6.1 The Equilibrium Condition. Chapter 6. Equilibrium Systems. Reaching Equilibrium 10/2/2015

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1 10// The Equilibrium Condition Chapter 6 Describe equilibrium systems and write the equilibrium constant expression for any chemical reaction. 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. Equilibrium Systems Chemical Equilibrium N (g) + 3H (g) NH 3 (g) The concentrations of all species become constant before all the reactants are consumed. Reaching Equilibrium Consider the reaction NO (g) N O 4 (g) We start with brown NO in a flask and observe as colorless N O 4 forms. NO (g) N O 4 (g) Starting with NO Starting with N O 4 1

2 10// The Equilibrium Constant Evaluate the equilibrium constant from experimental data. Relate the expression for the equilibrium constant to the form of the balanced equation. The Equilibrium Constant NO (g) N O 4 (g) Concentrations of Nitrogen Dioxide and Dinitrogen Tetroxide and the Equilibrium Constant at 317 K Initial Concentrations, M Equilibrium Concentrations, M [NO4 ] Keq [NO ] [N O 4 ] [NO ] [N O 4 ] [NO ] The Equilibrium Constant The Law of Mass Action NO (g) N O 4 (g) [NO4] Equilibrium constant : Keq 45.9 [NO ] [X] is the concentration, in mol/l of species X For a general chemical reaction aa + bb cc + dd the equilibrium constant is given by K eq c d [C] [D] a b [A] [B] O 3 (g) 3O (g) K eq [O [O 3 ] 3 ] CO(g) + H O(g) H (g) + CO (g) K [H ][CO ] eq [CO][H O] Examples K eq and the Chemical Equation K eq refers to a specific chemical equation. Reaction 1 H (g) + O (g) H O(g) [HO] K 1 [H ] [O Reaction H (g) + 1/O (g) H O(g) K [H O] [H ][O ] 1/ ] K 1

3 10//015 K eq and the Chemical Equation H O(g) H (g) + O (g) K 3 ] [H [O [H O] ] 1 K The form of the equilibrium expression depends on the balanced equation. For any reaction, K rev = 1/K for 1 N O 4 (g) NO (g) K eq = (a) Determine K eq for the reaction NO (g) N O 4 (g) K = 1/K eq = 1/ = (b) determine K eq for the reaction NO (g) 1/N O 4 (g) N O 4 (g) NO (g) K eq = (a) Determine K eq for the reaction NO (g) N O 4 (g) K = 1/K eq = 1/ = Equilibria Involving Pressures Convert between pressures and molarities Convert between K p and K c (b) determine K eq for the reaction NO (g) 1/N O 4 (g) K 1/ Keq Equilibria Involving Pressures Pressure and Concentration P = (n/v) RT N O 4 NO K [NO ] c [N O ] Subscript c (and square brackets) indicates molar concentration. pno Define Kp Subscript p = pressure. p P = (n/v)rt =M RT 4 N O 4 3

4 10//015 K p p p NO NO4 n n For any gas P RT. Because [Conc] V V ([NO] RT ) Kp KcRT [N O ] RT Relating K p and K c 4 In general K p = K c(rt) n, where n = moles gaseous product moles gaseous reactant. When n = 0, K p = K c. Example: Converting K p and K c K c is at 700 K SO (g) + O (g) SO 3 (g) Calculate K p. Calculate K p for PCl 5 (g) PCl 3 (g) + Cl (g) Given K c = 4.00 at 45 C. 6.4 Activity Activity is the ratio of the concentration or pressure of a substance to its concentration or pressure in the standard state. For gases, the standard state is 1 atm pressure a x = P x /1 atm For solutes in solution, the standard state is 1 M. a y = [y]/1 M Because units cancel, equilibrium constants have no units. 6.5 Heterogeneous Equilibria We will defer equilibria that involve species in more than one phase to the end of this chapter. When I talk about the topic, I will cover a section of Chapter 8 as well. 6.6 Applications Not all reactions are at equilibrium, for example when you first add reactants together, they have not had enough time to mix and form products. Time scales can be limited by mixing, but can also be much longer. Think about rust formation on your father s hammer that you inconsiderately left in the back yard. 4

5 10//015 Reactions Not at Equilibrium For the reaction conditions shown NO (g) N 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 c [C] [D] Q a [A] [B] d b Determining Direction of Reaction 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 Place the symbols for Q and K c on a number line. The reaction will proceed in the direction that moves Q toward K c. Determining Direction of Reaction Determining Direction of Reaction Data refers to conditions where K eq = [N O 4 ]/[NO ] =

6 10//015 CH 4 (g) + H O(g) CO(g) + 3H (g) K c = 5.67 Initial concentrations: CH 4 (g) + H O(g) CO(g) + 3H (g) M 0.00 M M M Determine in which direction the reaction will proceed. 6.7 Solving Equilibrium Problems 5 step organized approach Write the equilibrium Fill in the ice table Write the algebraic expression for the equilibrium constant Substitute the information from the table into the algebraic expression Solve Equilibrium Calculations 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.. You are given starting concentrations and K eq and calculate equilibrium concentrations. The same 5 step approach works for both types of problems. The center of the approach is the ice table. Determining K from Experiment Consider the chemical reaction of hydrogen and sulfur to make hydrogen sulfide. Experiment shows that at equilibrium, there are.50 mol H, mol S, and 8.70 mol H S in a 1.0 L flask. Calculate K c. Determining K from Experiment H (g) + S (g) H S(g) At equilibrium, there are.50 mol H, 1.35 x 10 5 mol S, and 8.70 mol H Sin a 1.0 L flask. [H ] =.50 mol/1.0 L = 0.08 M [S ] = 1.35 x 10 5 mol/1.0 L = 1.1 x10 6 M [H S] = 8.70 mol/1.0 L = 0.75 M 1. Balanced equation.. Calculate equilibrium concentrations. An ice table is not needed in this first example. [HS] K c [H ] [S ] K c K c Determining K from Experiment [0.75] [0.08] [ ] 3. Expression for equilibrium constant. 4. Substitute concentrations into expression for equilibrium constant. 5. Solve. 6

7 10//015 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 is 0.00 M. Calculate K c for HI(g) H (g)+ I (g) The initial concentration of HI is 1.0mol/10.0 L or [HI] = 0.10 M. Determining K from Experiment II Introducing the ice table 1. Chemical equation. HI(g) H (g) + I (g) initial conc., M Change, M equil conc., M?? ice table with starting information. Determining K from Experiment II HI(g) H (g)+ I (g) Initial conc., M Change, M Equil. conc., M ice table, complete. [H ][I ] Kc = [HI] [0.0][0.0] Kc = [0.06] Kc = Algebraic expression for K. 4. Substitute concentrations from table. 5. Solve. Equilibrium Calculations Given K and Initial Concentrations If the initial concentration of CO is 0.08 M and H is 0.14 M, and K c = 0.50 for CO(g) + H (g) CH O(g) calculate the equilibrium concentrations of all species. We will use our 5 step approach. 1. Write the balanced equation Step. Set up ice table CO(g) + H (g) CH O(g) CO(g) + H (g) CH O(g) i, M C, M y y +y e, M 0.08 y 0.14 y y We have defined y as the change in concentration, and it is unknown. 7

8 10// Write algebraic expression for K c 4. Substitute from ice table CO(g) + H (g) CH O(g) [CHO] Kc = [CO][H ] CO(g) + H(g) CHO(g) i, M C, M y y +y e, M 0.08 y 0.14 y y K c [CHO] [CO][H ] y 0.50 (0.08 y)(0.14 y) Step 5. Solve Solve by quadratic formula or the solver function on a programmable calculator. y = or.1665 Only one root will give possible concentrations. Value of y [CO] = 0.08 y = 0.06 M.1385 M [H ] = 0.14 y = 0.14 M.065 M [CH O] = y = M.1665 M Step 6. Check Substitute numerical values to calculate the equilibrium constant. K c = [CH O] / [CO][H ] = / (0.06 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? PCl 5 (g) PCl 3 (g) + Cl (g) Write the ice table and substitute into the expression for the equilibrium constant for an initial concentration of M ethane decomposing to ethylene and hydrogen. C H 6 (g) C H (g) + H (g) i, M C, M y +y +y e, M y y y [CH[H] K eq y (y ) Keq [C H ] y 6 8

9 10//015 CH 4 (g) + H O(g) CO(g) + 3H (g) K c = 5.67 Initial concentrations: CH 4 (g) + H O(g) CO(g) + 3H (g) M 0.00 M M M Calculate the equilibrium concentrations of all species. 6.8 Le Chatelier s Principle Predict the response of an equilibrium system to changes in conditions by applying the Le Chatelier s principle. Determine how changes in temperature influence the equilibrium system. Le Chatelier s Principle Any change to a chemical reaction at equilibrium causes the reaction to proceed in the direction that reduces the effects of the change. Changing Concentration or Partial Pressure 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 SO 3 (g) SO (g) + O (g) An increase in SO 3 partial pressure causes the formation of more SO. The Effect of a Volume Change 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. 9

10 10//015 The Effect of a Volume Change Changing Pressure by Adding an Inert Gas SO 3 (g) SO (g) + O (g) Increasing the volume causes the reaction to proceed to the right. Decreasing the volume causes the reaction to proceed to the left. 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. Le Chatelier s Principle NO (g) N O 4 (g) brown colorless Changes in Temperature Heat is a product in an exothermic reaction and a reactant in an endothermic reaction. CO(g) + H (g) CH 3OH(g) H = 18 kj The reaction is exothermic (heat is a product). Increasing temperature favors reactants. Adding a Catalyst A catalyst increases the reaction rate but does not affect the equilibrium concentrations. Does not affect the value of K eq. 10

11 10//015 Heterogeneous Equilibria 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. Activities of pure solids and liquids The standard state of a pure liquid or solid is the pure substance Calculate the molar concentration of water. Moles in 1 L = 1000 ml x 1.0 g/ml x 1 mol/18 g = 55.6 mol/l Calculate the activity of water. a = 1.0 Calculate the molar concentration of silver chloride(s), density 5.56 g/cm 3. Mole in 1 L = 1000 cm 3 x 5.56 g/cm 3 x 1 mol/143.3 g = 38.8 mol Heterogeneous Equilibria CaCO 3 (s) CaO(s) + CO (g) [CaO][CO ] [CaCO ] but [CaO] and [CaCO 3 ] are solids, and their concentrations do not change, so K c = [CO ] and/or K c Kp P CO 3 NaHCO 3 (s) Na CO 3 (s) + H O(g) + CO (g) K c = Hg(l) + Cl (g) Hg Cl (s) K c = NH 3 (g) + HCl(g) NH 4 Cl(s) K p = 8.8 Solubility Equilibria Write the expression for the solubility product constant. Calculate K sp from experimental data. Calculate solubility of slightly soluble salts from K sp. Predict the solubility of a solid in a solution that contains a common ion. Use numerical methods to calculate roots of polynomial equations. Determine if a precipitate will form under a particular set of conditions. Solubility Equilibria Solubility equilibria: reactions that involve dissolving or forming of a solid from solution. AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) The net ionic equation is Ag + (aq) + Cl (aq) AgCl(s) 11

12 10//015 The Solubility Product Constant 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 ] Write Expressions for K sp Fe(OH) 3 (s) Fe 3+ (aq) + 3OH (aq) PbCl (s) Pb + (aq) + Cl (aq) Ca 3 (PO 4 ) (s) 3Ca + (aq) + PO 4 3 (aq) 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 10 3 M. Calculate K sp. The solubility of Pb(IO 3 ) is M. Calculate K sp. Balanced equation ice table Algebraic expression for Ksp Substitute Solve Calculate Solubility from K sp Calculate the solubility of Be(OH). K sp is Balanced equation ice table Algebraic expression for K sp Substitute Solve 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. 1

13 10//015 Number Line Representation of Common Ion Effect The Common Ion Effect What is the solubility of Mg(OH) in a solution of M NaOH(aq)? K sp is for Mg(OH) The Common Ion Effect What is the solubility of Mg(OH) in a solution of M NaOH(aq)? K sp is for Mg(OH) 1. Eqn Mg(OH) (s) Mg + (aq) + OH (aq) Numerical Approximations Equation = (s)( s) 5. Solve Solve Assume s<< (s is much smaller than 0.100). Substitute: If s<< then s is approximately equal to (s)(0.100) = s Solve simplified expression. s = M Check approximation: Is 8.9 x less than 5% of 0.100? Yes, so we accept the approximation. The Common Ion Effect Calculate the solubility for the same solute in water with no added OH. Mg(OH) (s) Mg + (aq) + OH (aq) The solubility of Mg(OH) is M in water M in M OH. As expected, the calculations are consistent with the Le Chatelier s principle. 13

14 10//015 Complications with real solutions: Anion is a weak base Many anions are weak bases and react with water. Calculate the solubility of MnS, K sp = MnS(s) Mn + (aq) + S (aq) But sulfide ion is a weak base: S (aq) + H O(aq) HS (aq) + OH (aq) So the concentration of sulfide ion is actually smaller than expected and the solubility is larger than expected. Complications with real solutions: Anion is a weak base In general, the ph is important. Mg(OH) (s) Mg + (aq) + OH (aq) Consider two cases the salt contains the anion of a strong acid and a weak acid. AgI(s) Ag + (aq) + I (aq) There is no acidity effect on insoluble iodides because there is no tendency for I (aq) to associate with H + (aq) to form HI(aq) CaF (s) Ca + (aq) + F (aq) Increasing the acidity will decrease the F (aq) concentration. F (aq) + H O(l) HF(aq) + OH (aq) So the solubility is larger than expected in acidic solutions. Complications with real solutions Complex Ion Formation Complex ions can form. Al 3+ (aq) + 3OH (aq) Al(OH) 3 (s) Expected Al(OH) 3 (s) + OH (aq) Al(OH) 4(s) Forms too So the solubility is larger than expected. Complications with real solutions There are 8 species for which the solubility can be accurately predicted from the solubility product constant. 8.9 Applications of Precipitation Selective precipitation Qualitative analysis Forming a Precipitate Predict if a precipitate forms when two solutions are mixed. Calculate Q and compare it to K sp to determine if a precipitate will form when mixing solutions of soluble compounds. 14

15 10//015 Example: Ca(NO 3 ) + Na CO ml of M Na CO 3 and ml of Ca(NO 3 ) are mixed. Will a precipitate form? Selective Precipitation Propose a method to separate silver ions, barium ions, and sodium ions The Classical Qual Scheme Many salts are separated as sulfides, by controlling the ph. Highly acidic: Most sulfide is H S and the concentration of S (aq) is small, so only the most highly insoluble sulfides precipitated. (Cu + and Hg + have K sp values of and and precipitate under acidic conditions.) Insoluble in basic solution. Nickel(II) and iron(ii) sulfide precipitate. K sp = and respectively. H S is highly toxic, which is why we don t do this scheme here. The USC Qual Scheme Separate insoluble chlorides (Ag +, Pb +, Hg + ). Confirm Ag+ by forming ammine complex by adding NH3 then destroy complex with HNO3 AgCl(s) + NH 3 Ag(NH 3 ) + + Cl Ag(NH 3 ) + + Cl +HNO 3 (aq) AgCl(s) + NH 4 + Separate highly insoluble hydroxides with ammonia. [OH ] is about M and Fe 3+ and Al 3+ (K sp and 10 3 ) precipitate. Complex Ni 3+ and Cu + with ammonia forming Ni(NH 3 ) 6 + and Cu(NH 3 ) 4 +. Remove Ba + by adding SO 4. Increase hydroxide concentration with 0.1 M NaOH and precipitate Mg(OH) (K sp = ) 15