Chemical Equilibrium

1. Chemical Equilibrium

2. A + B C Reversible Reactions • If you recall, we mentioned that there are rxns that bounce back and forth from forming products to reforming reactants • A.K.A. reversible rxns • Reversible reactions occur simultan-eously in both directions • An example of a reversible rxn involves reactants A and B producing C.

4. Chemical Equilibrium • At chemical equilibrium there is no net change in the actual amounts of the components of the system. • And although the rates of the forward & reverse rxns are equalat chemical equilibrium, the concentrations of the components on both sides of the chem-icaleqn are not necessarily the same. • In fact they can be dramatically different.

5. Consider a set of escalators as being like the double arrows in a dynamic equilibrium. • The # of people using the up escalator must be the same as the # of people using the down escalator for equilibrium to have been established • However, the # of people upstairs do not have to equal the # of people downstairs • Just the transfer between floors must be consistent

6. Which Direction is Favored? • The equilibrium position of a rxn is given by the concentrations of the system’s components at equilibrium • The equilibrium position indicates whether the components on the left or right side of a reversible rxn are at a higher concentration. • If A reacts to give B and the mixture at equilibrium contains more of B – say 1% of A vs. 99% of B –the formation of B is said to be favored.

7. Forward direction is favored! A B 1% 99% Reverse direction is favored! A B 99% 1% Which Direction is Favored? • On the other hand, if the mixture contains 99% of A and 1% of B at equilibrium then the formation of A is favored.

8. Reversibility vs. Reality • In principle, almost all rxns are reversible to some extent under the right conditions • In practice, one set of components is often so favored at equilibrium that the other set cannot be detected. • If one set of components has established equilibrium by converting mostly into products, the rxn has gone to completion • When no products can be detected, you can say there is no rxn

9. Reversibility vs. Reality • Reversible rxns occupy a middle ground between the theoretical extremes of irreversibility and no rxn. • The addition of a catalyst will speed up forward and reverse rxns equally • By reducing the energy needed to activate the rxn in both forward and reverse directions. • Does not effect the amount of reactants and products present at equilibrium; simply decreases the time it takes to establish equilibrium

10. [C]c [D]d [A]a [B]b aA + bB cC + dD Equilibrium Expression • Chemist’s can express the equilibrium position in terms of a numerical constant • The equilibrium constant shows the relationship between the amount of product and reactant at equilibrium • Consider this hypothetical rxn… • We can write an expression to show the ratio of product concentrations to reactant concs called amass action expression

11. Equilibrium Expression • The conc of each substance is raised to a power equal to the # of mols of that substance in the balanced rxn eqn. • The square brackets indicate concentration in Molarity(mol/L)

12. K= [C]c [D]d [A]a [B]b • Molarity is a measure of how much “stuff” is dissolved in water. • The more stuff dissolved, the more concentrated the solution • The higher the molarity • The quotient ratio of the equilibrium is called theequilibrium constant or K • When the reactants and products amnts are in molarity the constant is called a Kc • When the reactants and products amounts are in pressure units is called a Kp

13. Equilibrium Expression • The constant is dependent on the temp • If the temp changes so does the constant NOTE: water and solid materials are not included in mass action expressions. • Write the mass action expression for each of the following reactions: • 2SO2(g) + O2(g) <==> 2SO3(g) • Bi2S3(s) <==> 2Bi+3(aq) + 3S-2(aq)

14. Classwork: • Balance and write the mass action expression for each of the following reactions: • C4H10(g) + O2(g) <==> CO2(g) + H2O(g) • Al(s) + O2(g) <==> Al2O3(s) • Mn+2(aq)+ BiO3-1(aq) <==> • MnO4-1(aq) + Bi+3(aq)

15. Equilibrium Constant • Equilibrium constants provide valuable chemical information • They show whether the products or the reactantsare favored in a rxn(spontaneus or nonspontaneous) • always written as a ratio ofproducts over reactants • a value ofK > 1means that products are favored • K < 1than reactants are favored

16. K > 1 products favored at equil K < 1 reactants favored at equil

17. N2O4(g) 2NO2(g) Sample Problem 1 Dinitrogentetroxide (N2O4), a colorless gas, and nitrogen dioxide (NO2), a brown gas, exist in equilibrium with each other according to the following eqn: A 1.0 liter of gas mixture at 10C at equilibrium contains .0045 mol N2O4 & .030 mol NO2. Write the mass action expression and calculate K for the rxn.

18. Analyze: list what we know • Known: • [N2O4] =.0045 mol/1.0 L • [NO2] =.030 mol/1.0 L • Unknown: • Mass action expression = ? • K = ? • At equil, there is no net change in the amount of N2O4 or NO2 at any given instant

19. [N2O4]1 [.030M]2 K= [.0045M]1 Calculate: solve for unknowns • The only product of the rxn is NO2, which has a coefficient of 2 in the balanced eqn • The only reactant N2O4 has a coefficient of 1 in the balanced eqn • The mass action expression is: [NO2]2 K= • K is equal to: K=0.20 • K< 1,therefore rxn doesn’t favor products

20. Classwork: • Find the equilibrium constant if [SO2] = 1.0 M; [O2]=1.0 M; [SO3]=2.0 M; using the mass action expression written in the examples • Find the equilibrium constant if [Bi+3] = 0.00058 M; [S-2] = 0.00087 M; using the mass action expression written in the examples

21. AgCl(s) Ag+(aq) + Cl-(aq) Solubility Product • Another type of equilibrium is the equilibrium of dissolving • This is the equilibrium of a solid and its aqueous form • Even the most insoluble salts will dissolve to some extent in water • For example, when AgCl is mixed with water a tiny amount of Ag+1 and Cl-1 exist

22. [AgCl] Solubility Product [Ag+][Cl-] Ksp= • If we exclude the solid “reactant” because of its constant conc. • We get a special kind of equilibrium constant called the solubility product constant (Ksp) • The lower the solubility of a substance the smaller the Ksp

23. Solubility Product • If there are coefficients in the dissociation equation (from balancing) they become exponents that the conc are raised to. Ksp = [Ag+]1[Cl-]1 • The solubility product (Ksp) for AgCl at 25C is 1.8x10-10 M2 • 10-10 indicates a very small conc of silver and chloride ions

24. Solubility Product: example 1 The equilibrium concentration of hydroxide ions in a saturated solution of iron(II) hydroxide is 1.2 x 10-5 M at a certain temperature. Calculate the Ksp of Fe(OH)2 at this temperature.

25. Solubility Product: example 1

26. Solubility Product: example 2 What is the concentration of silver ions in a saturated solution of silver carbonate? The Ksp of Ag2CO3 is 8.1 x 10-12

27. Reaction Quotient • We can also determine if a reaction has reached equilibrium by calculating a reaction quotient (Q). • It’s like taking a snapshot of a reaction at a given time and interpreting how far along the reaction is. • Once the reaction quotient is solved, it is compared to the equilibrium constant • The following picture helps us decide how to interpret the direction the reaction will continue.

29. Sample Problem 3 Will a precipitate of PbSO4 form when 400.0 ml of 0.0050M MgSO4 is mixed with 600.0 ml of 0.0020M Pb(NO3)2? The Ksp of PbSO4 = 6.3x10-7.

30. Sample Problem 3

31. Classwork: • 0.035 moles of SO2, 0.500 moles of SO2Cl2, and 0.080 moles of Cl2 are combined in an evacuated 5.00 L flask and heated to 100°C. What is Q before the reaction begins? Which direction will the reaction proceed in order to establish equilibrium? SO2Cl2(g) <==> SO2(g) + Cl2(g) Kc = 0.078 at 100°C

32. Manipulating the Equilibrium… • There is a principle that can be studied to govern changes in equilibrium Le Chatelier’sPrinciple. • Le Chatelier’s Principle states: • “If a stress is applied to a system in dynamic equilibrium, thesystem changes to relieve the stress.” • Stresses are changes in temperature, pressure, concentration of reactants, or concentration of products

33. Concentration & Equilibrium • Adjusting the concentrations of either reactants or products can have dramatic impact on the equilibrium • If we add more of reactant A to a system at equilibrium the system will strive to reestablish equilibrium at a new equilibrium position. • The reaction will push to use up the extra A and generate more C [A]↑, rxn will shift toward products

34. Concentration & Equilibrium • Adjusting the concentrations of either reactants or products can have dramatic impact on the equilibrium • If we add more of product C to a system at equilibrium the system will strive to reestablish equilibrium at a new equilibrium position. • The reaction will push to use up the extra C and generate more A and B [C]↑, rxn will shift toward reactants

36. Temp effects on Equilibrium • The impact of temperature changes on an equilibrium is dependent on if the process is endothermic or exothermic • Endothermic processes use energy as a reactant, while exothermic processes produce energy • Keq is temperature dependent • 250 kJ is a product If T↑, the equilibrium shifts left

37. Temp effects on Equilibrium • The impact of temperature changes on an equilibrium is dependent on if the process is endothermic or exothermic • Endothermic processes use energy as a reactant, while exothermic processes produce energy energy is a reactant If T↑, the equilibrium shifts right

38. Keq is TEMP dependent Endothermic or Exothermic?

39. Pressure & Equilibrium • If A, B, and C are all gases, then the equil they establish is pressure dependent • When the pressure is increased, the system relieves the pressure by favoring the direction thatproduces fewer gas molecules. • Pressure is # of particles dependent, the more particles the higher the pressure • Fewer gas molecules will exertless pressure. • So, more product is formed, which overall reduces the pressure, this is a shift right

40. Pressure & Equilibrium • Conversely, a decrease in pressure will favor the rxn that produces the most molecules • So we have a shift to theleft • P↑, this equilibrium shifts right If P↓, this equilibrium shifts left

43. Classwork Predict the effect of the following changes on the reaction in which SO3 decomposes to form SO2 and O2. 2SO3(g) <=> 2SO2 (g) + O2 (g) Ho = 197.78 kJ (a) Increasing the temperature of the reaction. (b) Increasing the pressure on the reaction. (c) Adding more O2 when the reaction is at equilibrium. (d) Removing O2 from the system when the reaction is at equilibrium.