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# UNIT 4 Equilibria. Things to Review for Unit 4 1.Solving quadratic equations: ax 2 + bx + c = 0 x = -b ± √ b 2 – 4ac 2a 2.Common logarithms log (base.

## Presentation on theme: "UNIT 4 Equilibria. Things to Review for Unit 4 1.Solving quadratic equations: ax 2 + bx + c = 0 x = -b ± √ b 2 – 4ac 2a 2.Common logarithms log (base."— Presentation transcript:

UNIT 4 Equilibria

Things to Review for Unit 4 1.Solving quadratic equations: ax 2 + bx + c = 0 x = -b ± √ b 2 – 4ac 2a 2.Common logarithms log (base 10 logarithms) 3.Calculating the molarity of a solution obtained by dilution…M 1 V 1 = M 2 V 2 4.Acid-base neutralization reactions (a type of double replacement reaction) 5.Titration problems (the stoichiometry of acid-base reactions) 6.Equation showing an ionic solid dissolving in water.

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”.

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.

Chemical Equilibrium

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. Chemical Equilibrium time (s)P A (bar) 00.612 10.604 20.596 50.574 100.539 200.475 500.334 1000.201 2000.109 5000.082 A(g) B(g)

As the partial pressure of A drops, the partial pressure of B increases until the partial pressures each no longer change with time. Chemical Equilibrium time (s) P A (bar)P B (bar) 00.6120 10.6040.008 20.5960.016 50.5740.038 100.5390.073 200.4750.136 500.3340.278 1000.2010.410 2000.1090.503 5000.0820.530 PAPA PBPB

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. Chemical Equilibrium PBPB PAPA A(g) B(g) Once the reaction reaches equilibrium, it appears to stop, because the partial pressure of product stops increasing. PAPA PBPB

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

Chemical Equilibrium A(g) B(g) Rate B  A Rate A  B At equilibrium, the forward and reverse rates are equal: k f P A = k r P B RT RT k f = P B k r P A The term “dynamic equilibrium” may be used instead of simply “equilibrium.” This reminds us that both forward and reverse reactions continue to occur. equilibrium starts here

Chemical Equilibrium At equilibrium, the forward and reverse rates are equal: k f P A = k r P B RT RT k f = P B k r P A 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 K P. A(g) B(g) k f = P B = K P k r P A If we were dealing with solutions, at equilibrium the ratio of the concentrations would be a constant K C.

The Equilibrium Constant K P The law of mass action states that, at equilibrium, the following ratio is a constant: K P = P C c P D d P A a P B b 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 K C, which has a different value than K P. aA(g) + bB(g) cC(g) + dD(g)

The Equilibrium Constant K C The law of mass action states that, at equilibrium, the following ratio is a constant: K C = [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. aA(aq) + bB(aq) cC(aq) + dD(aq)

Properties of the Equilibrium Constant K P or K C 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. Be careful in writing these two constants.

Examples of K K P = (P HI ) 2 (P H2 )(P I2 ) H 2 (g) + I 2 (g) 2HI(g) K C = [Ag(NH 3 ) 2 + ] [Ag + ] [NH 3 ] 2 Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2 + (aq) K C = [CH 3 COO - ] [H + ] [CH 3 COOH] CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq)

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

Calculating K C An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25°C: [CH 3 COOH] = 0.0165 M, [H + ] = 5.44 x 10 -4 M, and [CH 3 COO - ] = 5.44 x 10 -4 M. Calculate K C for the ionization of acetic acid at 25°C. K C (25°C) = (5.44 x 10 -4 ) (5.44 x 10 -4 ) = 1.79 x 10 -5 (0.0165) K C = [CH 3 COO - ] [H + ] [CH 3 COOH] CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq)

What Does K Tell Us About a Reaction? The small value of K C says that, at equilibrium, the ratio of products (acetate and hydrogen 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.” K C = [CH 3 COO - ] [H + ] = 1.79 x 10 -5 [CH 3 COOH] CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq)

What Does K Tell Us About a Reaction? 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? K P = (P HI ) 2 (P H2 )(P I2 ) H 2 (g) + I 2 (g) 2HI(g) K P (298 K) = 794 and K P (700 K) = 54 These values of K P 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.

K for the Reverse Reaction? K P,forward = (P HI ) 2 (P H2 )(P I2 ) H 2 (g) + I 2 (g) 2HI(g) K P,forward (298 K) = 794 and K P,forward (700 K) = 54 K P,reverse = (P H2 )(P I2 ) (P HI ) 2 2HI(g) H 2 (g) + I 2 (g) K (reverse rxn) = 1 K (forward rxn) K P,reverse (298 K) = 0.00126 and K P,reverse (700 K) = 0.019

K Depends on the Stoichiometry N 2 O 4 (g) 2NO 2 (g)K P,stoiA = (P NO2 ) 2 (P N2O4 ) 2N 2 O 4 (g) 4NO 2 (g)K P,stoiB = (P NO2 ) 4 (P N2O4 ) 2 stoichiometry A stoichiometry B K P,stoiB = ( K P,stoiA ) 2

K and Hess’s Law 2NOBr(g) 2NO(g) + Br 2 (g) K P,#1 = (P NO ) 2 (P Br2 ) (P NOBr ) 2 Br 2 (g) + Cl 2 (g) 2BrCl(g) K P,#2 = (P BrCl ) 2 (P Br2 )(P Cl2 ) reaction #1 reaction #2 K P = (P NO ) 2 (P BrCl ) 2 = (K P,#1 )( K P,#2 ) (P NOBr ) 2 (P Cl2 ) 2NOBr(g) + Cl 2 (g) 2NO(g) + 2BrCl(g) The equilibrium constant expression is the product of the expressions for the individual steps.

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). Review how ΔH is treated for each of these three and compare to K’s treatment.

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