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UNIT 4

UNIT 4. Equilibria. Things to Review for Unit 4. Solving quadratic equations: ax 2 + bx + c = 0 x = -b ± √ b 2 – 4ac 2a Common logarithms log (base 10 logarithms) Calculating the molarity of a solution obtained by dilution…M 1 V 1 = M 2 V 2

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UNIT 4

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  1. UNIT 4 Equilibria

  2. Things to Review for Unit 4 • Solving quadratic equations: ax2 + bx + c = 0 • x = -b ± √ b2 – 4ac • 2a • Common logarithms log (base 10 logarithms) • Calculating the molarity of a solution obtained by dilution…M1V1 = M2V2 • Acid-base neutralization reactions (a type of double replacement reaction) • Titration problems (the stoichiometry of acid-base reactions) • Equation showing an ionic solid dissolving in water.

  3. Chemical Equilibrium It is well-known among people who want to make money from synthesizing a chemical that reactions do not often go to 100% completion. Why? Because reactions involve collisions between molecules and the subsequent making and breaking of bonds. “Products” are just as susceptible to such collisions as “reactants”.

  4. Chemical Equilibrium As soon as a reaction A  B yields any B at all, the process B  A may occur. Remember that the rate of a reaction can depend on the concentration of reactants? As time goes on and more B is formed, the rate of B  A increases while the rate of A  B decreases until the rates are equal. Equilibrium occurs when opposing reactions proceed at equal rates.

  5. Chemical Equilibrium

  6. Chemical Equilibrium A(g) B(g) If this reaction were first order in both directions and had a forward reaction rate constant of 0.0129 s-1, then the following data could apply at 100°C.

  7. Chemical Equilibrium As the partial pressure of A drops, the partial pressure of B increases until the partial pressures each no longer change with time. PB PA

  8. Chemical Equilibrium Notice that the partial pressure of A does not drop all the way to zero. There is A left over when the partial pressure of the product B has leveled. A(g) B(g) Once the reaction reaches equilibrium, it appears to stop, because the partial pressure of product stops increasing. PB PB PA PA

  9. Chemical Equilibrium A(g) B(g) Let’s look at the reaction in terms of rates. Rate of forward reaction = kf [A] = kf PA RT Rate of reverse reaction = kr [B] = kr PB RT for an ideal gas P = MRT At equilibrium, the forward and reverse rates are equal: kf PA = kr PB RT RT kf = PB kr PA The k’s are constants, so at equilibrium, the ratio of the partial pressures is fixed.

  10. Chemical Equilibrium A(g) B(g) The term “dynamic equilibrium” may be used instead of simply “equilibrium.” This reminds us that both forward and reverse reactions continue to occur. At equilibrium, the forward and reverse rates are equal: kf PA = kr PB RT RT kf = PB kr PA RateAB equilibrium starts here RateBA

  11. Chemical Equilibrium A(g) B(g) At equilibrium, the forward and reverse rates are equal: kf PA = kr PB RT RT kf = PB kr PA The k’s are constants. At equilibrium, the ratio of the partial pressures must also be a constant. The value of this ratio is called the equilibrium constant KP. kf = PB = KP kr PA If we were dealing with solutions, at equilibrium the ratio of the concentrations would be a constant KC.

  12. The Equilibrium Constant KP aA(g) + bB(g) cC(g) + dD(g) The law of mass action states that, at equilibrium, the following ratio is a constant: KP = PCc PDd PAa PBb The rate at which the reaction proceeds does not matter. At equilibrium, this ratio holds. Note the use of partial pressures for gases. Concentrations (molarity) may be used, but they will lead to KC, which has a different value than KP.

  13. The Equilibrium Constant KC aA(aq) + bB(aq) cC(aq) + dD(aq) The law of mass action states that, at equilibrium, the following ratio is a constant: KC = [C]c [D]d [A]a [B]b Whether K is based on partial pressures or on molarities, K is unitless. This is because every pressure is ratioed to a standard pressure of 1 bar and every concentration is ratioed to a standard concentration of 1M.

  14. Properties of the Equilibrium Constant KP or KC • K is unitless. • K is a function of temperature. It will change as the temperature changes. • The expression for K depends on the stoichiometry of the reaction. No knowledge of the reaction rate is necessary. • K is NOT the rate constant k.

  15. Examples of K H2(g) + I2(g) 2HI(g) KP = (PHI)2 (PH2)(PI2) CH3COOH(aq) CH3COO-(aq) + H+(aq) KC = [CH3COO-] [H+] [CH3COOH] Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) KC = [Ag(NH3)2+] [Ag+] [NH3]2

  16. Calculating KP A mixture of hydrogen and nitrogen react to form ammonia. At 472°C, the equilibrium mixture of gases contains 7.38 bar H2, 2.46 bar N2, and 0.166 bar NH3. Calculate KP for this reaction. N2(g) + 3H2(g) 2NH3(g) KP = (PNH3)2 (PN2)(PH2)3 KP (472°C) = (0.166)2 = 2.79 x 10-5 (2.46)(7.38)3 We could also calculate a KC for this reaction.

  17. Calculating KC An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25°C: [CH3COOH] = 0.0165 M, [H+] = 5.44 x 10-4 M, and [CH3COO-] = 5.44 x 10-4 M. Calculate KC for the ionization of acetic acid at 25°C. CH3COOH(aq) CH3COO-(aq) + H+(aq) KC = [CH3COO-] [H+] [CH3COOH] KC(25°C) = (5.44 x 10-4) (5.44 x 10-4) = 1.79 x 10-5 (0.0165)

  18. What Does K Tell Us About a Reaction? CH3COOH(aq) CH3COO-(aq) + H+(aq) KC = [CH3COO-] [H+] = 1.79 x 10-5 [CH3COOH] The small value of KC says that the ratio of products (acetate and hydronium ions) to reactants (acetic acid) is small. In other words, acetic acid does not ionize to a great extent in water at 25°C. When K is much less than 1, there are fewer products than reactants, and the forward reaction is not favored. This means the reverse reaction IS favored. We say, “The equilibrium lies to the left.”

  19. What Does K Tell Us About a Reaction? H2(g) + I2(g) 2HI(g) KP = (PHI)2 (PH2)(PI2) KP (298 K) = 794 and KP (700 K) = 54 These values of KP tell us that the forward reaction is favored at both room temperature and at 700 K. However, the values also tell us that the forward reaction is more favored at room temperature. When K is much greater than 1, there are more products than reactants, and the forward reaction is favored. We say, “The equilibrium lies to the right.” How do we interpret K ≈ 1?

  20. K for the Reverse Reaction? H2(g) + I2(g) 2HI(g) KP = (PHI)2 (PH2)(PI2) KP (298 K) = 794 and KP (700 K) = 54 2HI(g) H2(g) + I2(g) KP = (PH2)(PI2) (PHI)2 K (reverse rxn) = 1 K (forward rxn) KP (298 K) = 1.26 x 10-3 and KP (700 K) = 1.9 x 10-2

  21. K Depends on the Stoichiometry N2O4(g) 2NO2(g) KP (stoi A) = (PNO2)2 (PN2O4) stoichiometry A 2N2O4(g) 4NO2(g) KP (stoi B) = (PNO2)4 (PN2O4)2 stoichiometry B KP (stoi B) = ( KP(stoi A) )2

  22. K and Hess’s Law 2NOBr(g) 2NO(g) + Br2(g) KP (#1) = (PNO)2(PBr2) (PNOBr)2 reaction 1 Br2(g) + Cl2(g) 2BrCl(g) KP (#2) = (PBrCl)2 (PBr2)(PCl2) reaction 2 2NOBr(g) + Cl2(g) 2NO(g) + 2BrCl(g) KP = (PNO)2 (PBrCl)2 = KP(#1) x KP(#2) (PNOBr)2(PCl2) The equilibrium constant expression is the product of the expressions for the individual steps.

  23. More Properties of K • K of a reaction in the reverse direction is the inverse of K for the reaction in the forward direction. • K for a reaction multiplied by a number is K raised to a power equal to that number. • K for a reaction made up of two or more steps is the product of the K’s for the individual steps (Hess’s Law).

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