Gas Phase Equilibrium Chemical Equilibrium Equilibrium Constant K eq Equilibrium constant expression Relationship between K p and K c Heterogeneous Equilibria Meaning of K eq Calculations of K c Solving Equilibrium Problems USING AN EQUILIBRIUM CONSTANT Information obtained from K c 1. Predict direction of a reaction Q reaction quotien 2. Obtaining equilibrium concentrations of reactants and products. Mary J. Bojan Chem 110 1
Equilibrium is achieved when opposing processes proceed at equal rates. equilibria previously discussed Eg. Mary J. Bojan Chem 110 2
In chemical equilibria, forward and reverse reactions occur at equal rates: at a macroscopic level, it looks like nothing is happening. N 2 O 4 (g) colorless 2NO 2 (g) brown Mary J. Bojan Chem 110 3
Achieving equilibrium: The relationship between the concentrations of products and reactants at equilibrium will be the same regardless of starting conditions. Initial State: reactants only Initial state: products only Mary J. Bojan Chem 110 4
Equilibrium point of any reaction is characterized by a single number called the Equilibrium Constant. Example: In general: ja + kb pr + qs B(g) 2 A (g) N 2 O 4 2 NO (g) Mary J. Bojan Chem 110 5
What is the equilibrium constant EXPRESSION (K eq ) for the Haber process? N 2 (g) + 3H 2 (g) 2NH 3 (g) Mary J. Bojan Chem 110 6
Summarize what we know about K eq. The ratio of equilibrium concentrations will be constant: K eq is the equilibrium constant for the reaction. It is a NUMBER. K eq (the number) DOES NOT depend on concentration It s a function of temperature only. Notation: K eq = K c K c R A p S q j B k concentration [ ] is expressed in M (mol/l) K eq = K p K p p p R p j A p p q S B k concentration is expressed using partial pressures in atm Mary J. Bojan Chem 110 7
There is a relationship between K p and K c. K p K c (RT) n where n = n prod n react When given K eq how do I know if it is K c or K p? Use context of problem. If concentrations are given in M, then K eq = K c If concentrations are given in P, then K eq = K p Mary J. Bojan Chem 110 8
If equilibrium concentrations are known, we can calculate K p and K c. N 2 (g) + 3H 2 (g) 2NH 3 (g) Equilibrium concentrations of NH 3, N 2, and H 2 were determined at 472 C. [H 2 ] = 0.1207M [N 2 ] = 0.0402M [NH 3 ] = 0.00272M Find K c and K p. Mary J. Bojan Chem 110 9
Heterogeneous Equilibria involves reactants and products in more than one phase. 3 Fe(s) + 4 H 2 O(g) Fe 3 O 4 (s) + 4 H 2 (g) [ K eq = Fe 3O ][ 4 H ] 4 2 What is [Fe]? [Fe 3 O 4 ]? [ Fe] 3 [ H 2 O] 4 Activities of pure solids and liquids = 1. Mary J. Bojan Chem 110 10
What is the equilibrium constant expression for these reactions? AgCl(s) Ag + (aq) + Cl (aq) C 6 H 6 (g) + 3H 2 (g) C 6 H 12 (g) CaCO 3 (s) CaO(s) + CO 2 (g) Mary J. Bojan Chem 110 11
Is K eq large or small??? What does that mean? Reactants Products reactants products reactants products reactants products K 1 Mary J. Bojan Chem 110 12
Try this problem. Which one of the following has the greatest tendency to proceed as written? 1. 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(g) K p = 1 10 22 2. N 2 (g) + O 2 (g) 2NO(g) K p = 5 10 31 3. 2HF(g) F 2 (g) + H 2 (g) K p = 1 10 13 4. 2NOCl(g) 2NO(g) + Cl 2 (g) K p = 4.7 10 4 Mary J. Bojan Chem 110 13
Meaning of K eq Is K eq large or small??? Reactants Products Cu +2 (aq) + 4NH 3 (aq) Cu(NH 3 ) 4 +2 (aq) K eq = Ni +2 (aq) + 6NH 3 (aq) Ni(NH 3 ) 6 +2 (aq) K eq = Mary J. Bojan Chem 110 14
Meaning of K eq Is K eq large or small??? Reactants Products PbI 2 (s) Pb +2 (aq) + 2I (aq) K eq = reactants products reactants products Mary J. Bojan Chem 110 15
More Relationships H 2 (g) + I 2 (g) 2HI(g) A K eq = What is K eq if the reaction is doubled?? 2H 2 (g) + 2 I 2 (g) 4HI(g) B K eq = What is K eq for the reverse reaction? 2HI(g) H 2 (g) + I 2 (g) C K eq = Mary J. Bojan Chem 110 16
At a certain temperature, K c for the following reaction is 16. H 2 (g) + I 2 (g) 2 HI(g) At the same temperature, what is K c for this reaction? HI(g) 1. 1/16 2. 4 3. 1/4 4. 16 ½ H 2 (g) + ½ I 2 (g) 5. there is not enough information to answer this question. Mary J. Bojan Chem 110 17
General Approach to Equilibrium Constant Problems 1. Write the balanced reaction. 2. Write the general form for K eq. 3. Set up a data table: (may need algebraic unknowns) initial conditions changes in concentrations equilibrium concentrations 4. Substitute equilibrium concentrations into the expression for K eq and solve. Mary J. Bojan Chem 110 18
Equilibrium problem solving: calculating K eq 1.00 mole of SO 2 (g) and 1.00 mole of O 2 (g) are added to a 1.00 L container and react until equilibrium is achieved. At equilibrium, the container has 0.919 moles of SO 3 (g). Find K c at 1000K. Mary J. Bojan Chem 110 19
Equilibrium problem solving: calculating K eq A one L container holds 224g of Fe and 5.00 mole of H 2 O( ). It is heated to 1000K and reaches equilibrium. 56g of Fe are left unreacted. What is K c at 1000K? I. initial C. change E. equilibrium 3Fe(s) + 4H 2 O(g) Fe 3 O 4 (s) + 4H 2 (g) Mary J. Bojan Chem 110 20
The reaction quotient Q can be used to predicting the direction of a reaction. aa + bb cc + dd Reaction quotient Q Q = [ C]c D A [ ] d [ ] a [ B] b Note: the concentrations used are NOT equilibrium concentrations. Mary J. Bojan Chem 110 21
We use Q to predict the direction of a reaction. Q When Q = K c system IS at equilibrium K K Q K Q Q < K Q = K Q > K When Q < K c reaction moves to right (produces product) When Q > K c reaction moves to left (produces reactant) Mary J. Bojan Chem 110 22
2HI(g) H 2 (g) + I 2 (g) If we put 0.1 mole of HI in a 1L container, what will happen? 1 reaction shifts to right 2 reaction shifts to left 3 no change occurs [ ][ I ] 2 K c = H 2 HI [ ] 2 =1.25 10-3 If the initial concentrations of all three gases in the vessel are 0.1 mol/l, what will happen? Mary J. Bojan Chem 110 23
We can use K eq to obtain equilibrium concentrations of reactants and products. Initially [IBr] = [I 2 ] = [Br 2 ] = 0.0500 M What are the final concentrations of reactants and products? K c = 2.50 10 3 I. C. E. 2IBr(g) Br 2 (g) + I 2 (g) Mary J. Bojan Chem 110 24
The Haber process is the industrial process used to make ammonia N 2 (g) + 3H 2 (g) 2NH 3 (g) + heat Mary J. Bojan Chem 110 25
Factors that affect equilibrium What happens to a system at equilibrium when it is disturbed by changing concentration volume pressure temperature Answer: Le Chatelier s Principle If a system at equilibrium is disturbed, the system will shift its equilibrium position to minimize the effect of the disturbance. Mary J. Bojan Chem 110 26
Le Chatelier s Principle: To illustrate this principle use the following chemical system in a closed 1L container. N 2 (g) + 3H 2 (g) 2NH 3 (g) + heat H rxn = 92.4kJ/mol T = 700 C Initial conditions: [N 2 ] = 3.00 M [H 2 ] = 4.50 M [NH 3 ] = 0 At equilibrium: [N 2 ] = 2.00M [H 2 ] = 1.50 M [NH 3 ] = 1.41 M What is K c? Mary J. Bojan Chem 110 27
Use LeChatelier s principle to answer the following questions 1. What will happen to [NH 3 ] if 3.0 moles of H 2 are added to the system at equilibrium? Increase the amount of a reactant or product 2. What will happen if the volume decreases from 1.0L to 0.5L? Increase P by decreasing volume 3. What will happen if 5.0 moles of inert gas is added to the system? Increase pressure by adding an inert gas. 4. What will happen if the temperature is increased to 1000K? Mary J. Bojan Chem 110 28
1. What will happen to [NH 3 ] if 3.0 moles of H 2 are added to the system at equilibrium? System is no longer at equilibrium. Which direction will it go to get to equilibrium? Using LeChatelier s principle: Reactant or product added to a mixture at equilibrium will cause reaction to shift in the direction that consumes part of the added material. N 2 (g) + 3H 2 (g) 2NH 3 (g) + heat Mary J. Bojan Chem 110 29
2. Which direction will the reaction proceed if the volume is decreased to 0.5L? [ ] n V as V, [] so P [ ] RT LeChatelier s principle: Since pressure is increased by disturbance, reaction will shift in direction that will reduce pressure N 2 (g) + 3H 2 (g) 4 moles 2NH 3 (g) 2 moles Mary J. Bojan Chem 110 30
3. What happens when 5.0 moles of an inert gas are added? Inert gas: gas that does not participate in the reaction. How does an inert gas effect the concentrations of reactants and products? N 2 (g) + 3H 2 (g) 4 moles 2NH 3 (g) 2 moles LeChatelier s principle: Mary J. Bojan Chem 110 31
Examples Change concentration Fe +3 (aq) + SCN (aq) FeSCN +2 (aq) Yellow colorless red Change pressure 2NO 2 (g) brown N 2 O 2 (g) colorless Mary J. Bojan Chem 110 32
changing concentration (or V so that [ ] changes) puts a stress on the system. Stresses do not change K eq! Q changes; system shifts to re-establish equilibrium Q K Summary WHAT IF TEMPERATURE CHANGES? K eq changes change depends on whether the reaction is exothermic or endothermic. H + H Mary J. Bojan Chem 110 33
4. What happens when the temperature is increased to 1000K? N 2 (g) + 3H 2 (g) 2NH 3 (g) + heat H rxn = 92.4 kj/mol Treat heat like a product (exothermic) or reactant (endothermic). Le Chatelier s Principle: heat is added to a system, the reaction will shift in the direction that absorbs heat Reaction is exothermic (heat is a product) so adding heat will cause reaction to shift to left. Mary J. Bojan Chem 110 34
4. What happens when the temperature is decreased? Treat heat like a product (exothermic) or reactant (endothermic). Example: endothermic reaction heat + Co[H 2 O] 6 2+ (aq) + 4Cl (aq) pink CoCl 4 2 (aq)+ 6H 2 O( ) blue Mary J. Bojan Chem 110 35
To produce ammonia: Summarize N 2 (g) + 3H 2 (g) 2NH 3 (g) + heat Do we want high or low temperature? Do we want high or low pressure? Liquefy ammonia as process proceeds. WHY? Problem: rate of reaction increases as T increases, BUT equilibrium constant decreases at higher T. Mary J. Bojan Chem 110 36