Daniel L. Reger Scott R. Goode David W. Ball Chapter 14 Chemical Equilibrium

Chemical Equilibrium Chemical Equilibrium: A state in which the tendency of the reactants to form products is balanced by the tendency of the products to form reactants. Could also be defined as a system in which the rates of the forward and reverse reactions are the same.  No observable changes occur at equilibrium.

Phase Change H 2 O (g) ⇌ H 2 O ( l ) At equilibrium, liquid water evaporates at the same rate the water vapor condenses. Chemical Equilibrium N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) The concentrations of all species become constant before all the reactants are consumed. Equilibrium Systems

Consider the reaction 2NO 2 (g) ⇌ N 2 O 4 (g) We start with brown NO 2 in a flask and observe as colorless N 2 O 4 forms. Reaching Equilibrium

Starting with NO 2 Starting with N 2 O 4 2NO 2 (g) ⇌ N 2 O 4 (g)

For a general chemical reaction aA + bB ⇌ cC + dD the equilibrium constant is given by The Law of Mass Action

2NO 2 (g) ⇌ N 2 O 4 (g) The Equilibrium Constant Concentrations of Nitrogen Dioxide and Dinitrogen Tetroxide and the Equilibrium Constant at 317 K Initial Concentrations, MEquilibrium Concentrations, M [NO 2 ][N 2 O 4 ][NO 2 ][N 2 O 4 ] 2.00 × × × × × × × × × × × × ×

2NO 2 (g) ⇌ N 2 O 4 (g) The Equilibrium Constant

2O 3 (g) ⇌ 3O 2 (g) CO(g) + H 2 O(g) ⇌ H 2 (g) + CO 2 (g) Examples

For any elementary reaction that reaches equilibrium aA + bB ⇌ cC + dD  the forward rate = k f [A] a [B] b  the reverse rate = k r [C] c [D] d At equilibrium the forward rate is equal to the reverse rate so k f [A] a [B] b = k r [C] c [D] d The ratio of the two rate constants is a constant that we call K eq. Kinetics and Equilibrium

K eq refers to a specific chemical equation. Reaction 1 2H 2 (g) + O 2 (g) ⇌ 2H 2 O(g) Reaction 2 H 2 (g) + 1/2O 2 (g) ⇌ H 2 O(g) K eq and the Chemical Equation

2H 2 O(g) ⇌ 2H 2 (g) + O 2 (g) The form of the equilibrium expression depends on the balanced equation. For any reaction, K rev = 1/K for K eq and the Chemical Equation

N 2 O 4 (g) ⇌ 2NO 2 (g) K eq = 4.63 × (a) Determine K eq for the reaction 2NO 2 (g) ⇌ N 2 O 4 (g) (b) determine K eq for the reaction NO 2 (g) ⇌ 1/2N 2 O 4 (g) Test Your Skill

N 2 O 4 ⇌ 2NO 2 Subscript c indicates molar concentration. Subscript p indicates pressure. P = (n/V)RT =M RT Pressure and Concentration

In general K p = K c (RT)  n, where   n = moles gaseous product – moles gaseous reactant. When  n = 0, K p = K c. Relating K p and K c

K c is 5.0 × 10 6 at 700 K 2SO 2 (g) + O 2 (g) ⇌ 2SO 3 (g) Calculate K p. Example: Converting K p and K c

Calculate K p for PCl 5 (g) ⇌ PCl 3 (g) + Cl 2 (g) Given K c = 4.00 at 425° C. Test Your Skill

Reactions Not at Equilibrium For the reaction conditions shown 2NO 2 (g) ⇌ N 2 O 4 (g) it takes time to reach equilibrium.

The Reaction Quotient Reaction Quotient, Q, has the same algebraic form as K eq, but is evaluated with current concentrations, rather than equilibrium concentrations. aA + bB ⇌ cC + dD

Q < K c :ratio of products to reactants is too small, reaction will proceed in forward direction to reach equilibrium. Q = K c :the system is at equilibrium. Q > K c :ratio of products to reactants is too large, reaction will proceed in reverse direction to reach equilibrium. Determining Direction of Reaction

Data refers to conditions where K eq = [N 2 O 4 ]/[NO 2 ] 2 = 0.45 Determining Direction of Reaction

CH 4 (g) + H 2 O(g) ⇌ CO(g) + 3H 2 (g) K c = 5.67 Initial concentrations: CH 4 (g) + H 2 O(g) ⇌ CO(g) + 3H 2 (g) M M M M Determine in which direction the reaction will proceed. Test Your Skill

Any change to a chemical reaction at equilibrium causes the reaction to proceed in the direction that reduces the effects of the change. The Le Chatelier’s Principle

Adding a reactant or product causes the reaction to proceed in the direction that consumes the added substance. Removing a reactant or product causes the reaction to proceed in the direction that produces the missing substance. Changing Concentration or Partial Pressure

2SO 3 (g) ⇌ 2SO 2 (g) + O 2 (g) An increase in SO 3 partial pressure causes the formation of more SO 2. Changing Concentration or Partial Pressure

A decrease in volume causes the reaction to proceed in the direction that decreases the number of moles of gas. An increase in volume causes the reaction to proceed in the direction that increases the number of moles of gas. The Effect of a Volume Change

2SO 3 (g) ⇌ 2SO 2 (g) + O 2 (g) Increasing the volume causes the reaction to proceed to the right. Decreasing the volume causes the reaction to proceed to the left. The Effect of a Volume Change

Changing the pressure of a system by adding an inert gas does not change the concentration or partial pressure of the reactants or products. The equilibrium does not change when an inert gas is added. Changing Pressure by Adding an Inert Gas

Heat is a “product” in an exothermic reaction and a “reactant” in an endothermic reaction. CO(g) + 2H 2 (g) ⇌ CH 3 OH(g)  H = -18 kJ The reaction is exothermic (heat is a product). Increasing temperature favors reactants. Changes in Temperature

Adding a Catalyst A catalyst increases the reaction rate but does not affect the equilibrium concentrations. Does not affect the value of K eq.

Equilibrium Calculations We will use a 5-step approach: 1.Write the balanced chemical equation. 2.Fill in a table, which we will call an iCe table, with the concentrations of the various species. 3.Write the algebraic expression for the equilibrium constant. 4.Substitute concentrations from the iCe table into the algebraic expression. 5.Solve the expression for the unknown quantity or quantities.

You will likely see two kinds of equilibrium calculations: 1.Equilibrium concentrations are given or computed from other data and you determine the value for K eq. 2.You are given starting concentrations and K eq and calculate equilibrium concentrations. The same 5-step approach works for both types of problems. Equilibrium Calculations

Consider the chemical reaction of hydrogen and sulfur to make hydrogen sulfide. Experiment shows that at equilibrium, there are 2.50 mol H 2, 1.35 × mol S 2, and 8.70 mol H 2 S in a 12.0 L flask. Calculate K c. Determining K from Experiment

2H 2 (g) + S 2 (g) ⇌ 2H 2 S(g) At equilibrium, there are 2.50 mol H 2, 1.35 x mol S 2, and 8.70 mol H 2 S in a 12.0 L flask.  [H 2 ] = 2.50 mol/12.0 L = M  [S 2 ] = 1.35 x mol/12.0 L = 1.12 x10 -6 M  [H 2 S] = 8.70 mol/12.0 L = M 1. Balanced equation. 2. Calculate equilibrium concentrations. An iCe table is not needed in this first example. Determining K from Experiment

3. Expression for equilibrium constant. 4. Substitute concentrations into expression for equilibrium constant. 5. Solve. Determining K from Experiment

Determining K from Experiment II A scientist places 1.0 mol of HI in a 10.0-L flask. Experiments show that the equilibrium concentration of I 2 is M. Calculate K c for 2HI(g) ⇌ H 2 (g)+ I 2 (g) The initial concentration of HI is 1.0mol/10.0 L or [HI] = 0.10 M.

1. Chemical equation. 2. iCe table with starting information. Determining K from Experiment II Introducing the iCe table 2HI(g) ⇌ H 2 (g) + I 2 (g) initial conc., M Change, M equil conc., M ? ? 0.020

2HI(g) ⇌ H 2 (g)+ I 2 (g) Initial conc., M Change, M Equil. conc., M iCe table, complete. 3. Algebraic expression for K. 4. Substitute concentrations from table. 5. Solve. Determining K from Experiment II

Equilibrium Calculations Given K and Initial Concentrations If the initial concentration of CO is M and H 2 is 0.14 M, and K c = 0.50 for CO(g) + H 2 (g) ⇌ CH 2 O(g) calculate the equilibrium concentrations of all species. We will use our 5-step approach.

1. Write the balanced equation CO(g) + H 2 (g) ⇌ CH 2 O(g)

Step 2. Set up iCe table CO(g) + H 2 (g) ⇌ CH 2 O(g) i, M C, M -y -y+y e, M0.028-y0.14-y y We have defined y as the change in concentration, and it is unknown.

CO(g) + H 2 (g) ⇌ CH 2 O(g) 3. Write algebraic expression for K c

4. Substitute from iCe table CO(g) + H 2 (g) ⇌ CH 2 O(g) i, M C, M -y -y+y e, M0.028-y0.14-y y

Step 5. Solve Solve by quadratic formula y = or Only one root will give possible concentrations. Value of y [CO] = y = M M [H 2 ] = y = 0.14 M M [CH 2 O] = y = M M

Step 6. Check Substitute numerical values to calculate the equilibrium constant. K c = [CH 2 O] / [CO][H 2 ] = / (0.026 x 0.14) = is quite close to the expected 0.50, so we can be confident that we did the problem correctly.

If the initial concentration of PCl 5 is M, and K c = 0.60, what are the equilibrium concentrations for the reaction? First, calculate y, the change in concentration. PCl 5 (g) ⇌ PCl 3 (g) + Cl 2 (g) Test Your Skill

If the initial concentration of PCl 5 is M, and K c = 0.60, what are the equilibrium concentrations for the reaction? First, calculate y, the change in concentration. PCl 5 (g) ⇌ PCl 3 (g) + Cl 2 (g) y = or  Choose to get a result in which all the concentrations are positive. [PCl 5 ] = = M [PCl 3 ] = [Cl 2 ] = M Test Your Skill

A heterogeneous system is one in which the reactants and products are present in more than one phase. The concentration of a pure solid or liquid is a constant and is not included in the equilibrium expression. Heterogeneous Equilibria

CaCO 3 (s) ⇌ CaO(s) + CO 2 (g) but [CaO] and [CaCO 3 ] are solids, and their concentrations do not change, so K c = [CO 2 ] and/or Heterogeneous Equilibria

Test Your Skill 2NaHCO 3 (s) ⇌ Na 2 CO 3 (s) + H 2 O(g) + CO 2 (g) K c = 2Hg( l ) + Cl 2 (g) ⇌ Hg 2 Cl 2 (s) K c = NH 3 (g) + HCl(g) ⇌ NH 4 Cl(s) K p =

Solubility Equilibria Solubility equilibria: reactions that involve dissolving or forming of a solid from solution. AgNO 3 (aq) + NaCl(aq) ⇌ AgCl(s) + AgNO 3 (aq) The net ionic equation is Ag + (aq) + Cl - (aq) ⇌ AgCl(s)

For a partly soluble or insoluble solid such as AgCl, AgCl(s) ⇌ Ag + (aq) + Cl - (aq) we define K sp, the solubility product constant, as K sp = [Ag + ][Cl - ] The Solubility Product Constant

Fe(OH) 3 (s) ⇌ Fe 3+ (aq) + 3OH - (aq) K sp = PbCl 2 (s) ⇌ Pb 2+ (aq) + 2Cl - (aq) K sp = Ca 3 (PO 4 ) 2 (s) ⇌ 3Ca 2+ (aq) + 2 PO 4 3- (aq) K sp = Write Expressions for K sp

Calculating the Solubility Product Constant A scientist prepares a saturated solution of lead(II) iodide. Independent measurements show that the concentration of lead is 1.3 x M. Calculate K sp.

Test Your Skill The solubility of Pb(IO 3 ) 2 is 4.5 × M. Calculate K sp.

Solubility Calculations K sp for Ag 2 S is 1.6 × Calculate the solubility in mol/L. K sp = 1.6 × Ag 2 S(s) ⇌ 2 Ag + (aq) + S 2- (aq)

Test Your Skill K sp for Ca(OH) 2 is 1.3 × Calculate the solubility in mol/L.

Solubility Solubilities of different compounds cannot be predicted by ranking them in order of K sp.

Solubility and the Common Ion Effect Common ion effect: the effect of adding a solute to a solution that contains an ion in common. In a precipitation reaction, the common ion effect decreases the solubility of the solid. The effect is consistent with Le Chatelier’s principle: AgCl(s) ⇌ Ag + (aq) + Cl - (aq) adding NaCl decreases the solubility of AgCl.

The Common Ion Effect What is the solubility of Mg(OH) 2 in a solution of M NaOH(aq)? K sp is 8.9 × for Mg(OH) 2 Mg(OH) 2 (s) ⇌ Mg 2+ (aq) + 2OH - (aq)

The Common Ion Effect Calculate the solubility for the same solute in water with no added OH -. Mg(OH) 2 (s) ⇌ Mg 2+ (aq) + 2OH - (aq)

Example: Ca(NO 3 ) 2 + Na 2 CO mL of M Na 2 CO 3 and mL of Ca(NO 3 ) 2 are mixed. Will a precipitate form? Using the solubility rules, the likely precipitate is CaCO 3. K sp (CaCO 3 ) = 8.7 × 10 -9