionI. Introduction• Equilibrium will be the focus for the next

several chapters.• Most reactions are reversible, meaning

they can proceed in both forward and reverse directions.

• This means that as products build up, they will react and reform reactants.

• At equilibrium, the forward and backward reaction rates are equal.

I. Example EquilibriumI. Example Equilibrium

II. Equilibrium ConcentrationsII. Equilibrium Concentrations

• Equilibrium does not mean that concentrations are all equal!!

• However, we can quantify concentrations at equilibrium.

• Every equilibrium has its own equilibrium constant.

II. The Equilibrium ConstantII. The Equilibrium Constant

• equilibrium constant: the ratio at equilibrium of the [ ]’s of products raised to their stoichiometric coefficients divided by the [ ]’s of reactants raised to their stoichiometric coefficients.

• The relationship between a balanced equation and equilibrium constant expression is the law of mass action.

II. The Equilibrium ConstantII. The Equilibrium Constant

• For a general equilibrium aA + bB cC + dD, the equilibrium expression is:

II. Sample “Problem”II. Sample “Problem”

• Write the equilibrium constant expression for the reaction:

2H2(g) + O2(g) 2H2O(g).

II. Physical Meaning of KII. Physical Meaning of K

• Large values of K mean that the equilibrium favors products, i.e. there are high [ ]’s of products and low [ ]’s of reactants at equilibrium.

• Small values of K mean that the equilibrium favors reactants, i.e. there are low [ ]’s of products and high [ ]’s of reactants at equilibrium.

II. Rules for Manipulating KII. Rules for Manipulating K

• If the equation is reversed, the equilibrium constant is inverted.

II. Rules for Manipulating KII. Rules for Manipulating K

• If the equation is multiplied by a factor, the equilibrium constant is raised to the same factor.

II. Rules for Manipulating KII. Rules for Manipulating K

• When chemical equations are added, their equilibrium constants are multiplied together to get the overall equilibrium constant.

II. Sample ProblemII. Sample Problem

• Predict the equilibrium constant for the first reaction given the equilibrium constants for the second and third reactions.

CO2(g) + 3H2(g) CH3OH(g) + H2O(g) K1 = ?

CO(g) + H2O(g) CO2(g) + H2(g) K2 = 1.0 x 105

CO(g) + 2H2(g) CH3OH(g) K3 = 1.4 x 107

II. K in Terms of PressureII. K in Terms of Pressure

• Up to this point, we’ve been using concentration exclusively in the equilibrium expressions.

• Partial pressures are proportional to concentration via PV = nRT.

• Thus, for gas reactions, partial pressures can be used in place of concentrations.

II. Two Different K’sII. Two Different K’s

• For the reaction 2SO3(g) 2SO2(g) + O2(g), we can write two equilibrium expressions.

II. Relationship Between II. Relationship Between Concentration and PressureConcentration and Pressure

• To be able to convert between Kc and Kp, we need a relationship between concentration and pressure.

II. Converting Between KII. Converting Between Kcc and K and Kpp

II. Converting Between KII. Converting Between Kcc and K and Kpp

• The Δn is the change in the number of moles of gas when going from reactants to products.

• When does Kp equal Kc?

II. Sample ProblemII. Sample Problem

• Methanol can be synthesized via the reaction CO(g) + 2H2(g) CH3OH(g). If Kp of this reaction equals 3.8 x 10-2 at 200 °C, what’s the value of Kc?

II. Heterogeneous EquilibriaII. Heterogeneous Equilibria

• If an equilibrium contains pure solids or pure liquids, they are not included in the equilibrium constant expression.

III. Values of KIII. Values of K

• Values of K are most easily calculated by allowing a system to come to equilibrium and measuring [ ]’s of the components.

• For the equilibrium H2(g) + I2(g) 2HI(g), let’s say equilibrium [ ]’s at 445 °C were found to be 0.11 M, 0.11 M, and 0.78 M for molecular hydrogen, molecular iodine, and hydrogen iodide, respectively.

III. KIII. Kcc for a H for a H22/I/I22 Mixture Mixture

• Note that units are not included when calculating K’s.

• Thus, equilibrium constants are unitless.

III. Equilibrium [ ]’s Vs. KIII. Equilibrium [ ]’s Vs. K

• For any reaction, the equilibrium [ ]’s will depend on the initial [ ]’s of reactants or products.

• However, no matter how you set up the reaction, the value of the equilibrium constant will be the same if the temperature is the same.

III. Equilibrium [ ]’s Vs. KIII. Equilibrium [ ]’s Vs. K

IV. The Reaction QuotientIV. The Reaction Quotient

• What happens when we mix reactants together and wait?

• Can we predict what will happen when we have a mixture of reactants and products?

• The reaction quotient, Qc or Qp, is used to predict in which direction an equilibrium will move.

IV. Formula for QIV. Formula for Qcc or Q or Qpp

• You already know the formula because it’s the same as for Kc or Kp!!

• The difference is, we don’t know if the reaction is at equilibrium, thus, we cannot set the ratio equal to K!

• For the reaction aA + bB cC + dD:

IV. Using QIV. Using Q

• The value of Q relative to K tells you whether the reaction will form more products or more reactants to reach equilibrium. Q < K means reaction forms products. Q > K means reaction form reactants. Q = K means reaction is at equilibrium.

IV. Sample ProblemIV. Sample Problem

• Consider the reaction N2O4(g) 2NO2(g) with Kc = 5.85 x 10-3. If a reaction mixture contains [NO2] = 0.0255 M and [N2O4] = 0.0331 M, which way will the reaction proceed?