Equilibrium. What is equilibrium? Hebden Unit 2 (page 37 69) Dynamic Equilibrium

Equilibrium What is equilibrium? Hebden Unit (page 37 69) Dynamic Equilibrium Hebden Unit (page 37 69) Experiments show that most reactions, when carried out in a closed system, do NOT undergo complete conversion from reactants to products regardless of the time allowed for the reaction to take place. 1

Dynamic Equilibrium Hebden Unit (page 37 69) What happens is, products begin to react with each other to re form the reactants. In time, a STEADY STATE situation results such that Rate (formation of products) = Rate (formation of reactants) 3 Dynamic Equilibrium Hebden Unit (page 37 69) Rate (formation of products) = Rate (formation of reactants) A + B C + D (reactants) (products) At this point, the concentrations of all the reactants and products remain constant. The reaction has reached CHEMICAL EQUILIBRIUM. 4

Dynamic Equilibrium Hebden Unit (page 37 69) H (g) + I (g) HI (g) (reactants) (products) After some time, equilibrium state is reached. The colour of the vapour remains constant. 5 Dynamic Equilibrium Hebden Unit (page 37 69) Concentration VS. Time H (g) + I (g) HI (g) (reactants) (products) Concentrations of HI remain constant. Concentrations of H and I remain constant. The reaction is at CHEMICAL EQUILIBRIUM. NOTE: Concentrations of reactants and products do NOT have to be equal. 6 3

Dynamic Equilibrium Hebden Unit (page 37 69) Reaction Rate VS. Time H (g) + I (g) HI (g) (reactants) (products) Forward and reverse reaction rates are EQUAL. The reaction is at CHEMICAL EQUILIBRIUM. 7 Dynamic Equilibrium Hebden Unit (page 37 69) A + B C + D (reactants) (products) Chemical equilibrium is the process wherein two opposing chemical reactions occur simultaneously at the same rate. 8 4

Dynamic Equilibrium Hebden Unit (page 37 69) A + B C + D (reactants) (products) Characteristics of Equilibrium: 1. Forward rate = Reverse rate.. [Reactants] and the [Products] are constant. 3. In general, [Reactants] does not equal [Products]. 4. Equilibrium is temperature dependent. 5. If an equilibrium is disturbed, it will re-establish itself. 6. A state of equilibrium can be established from any starting point. 9 Three types of equilibrium: Dynamic Equilibrium Hebden Unit (page 37 69) A + B C + D (reactants) (products) 1. Phase equilibrium - A substance undergoes phase changes. Solubility equilibrium A solute interacts with a solvent 3. Chemical reaction equilibrium For each of the above, we need to identify what is actually in equilibrium. 10 5

Dynamic Equilibrium Hebden Unit (page 37 69) 1. Phase Equilibrium Liquid Vapour Two process in opposition: Evaporation is the forward reaction. Condensation is the reverse reaction. Rate of evaporation = Rate of condensation 11 Dynamic Equilibrium Hebden Unit (page 37 69) 1. Phase Equilibrium H O () + heat H O (g) 1 6

1. Phase Equilibrium Dynamic Equilibrium Hebden Unit (page 37 69) H O () + heat H O (g) Evaporation is the forward reaction. Condensation is the reverse reaction. Rate of evaporation = Rate of condensation 13. Solubility Equilibrium Dynamic Equilibrium Hebden Unit (page 37 69) dissolution solid dissolved particles crystallization Dissolution is the forward reaction. Crystallization is the reverse reaction. The opposing process are in dynamic equilibrium when the solution is saturated. That is, the solution is holding all the solute it can at a given temperature. 14 7

. Solubility Equilibrium A saturated sodium chloride solution Dynamic Equilibrium Hebden Unit (page 37 69) NaCl (s) Na + (aq) + Cl - (aq) A saturated solution has an established equilibrium between dissolved and un-dissolved particles. Rate of dissolution = Rate of crystallization 15 Dynamic Equilibrium Hebden Unit (page 37 69) 3. Chemical Reaction Equilibrium H (g) + I (g) HI (g) (reactants) (products) Rate (forward) = Rate (reverse) 16 8

Dynamic Equilibrium Hebden Unit (page 37 69) Summary A state of equilibrium can exist when: 1. There needs to be a closed system.. The reaction is REVERSIBLE. 3. No observable change in the macroscopic property (i.e. concentrations) 4. The temperature is constant We will need to look for the macroscopic property. 17 Equilibrium Recognizing Equilibrium States Hebden Unit (page 37 69) 9

Dynamic Equilibrium Hebden Unit (page 37 69) H (g) + I (g) HI (g) (reactants) (products) After some time, equilibrium state is reached. The colour of the vapour remains constant. 19 Dynamic Equilibrium Hebden Unit (page 37 69) 0 10

Dynamic Equilibrium Hebden Unit (page 37 69) We need to look for a property in the chemical reaction in order to determine whether a reaction is in a state of equilibrium. SO (g) + O (g) SO 3 (g) Eg: 3 moles moles gas molecules gas molecules Macroscopic property: Pressure in the forward reaction. Pressure in the reverse reaction. When equilibrium state is reached, the pressure of the system is constant. 1 Eg; Dynamic Equilibrium Hebden Unit (page 37 69) We need to look for a property in the chemical reaction in order to determine whether a reaction is in a state of equilibrium. H O () H O (g) Macroscopic property: Pressure in the forward reaction. Pressure in the reverse reaction. When equilibrium state is reached, the pressure of the system is constant. 11

Eg; Dynamic Equilibrium Hebden Unit (page 37 69) We need to look for a property in the chemical reaction in order to determine whether a reaction is in a state of equilibrium. Cl (g) + I (g) ICl (g) purple Macroscopic property: Colour in the forward reaction. Colour in the reverse reaction. When equilibrium state is reached, the colour of the system is constant. 3 Eg; Dynamic Equilibrium Hebden Unit (page 37 69) We need to look for a property in the chemical reaction in order to determine whether a reaction is in a state of equilibrium. Cl (g) + HI (g) I (s) + HCl (g) Macroscopic properties First property purple Colour in the forward reaction. Colour in the reverse reaction. When equilibrium state is reached, the colour of the system is constant. 4 1

Eg; Dynamic Equilibrium Hebden Unit (page 37 69) We need to look for a property in the chemical reaction in order to determine whether a reaction is in a state of equilibrium. Cl (g) + HI (g) I (s) + HCl (g) purple Macroscopic properties: Second property Pressure in the forward reaction. Pressure in the reverse reaction. When equilibrium state is reached, the pressure of the system is constant. 5 Eg; Dynamic Equilibrium Hebden Unit (page 37 69) We need to look for a property in the chemical reaction in order to determine whether a reaction is in a state of equilibrium. CaCO 3 (s) CaO (s) + CO (g) Macroscopic properties: Pressure in the forward reaction. Pressure in the reverse reaction. When equilibrium state is reached, the pressure of the system is constant. 6 13

Equilibrium Endothermic / Exothermic Hebden Unit (page 37 69) Dynamic Equilibrium Hebden Unit (page 37 69) Endothermic reaction: reactions requiring input of energy Eg. Photosynthesis Plants use the energy from the sun to convert CO and water to glucose and oxygen. sunlight + 6CO (g) + 6H O () C 6 H 1 O 6 (aq) + 6O (g) 8 14

Dynamic Equilibrium Hebden Unit (page 37 69) Endothermic reaction: reactions requiring input of energy Eg. Dissolving compounds in water: Ammonium chloride Ammonium nitrate Potassium chloride 9 Dynamic Equilibrium Hebden Unit (page 37 69) Endothermic reaction: reactions requiring input of energy Eg. - Melting ice - Boiling water - Splitting water - electrolysis H 0 () + energy H (g) + O (g) 30 15

Dynamic Equilibrium Hebden Unit (page 37 69) Exothermic reaction: reactions produce energy Eg. Combustion reactions C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O () + energy (propane) Eg. Synthesis of water from its elements H (g) + O (g) H O() + energy 31 Dynamic Equilibrium Hebden Unit (page 37 69) What determines whether a chemical reaction is ENDOTHERMIC or EXOTHERMIC? 3 16

Dynamic Equilibrium Hebden Unit (page 37 69) The strength of the chemical bonds before and after the chemical reaction. When bonds are broken in the reactants, energy is required. When new bonds are formed as products, energy is released. 33 Dynamic Equilibrium Hebden Unit (page 37 69) The change in energy observed for a chemical reaction is called Enthalpy, H. energy used in bond H = - breaking the reactants energy released in bond forming products H = H products -H reactants 34 17

Dynamic Equilibrium Hebden Unit (page 37 69) The change in energy observed for a chemical reaction is called Enthalpy, H. energy used in bond H = - breaking the reactants energy released in bond forming products Endothermic reaction: 1. energy used in bond breaking the reactants > energy released in bond forming products. H reactants <H products 3. H = H products -H reactants > 0 35 Dynamic Equilibrium Hebden Unit (page 37 69) The change in energy observed for a chemical reaction is called Enthalpy, H. energy used in bond H = - breaking the reactants energy released in bond forming products Exothermic reaction: 1. energy used in bond breaking the reactants < energy released in bond forming products. H reactants > H products 3. H = H products -H reactants < 0 36 18

Dynamic Equilibrium Hebden Unit (page 37 69) Let s take a look at the reaction profiles of endothermic and exothermic reactions. 37 Endothermic reaction: Dynamic Equilibrium Hebden Unit (page 37 69) H products > H reactants H = H products -H reactants H > 0 38 19

Exothermic reaction: Dynamic Equilibrium Hebden Unit (page 37 69) H products < H reactants H = H products -H reactants H < 0 39 Dynamic Equilibrium Hebden Unit (page 37 69) To include energy in the chemical reaction, Endothermic reaction: heat in; energy on the reactant side. Heat Enters the reaction. Endothermic. A + B + energy C + D Exothermic reaction: heat produced; energy on the product side. Heat Exits the reaction, Exothermic. A + B C + D + energy 40 0

Equilibrium Factors Controlling Equilibrium Enthalpy and Entropy Hebden Unit (page 37 69) Dynamic Equilibrium Hebden Unit (page 37 69) Factors Controlling Equilibrium Two driving factors (or tendencies) must be considered for all chemical reactions: 1. Enthalpy. Entropy 4 1

Dynamic Equilibrium Hebden Unit (page 37 69) 1. Enthalpy The tendency is to achieve an overall LOWER enthalpy. Endothermic Exothermic Uphill in energy NOT PREFERRED Downhill in energy PREFERRED 43 Dynamic Equilibrium Hebden Unit (page 37 69) 1. Enthalpy The tendency is to achieve an overall LOWER enthalpy. For an endothermic reaction: Reactants have LOWER energy than the products. The drive is towards the formation of reactants (the state of lower enthalpy) Uphill in energy NOT PREFERRED Therefore, reactant formation is favoured because it is downhill in energy. 44

Dynamic Equilibrium Hebden Unit (page 37 69) 1. Enthalpy The tendency is to achieve an overall LOWER enthalpy. For an exothermic reaction: Products have LOWER energy than the reactants. The drive is towards the formation of products (the state of lower enthalpy) Downhill in energy PREFERRED Therefore product formation is favoured because it is downhill in energy. 45 Dynamic Equilibrium Hebden Unit (page 37 69). Entropy Entropy is a measure of randomness or disorder. Qualitatively, we can approximate entropy. Gases > Solutions > Liquids > Solids (higher entropy) (lower entropy) The tendency is to achieve a HIGHER state of disorder or randomness. 46 3

Dynamic Equilibrium Hebden Unit (page 37 69). Entropy In a chemical reaction, if entropy is increased, product formation is favoured. If entropy is decreased, reactant formation is favoured. S = S products S reactants If S > 0, products have HIGHER entropy than reactants. {disorder} {randomness} If S < 0, products have LOWER entropy than reactants. {disorder} {randomness} 47 Dynamic Equilibrium Hebden Unit (page 37 69). Entropy The greater the degree of randomness or disorder in a system, the greater the entropy of the system Entropy increases are expected to accompany processes in which Pure liquids or liquids solutions are formed from solids Gases are formed, either from solids or liquids The number of molecules of gases increase during the reaction The temperature of a substance is increased (increased temperature means increased molecular motion) 48 4

. Entropy Dynamic Equilibrium Hebden Unit (page 37 69) Predict S for the following reactions: C (s) + H (g) C H 4 (g) H O (g) H (g) + O (g) NH 4 Cl (s) NH 4+ (aq) + Cl - (aq) C H 5 OH (g) C H 5 OH () BaCO 3 (s) BaO (s) + CO (g) S < 0 reactants favoured S > 0 products favoured S > 0 products favoured S < 0 reactants favoured S > 0 products favoured 49 Dynamic Equilibrium Hebden Unit (page 37 69) Putting it together and take a look at BOTH enthalpy and entropy at work. Case I: When BOTH enthalpy and entropy FAVOUR products. It means H < 0 (exothermic) and S > 0 Reaction can go to COMPLETION Reaction is SPONTANEOUS Eg Combustion of octane is spontaneous C 8 H 18 () + 5 O (g) 16 CO (g) + 18 H O (g) + heat Heat is produced. H < 0 exothermic 5 gas molecules 34 gas molecules S > 0 50 5

Dynamic Equilibrium Hebden Unit (page 37 69) Putting it together and take a look at BOTH enthalpy and entropy at work. Case II: When BOTH enthalpy and entropy FAVOUR reactants. It means H > 0 (endothermic) and S < 0 Reaction will not go. No reaction. Reaction will not occur spontaneously. Eg 4 Au (s) + 3 O (g) + 16 kj Au O 3 (s) Heat is required. H > 0 endothermic 3 gas molecules no gas molecules S < 0 51 Dynamic Equilibrium Hebden Unit (page 37 69) Putting it together and take a look at BOTH enthalpy and entropy at work. Case III: When enthalpy and entropy FAVOUR opposing sides (i.e. one favours products, the other favours reactants). H < 0 (exothermic, favours products) and S < 0 favours reactants H > 0 (endothermic, favours reactants) and S > 0 favours products Reactions form EQUILIBRIUM which can shift in the forward and reverse direction. 5 6

Dynamic Equilibrium Hebden Unit (page 37 69) Putting it together and take a look at BOTH enthalpy and entropy at work. Case III (cont d): When enthalpy and entropy FAVOUR opposing sides (i.e. one favours products, the other favours reactants). Eg NO (g) N O 4 (g) + heat H < 0 exothermic; products favoured S < 0 entropy favors reactants Eg H O (s) + heat H O () H > 0 endothermic; reactants favoured S > 0 entropy favors products 53 Dynamic Equilibrium Hebden Unit (page 37 69) Let s take a look at the reaction profiles for reversible reactions. A + B C + D + heat Write it separately, 1. Forward reaction: A + B C + D + heat (exothermic) reactants products. Reverse reaction: C + D + heat A + B (endothermic) reactants products 54 7

Dynamic Equilibrium Hebden Unit (page 37 69) A + B C + D + heat Forward reaction Reverse reaction A + B C + D + heat C + D + heat A + B Points to note: E a (forward) < E a (reverse) H forward < 0 - exothermic H reverse > 0 - endothermic 55 Dynamic Equilibrium Hebden Unit (page 37 69) It is important to understand the factors that control the position of a chemical equilibrium. A + B C + D + heat Example: Industrial chemical manufacturers The chemical engineers in charge of production WANT to choose conditions that FAVOUR formation of products. What conditions can be changed to effectively cause the position of the equilibrium to lie as far to the right as possible? 56 8

Dynamic Equilibrium Hebden Unit (page 37 69) It is important to understand the factors that control the position of a chemical equilibrium. A + B C + D + heat Let s take a look at the effects on the position of a chemical equilibrium when the following conditions are changed: 1. Concentration. Temperature 3. Pressure 4. Volume It turns out that we can predict how these conditions have on a system at equilibrium! 57 Equilibrium Le Chatelier s Principle Hebden Unit (page 37 69) 9

Le Chatelier s Principle Hebden Unit (page 50 55) We can use Le Chatelier s Principle to qualitatively predict the effects of changes by: 1. Changing concentration. Changing temperature 3. Adding an inert gas 4. Changing the pressure 5. Changing the volume on a system at equilibrium. 59 Le Chatelier s Principle Hebden Unit (page 50 55) Le Chatelier s Principle When a stress is applied to a closed system at equilibrium, the system will respond to REDUCE the stress and establish a new equilibrium. Photo from Wiki Henri Louis Le Chatelier, a French/Italian chemist and engineer. 60 30

Le Chatelier s Principle Hebden Unit (page 50 55) Le Chatelier s Principle When a stress is applied to a closed system at equilibrium, the system will respond to REDUCE the stress and establish a new equilibrium. Macroscopically, as a consequence of the stress on the equilibrium, we observe SHIFTS in the equilibrium in the FORWARD or REVERSE direction. We will call the observable as a shift to the RIGHT (forward) or a shift to the LEFT (reverse). Physically, we observe the shift by seeing a visible colour change, a pressure change, etc 61 Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the equilibrium reaction of N (g) + 3 H (g) NH 3 (g) + heat At t = 0 min, the initial concentrations are: [H ]=.400 M, [N ] = 0.800 M ([NH 3 ] = 0.000 M) N and H Let s keep the equilibrium reaction at constant temperature. 6 31

Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the equilibrium reaction of N (g) + 3 H (g) NH 3 (g) + heat At constant temperature Initial concentrations: [H ]=.400 M, [N ] = 0.800 M, [NH 3 ] = 0.000 M Equilibrium concentrations: [H ]=1.197 M, [N ] = 0.399 M, [NH 3 ] = 0.01 M N and H N, H, NH 3 t = 0 min t = 40 min 63 Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, At t = 0 min, the initial concentrations are: [H ]=.400 M, [N ] = 0.800 M At t=40 min, system reaches equilibrium. [H ]=1.197 M, [N ] = 0.399 M, [NH 3 ] = 0.01 M 64 3

Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, Equilibrium system is STRESSED at the 60 min mark! A concentration change occurred while keeping the temperature constant. 65 Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the equilibrium reaction of N (g) + 3 H (g) NH 3 (g) + heat Equilibrium concentrations achieved: [H ]=1.197 M, [N ] = 0.399 M, [NH 3 ] = 0.01 M STRESS EQUILIBRIUM ADD 1.000 M N at the 60 min mark. N, H, NH 3 ADD 1.000 M N N, H, NH 3 40 minutes later, a new equilibrium re establishes! New equilibrium concentrations: [H ]=1.044 M, [N ] = 1.348 M, [NH 3 ] = 0.304 M 66 33

Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, System responds to REDUCE the stress! 67 Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, 40 minutes after the stress, the system finds a new equilibrium. 68 34

Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, New equilibrium concentrations are established. 69 Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, Now that we know how the system reacts to the stress. Let s understand this behaviour. 70 35

Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat Equilibrium was stressed by the addition of 1.000 M N (g). Le Chatelier s Principle says the system will respond to REDUCE the stress and establish a new equilibrium. The system responds by reducing the amount of N added. The equilibrium responds by shifting to the RIGHT to remove the added N. At constant temperature, 71 Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Summarize: N (g) + 3 H (g) NH 3 (g) + heat Equilibrium is stressed by the addition of N. Le Chatelier s Principle says the system will respond to REDUCE the stress and establish a new equilibrium. The system responds by reducing the amount of N added. The equilibrium responds by shifting to the RIGHT to remove the added N. N, H, NH 3 N, H, NH 3 At constant temperature, add N (g). [N ] [H ] [NH 3 ] 7 36

Le Chatelier s Principle Effect of Temperature Change Hebden Unit (page 50 55) Let s examine the exothermic equilibrium reaction N (g) + 3 H (g) NH 3 (g) + heat Equilibrium is stressed by the addition of heat. Le Chatelier s Principle says the system will respond to REDUCE the stress and establish a new equilibrium. The system responds by reducing the amount of heat added. The exothermic equilibrium reaction responds by shifting to the LEFT to remove the added heat. N, H, NH 3 N, H, NH 3 Add heat [N ] [H ] [NH 3 ] 73 Le Chatelier s Principle Effect of Temperature Change Hebden Unit (page 50 55) Let s examine the endothermic equilibrium reaction N O 4 (g) + heat NO (g) Colourless Red-brown 0 C 100 C 74 37

Le Chatelier s Principle Effect of Pressure Change Hebden Unit (page 50 55) What happens if the equilibrium is stressed by a change in pressure? The greatest effects will be on gaseous substances. Let s examine the ways to change the pressure of a reaction system that involves gaseous components: 1. Add a gaseous product or reactant.. Remove a gaseous product or reactant. 3. Add an inert gas* a) while holding the volume constant b) while holding the pressure constant * An inert gas is a gas that does not take part in the reaction. 75 Le Chatelier s Principle Effect of Pressure Change Hebden Unit (page 50 55) 1. Predict what would happen to the following equilibrium if CO (g) is added while the volume and the temperature are held constant. As 4 O 6 (s) + 6 C (s) As 4 (g) + 6 CO (g) Equilibrium is stressed by adding CO(g) (i.e. pressure increased). Le Chatelier s Principle says the system will respond to REDUCE the stress and establish a new equilibrium. The system responds by reducing the pressure of the system. The equilibrium reaction responds by shifting to the LEFT, the side to remove some of the CO(g). 76 38

Le Chatelier s Principle Effect of Pressure Change Hebden Unit (page 50 55). Predict what would happen to the following equilibrium if As 4 (g) is removed while the volume and the temperature are held constant. As 4 O 6 (s) + 6 C (s) As 4 (g) + 6 CO (g) Equilibrium is stressed by removing As 4 (g) (i.e. pressure decreased). Le Chatelier s Principle says the system will respond to REDUCE the stress and establish a new equilibrium. The system responds by increasing the pressure of the system. The equilibrium reaction responds by shifting to the RIGHT, to produce more As 4 (g). * This reaction is used in the industrial extraction of Arsenic. As is continuously removed from the reaction system to increase the yield. 77 Le Chatelier s Principle Effect of Pressure Change Hebden Unit (page 50 55) 3(a). Predict what would happen to the following equilibrium if He (g), an inert gas, is added while the volume and temperature are held constant. As 4 O 6 (s) + 6 C (s) As 4 (g) + 6 CO (g) Addition of He (g), an inert gas, to the system without changing the system s volume. Addition of He gas increase the TOTAL pressure of the system. it has NO effect on the concentrations of the reactants and products. p CO n CO V RT p As Partial pressures of CO and As 4 4 n As 4 V RT No shift because the partial pressures of the gases are unchanged because volume is unchanged. 78 39

Le Chatelier s Principle Effect of Pressure Change Hebden Unit (page 50 55) 3 (b). Predict what would happen to the following equilibrium if He (g), an inert gas, is added while the pressure and temperature are held constant. As 4 O 6 (s) + 6 C (s) As 4 (g) + 6 CO (g) Equilibrium is stressed by the addition of He (g), an inert gas. Le Chatelier s Principle says the system will respond to REDUCE the stress and establish a new equilibrium. p CO ncort V The system s volume INCREASES in order to maintain constant pressure. This results in a DECREASE the partial pressures (or concentration) of As 4 and CO. p As 4 n Partial pressures of CO and As 4 As 4 V RT As p CO and p As4 decrease, equilibrium will shift to the RIGHT, to increase the pressure of the 79 system. Le Chatelier s Principle Effect of Volume Change Hebden Unit (page 50 55) Predict what would happen to the following equilibrium if the system s volume is decreased while temperature is constant. As 4 O 6 (s) + 6 C (s) As 4 (g) + 6 CO (g) Equilibrium is stressed because this will result in an increase in the system s pressure. Le Chatelier s Principle says the system will respond to REDUCE the stress and establish a new equilibrium. ncort nas RT 4 pco pas As p CO and p As4 is 4 V V increased, the Partial pressures of CO and As 4 equilibrium will The DECREASE in the system s volume causes an shift to the LEFT increase in the partial pressures (or concentration) to remove some of As 4 and CO. CO (g) and As 4 (g). 80 40

Le Chatelier s Principle Effect of Volume Change Hebden Unit (page 50 55) Predict what would happen to the following equilibrium if the system s volume is increased while temperature is constant. As 4 O 6 (s) + 6 C (s) As 4 (g) + 6 CO (g) Equilibrium is stressed because this will result in a decrease in the system s pressure. Le Chatelier s Principle says the system will respond to REDUCE the stress and establish a new equilibrium. ncort nas RT 4 pco pas 4 V V Partial pressures of CO and As 4 The INCREASE in the system s volume causes a decrease in the partial pressures (or concentration) of As 4 and CO. As p CO and p As4 is decreased, the equilibrium will shift to the RIGHT to produce more CO (g) and As 4 (g). 81 Le Chatelier s Principle Hebden Unit (page 50 55) We have seen how to use Le Chatelier s Principle to qualitatively predict the effects of changes by: 1. Changing concentration. Changing temperature 3. Adding an inert gas a) Changing the pressure b) Changing the volume on a system at equilibrium. 8 41

Le Chatelier s Principle Hebden Unit (page 50 55) Le Chatelier s Principle When a stress is applied to a closed system at equilibrium, the system will respond to REDUCE the stress and establish a new equilibrium. We can predict the shift in equilibrium position as a result of the stress applied. 83 Le Chatelier s Principle Hebden Unit (page 50 55) Le Chatelier s Principle View Animation We can predict the shift in equilibrium position as a result of the stress applied. 84 4

Le Châtellier s Principle Example Consider the following equilibrium processes: N (g) + 6H O(g) 4NH 3 (g) + 3O (g) H = 1486.4 kj/mol N (g) + 6H O() 4NH 3 (g) + 3O (g) H = 1530.4 kj/mol Indicate the effect on the mass of NH 3 and the value of K by: H O(g) H O () Action taken m NH3 m NH3 Some N removed Some H O added Volume increased Temperature decreased A catalyst is added Argon gas is added at constant volume Le Châtellier s Principle Example Consider the following equilibrium processes: N (g) + 6H O(g) 4NH 3 (g) + 3O (g) H = 1486.4 kj/mol N (g) + 6H O() 4NH 3 (g) + 3O (g) H = 1530.4 kj/mol Indicate the effect on the mass of NH 3 and the value of K by: H O(g) H O () Action taken m NH3 m NH3 Some N removed Some H O added NC Volume increased Temperature decreased A catalyst is added NC NC Argon gas is added at constant volume NC NC 43

Equilibrium Equilibrium Constant Expression and the Equilibrium Constant, K eq Hebden Unit (page 37 69) Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 1. Equilibrium Constant Expression. Equilibrium Constant, K eq 3. K c and K p 4. Heterogeneous equilibrium 5. Equilibrium laws for Combined equilibrium Scaled equilibrium Reversed equilibrium 6. Interpreting K eq values 7. Explain Le Chatelier s Principle mathematically with K eq. 88 44

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 1. Equilibrium Constant Expression is a mathematical relationship that is specific for the equilibrium reaction. Equilibrium Constant, K eq is a number that characterizes the equilibrium of the reaction is specific for the equilibrium reaction is temperature dependent 89 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) For a general equilibrium reaction a A (aq) + b B (aq) c C (aq) + d D (aq) where A and B are reactants, C and D are products, a, b, c, and d are the coefficients in the balanced chemical equilibrium. K eq [C] [A] c a [D] [B] d b Equilibrium expression 90 45

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) For a general equilibrium reaction a A (aq) + b B (aq) c C (aq) + d D (aq) K eq [C] [A] c a [D] [B] d b Where does the equilibrium expression come from? 91 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) For a general equilibrium reaction a A (aq) + b B (aq) c C (aq) + d D (aq) K eq [C] [A] c a [D] [B] Rate (forward reaction) = Rate (reverse reaction) Rate (forward reaction) = k forward [A] a [B] b Rate (reverse reaction) = k reverse [C] c [D] d d b Definition of equilibrium! 9 46

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) For a general equilibrium reaction a A (aq) + b B (aq) c C (aq) + d D (aq) K eq K eq c [C] [A] a [D] d [B] Rate (forward reaction) = Rate (reverse reaction) k forward [A] a [B] b k k forward reverse [C] [A] c b = k reverse [C] c [D] d a [D] d [B] b K eq [Pr oducts] [Reactants] 93 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) For a general equilibrium reaction a A (aq) + b B (aq) c C (aq) + d D (aq) Points to note about the Equilibrium Constant Expression: [A], [B], [C], [D] that is used in the Equilibrium Constant Expression must be the equilibrium concentrations of A, B, C, and D. For aqueous ions in solutions, the concentration unit is moles/l. c d [C] [D] K eq K eq is a b unitless. [A] [B] 94 47

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Let s re-examine the equilibrium N (g) + 3 H (g) NH 3 (g) + heat The equilibrium constant expression is K eq [NH3] [N ] [H If we use molar concentrations for [N ], [H ] and [NH 3 ], then [NH3] K K c is c 3 [N] [H ] unitless. ] 3 95 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63). For gaseous systems, we can express concentration of the gas in terms of partial pressure of gases rather than molar concentrations. Assuming ideal gas behaviour, PV = nrt Rearrange, P nrt V where P = pressure of the gas V = volume of the gas n = number of moles of the gas R = gas constant, T = temperature in Kelvin At constant temperature, P n V pressure is directly proportional to the concentration of the gas. 96 48

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Let s re-examine the equilibrium N (g) + 3 H (g) NH 3 (g) + heat [NH3] The equilibrium constant expression, K eq 3, [N ] [H ] can also be calculated using partial pressures of the gases. K p p N p NH 3 p H 3 where pnh, p 3 N, p H are partial pressures of NH 3, N, and H in units of atm. nnh RT n RT n 3 N H RT pnh p 3 N ph ( V V V ) K p is unitless. 97 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 3. K c and K p Let s re-examine the equilibrium N (g) + 3 H (g) NH 3 (g) + heat K p p N p NH 3 p H (Concentrations in atm.) 3 K c [NH3] [N ] [H (Concentrations in M.) ] 3 Both of these expressions give K eq. Are K p and K c equal? 98 49

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Are K p and K c equal? For some reactions, K p and K c are equal. For others, the constants have DIFFERENT values. n K g p Kc ( RT ) Where R = gas constant = 0.081 L atm mol -1 K -1 T = temperature in Kelvin n g = n products n reactants (moles of gaseous products moles of gaseous reactants) 99 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Let s re-examine the equilibrium N (g) + 3 H (g) NH 3 (g) + heat K c [ 3 o 6.0 x 10 at 500 3 ] NH ] [N ] [H Convert to K p : K p n g = n products n reactants = 4 = - K c ( RT ) n g 6.0 x 10 1.5 x 10 - -5 C (0.081773) 100 50

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Let s consider the equilibrium HI (g) H (g) + I (g) n g = n products n reactants = = 0 p c n 0 Kc ( RT ) g K K ( RT ) K c In this case, K p = K c. 101 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 4. Heterogeneous Equilibrium Equilibrium reaction has more than one phase Equilibrium reactions involving solids and liquids are omitted from the equilibrium constant expression because they have FIXED concentrations. Example: The thermal decomposition of sodium bicarbonate NaHCO 3 (s) Na CO 3 (s) + H O (g) + CO (g) Since NaHCO 3 (s) and Na CO 3 (s) are solids, their concentrations are unchanged at a given temperature. Kc [ HO] [CO] K p p H O pco 10 51

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 4. Heterogeneous Equilibrium Example: CaO (s) + CO (g) CaCO 3 (s) K c 1 [CO ] K p 1 p CO 103 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Write the proper expression for K c for each of the following equilbria. a) 4 NH 3 (g) + 3 O (g) N (g) + 6 H O(g) 6 N H O K c 4 NH O 3 b) NH 3 (g) 3 H (g) + N (g) 3 H N K c NH c) Co(OH) (s) + H (g) Co(s) + H O(g) H O(g) K c H (g) 3 3 104 5

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 5. Equilibrium laws for Combined Equilibrium Example: Adding Chemical Equilibria (1) N (g) + O (g) N O (g) () N O (g) + 3 O (g) 4 NO (g) ADD: N (g) + O (g) N O (g) N O (g) + 3 O (g) 4 NO (g) N (g) + 4 O (g) 4 NO (g) [N O] K C 3 [N] [O] [NO ] [N O] 4 [O ] 34 K C [N O] K C1 [N ] [O ] [NO ] K C [N O] 4 [O [NO] K C3 [N ] [O K K 3 C1 C 4 ] 4 ] 3 K C1 K C 105 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 5. Equilibrium laws for scaled equilibrium Example: Multiply the coefficients by a factor (1) PCl 3 (g) + Cl (g) PCl 5 (g) Multiply the coefficients by [PCl5] K C1 [PCl ] [Cl 3 ] () PCl 3 (g) + Cl (g) PCl 5 (g) K C K C 1 [PCl ] K C [PCl 5 3] [Cl ] 106 53

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 5. Equilibrium laws for scaled equilibrium Example: Multiply the coefficients by a factor [PCl5] (1) PCl 3 (g) + Cl (g) PCl 5 (g) K C1 [PCl3] [Cl In general, multiply the coefficients by n () n PCl 3 (g) + n Cl (g) n PCl 5 (g) n n 3 n KC K C 1 K C n [PCl5] [PCl ] [Cl ] ] 107 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 5. Equilibrium laws for reversed equilibrium Example: Changing the direction of an equilibrium [PCl5] (1) PCl 3 (g) + Cl (g) PCl 5 (g) K C1 [PCl ] [Cl Reverse the direction 3 ] () PCl 5 (g) PCl 3 (g) + Cl (g) K C 1 K C 1 [PCl3] [Cl] K C [PCl ] 5 108 54

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Given at 300 C 3 H (g) + N (g) NH 3 (g) What is K c for a) 6 H (g) + N (g) 4 NH 3 (g) K C 1 K C K NH 3 3 H N 9.5 9.5 90 C 1 b) NH 3 3 H (g) + N (g) K C 1 K C1 1 9.5 0.11 c) 3/ H (g) + 1/ N (g) NH 3 (g) K C KC 9.5 3.1 1 109 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 6. Interpreting K eq values (i) Since K eq represents the ratio between equilibrium concentrations of products to reactants K eq [Pr oducts] [Reactants] Large K eq means the equilibrium favours the products (i.e. more products than reactants) Small K eq means the equilibrium favours the reactants (i.e. more reactants than products) K eq 1 means reactants and products concentrations are nearly the same (midway) at equilibrium. 110 55

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 6. Interpreting K eq values Reaction K p H (g) + O (g) H O(l) 1.4x10 83 @ 5 C CaCO 3 (s) CaO(s) + CO (g) 1.9x10-3 @ 5 C 1.0 @ 900 C SO (s) + O (g) SO 3 (g) 3.4 @ 700 C K >> 1 [products]>>[reactants] Reaction is product-favored K << 1 [products]<<[reactants] Reaction is reactant-favored 111 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 6. Interpreting K eq values K eq (ii) ONLY TEMPERATURE affects the value of K eq. Increasing temperature for an exothermic reaction aa (aq) + bb (aq) cc (aq) + dd (aq) + heat shifts the equilibrium to the left, favouring the reactants. This results in increasing [Reactants], and K eq becomes smaller. Increasing temperature for an endothermic reaction aa (aq) + bb (aq) + heat cc (aq) + dd (aq) shifts the equilibrium to the right, favouring the products. This results in increasing [Products], and K eq becomes larger. [Products] [Reactants ] 11 56

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 6. Interpreting K eq values 3. The following will have NO effect on the chemical equilibrium. (i) Adding a catalyst No effect; A catalyst changes the rate of both the forward and reverse reactions to the same extent. (ii) Adding a solid No effect; Concentration is omitted from the equilibrium expression. (iii) Adding a pure liquid - No effect; Concentration is omitted from the equilibrium expression. (iv) Adding an inert gas under constant volume condition No effect; It affects the total pressure of the system. It has no effect on the partial pressures of the gases. 113 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 7. Explain Le Chatelier s Principle mathematically with K eq. Recall Le Chatelier s Principle: When a stress is applied to a closed system at equilibrium, the system will respond to REDUCE the stress and establish a new equilibrium. 114 57

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, At t = 0 min, the initial concentrations are: [H ]=.400 M, [N ] = 0.800 M At t=40 min, system reaches equilibrium. [H ]=1.197 M, [N ] = 0.399 M, [NH 3 ] = 0.01 M 115 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, K eq 1 [NH3] [N ] [H ] 3 0.03 0.3991.197 0.060 3 116 58

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, Equilibrium system is STRESSED! A concentration change occurred while keeping the temperature constant. 117 Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, System responds to REDUCE the stress! 118 59

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, 40 minutes later, the system finds a new equilibrium. 119 Le Chatelier s Principle Effect of Concentration Change Hebden Unit (page 50 55) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, New equilibrium concentrations are established. 10 60

Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) Let s examine the Concentration VS. Time graph for N (g) + 3 H (g) NH 3 (g) + heat At constant temperature, K eq 1 3 0.03 0.3991.197 0.060 K eq [NH3] [N ] [H ] 3 0.304 1.3481.044 0.060 3 K eq1 = K eq = 0.060 The system adjusts to reduce the stress such that the new equilibrium has the same K eq same 11 value. Equilibrium Constant Expression and K eq Hebden Unit (page 57 63) 1. Equilibrium Constant Expression. Equilibrium Constant, K eq 3. K c and K p 4. Heterogeneous equilibrium 5. Equilibrium laws for Combined equilibrium Scaled equilibrium Reversed equilibrium 6. Interpreting K eq values 7. Explain Le Chatelier s Principle mathematically with K eq. 1 61

Equilibrium Equilibrium Constant, K eq, Calculations Hebden Unit (page 37 69) Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I: Calculating the Equilibrium Constant Type II: Calculating the Equilibrium Concentrations Type III: Calculating the Equilibrium Concentrations when a reagent is added 14 6

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Consider the magnitude of K eq and the position of equilibrium. Reactants Products When K eq is very large - Reaction proceeds towards completion. The position (K eq >> 1) of equilibrium lies far to the right, toward the products. When K eq 1 - The concentrations of reactants and products are nearly the same at equilibrium. The position of equilibrium lies approximately midway between reactants and products. When K eq is very small - Extremely small amounts of products are formed. (K eq << 1) The position of equilibrium lies far to the left, toward the reactants. 15 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I: 1. Find the value of K eq for the following equilibrium CO (g) + H O () H + (aq) + HCO 3- (aq) when the equilibrium concentrations of: [CO ] = 0.16 M [H + ] = 0.38 M [HCO 3- ] = 3.0 M K eq [H ][HCO [CO ] 0.383.0-3 ] 0.16 7.1 16 63

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I:. Consider the equilibrium PCl 5 (g) PCl 3 (g) + Cl (g) Find K eq if the system initially contained only [PCl 5 ] = 0.70 M and the final [Cl ] = 0.15 M. 17 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I:. Consider the equilibrium PCl 5 (g) PCl 3 (g) + Cl (g) Find K eq if the system initially contained only [PCl 5 ] = 0.70 M and the final [Cl ] = 0.15 M. PCl 5 (g) PCl 3 (g) + Cl (g) [I] 0.70 0 0 [C] [E] 0.15 Given information 18 64

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I:. Consider the equilibrium PCl 5 (g) PCl 3 (g) + Cl (g) Find K eq if the system initially contained only [PCl 5 ] = 0.70 M and the final [Cl ] = 0.15 M. PCl 5 (g) PCl 3 (g) + Cl (g) [I] 0.70 0 0 [C] + 0.15 [E] 0.15 Solve for Change in Cl 19 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I:. Consider the equilibrium PCl 5 (g) PCl 3 (g) + Cl (g) Find K eq if the system initially contained only [PCl 5 ] = 0.70 M and the final [Cl ] = 0.15 M. PCl 5 (g) PCl 3 (g) + Cl (g) [I] 0.70 0 0 [C] 0.15 + 0.15 + 0.15 [E] 0.15 Solve for Change in PCl 5 and PCl 3 130 65

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I:. Consider the equilibrium PCl 5 (g) PCl 3 (g) + Cl (g) Find K eq if the system initially contained only [PCl 5 ] = 0.70 M and the final [Cl ] = 0.15 M. PCl 5 (g) PCl 3 (g) + Cl (g) [I] 0.70 0 0 [C] 0.15 + 0.15 + 0.15 [E] 0.55 0.15 0.15 Solve for Final Concentrations 131 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I:. Consider the equilibrium PCl 5 (g) PCl 3 (g) + Cl (g) Find K eq if the system initially contained only [PCl 5 ] = 0.70 M and the final [Cl ] = 0.15 M. K eq [Cl ] [PCl [PCl 5 ] 3 ] 0.150.15 0.55 0.041 13 66

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I: Equilibrium is established for the reaction N O 4 (g) NO (g) at 5 C. The quantities of the two gases present in a 3.00 liter vessel are 7.64 g N O 4 and 1.56 g NO. What is the value of K for this reaction? 133 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I: Equilibrium is established for the reaction N O 4 (g) NO (g) at 5 C. The quantities of the two gases present in a 3.00 liter vessel are 7.64 g N O 4 and 1.56 g NO. What is the value of K for this reaction? mol NO 4 n 7.64 4 0.0830 4 9.01 N O g N O mol N O g NO4 0.0830 mol NO4 NO4 0.077 M NO4 3.00 mol NO n 1.56 0.0339 46.01 NO g NO mol NO g NO 0.0830 mol NO NO 0.0113 M NO 3.00 K NO N O 0.0113 0.077 CHEM 4 001 Lecture Notes 4.61x10 3 4 134 67

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I: Equilibrium is established for the reaction SO 3 (g) SO (g) + O (g) at 900K When a 0.000 mol sample of SO 3 (g) is introduced into a 1.5 liter vessel, 0.014 mol of SO 3 (g) is found to be present at equilibrium. What is the value of K for this reaction? 135 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I: Equilibrium is established for the reaction Equilibrium is established for the reaction SO 3 (g) SO (g) + O (g) at 900K. When a 0.000 mol sample of SO 3 (g) is introduced into a 1.5 liter vessel, 0.014 mol of SO 3 (g) is found to be present at equilibrium. What is the value of K for this reaction? Reaction SO 3 (g) SO (g) + O (g) Initial amount 0.000 mol 0.00 mol 0.00 mol Change -0.0058 mol +0.0058 mol +0.009 mol Equil. Amt. 0.014 mol 0.0058 mol 0.009 mol Equil. Conc. 9.34x10-3 M 3.8x10-3 M 1.9x10-3 M K SO O SO 3 3 3.8x10 1.9 x10 3 9.34x10 3 4 3.1x10 136 68

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type II: 3. Consider the equilibrium H (g) + I (g) HI (g) The K eq = 50. at a certain temperature. Find the equilibrium concentration of HI, if at equilibrium [H ] = 0.50 M, [I ] = 0.50 M. H (g) + I (g) HI (g) [I] [C] [E] 0.50 0.50 x Equilibrium Concentrations 137 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type II: 3. Consider the equilibrium H (g) + I (g) HI (g) The K eq = 50. at a certain temperature. Find the equilibrium concentration of HI, if at equilibrium [H ] = 0.50 M, [I ] = 0.50 M. K eq [HI ] [H ][ I ] x (0.50 ) (0.50 ) 50 x (0.50)² (50) 3.5 Since x [HI], x 0 x [HI] 3.5 M 138 69

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type II: 4. Consider the equilibrium H (g) + I (g) HI (g) The K eq = 50. at a certain temperature. What are the equilibrium concentrations of each substance if a.0 L flask that initially contains only 1.0 mole H and 1.0 mole I. Assume the same temperature is maintained. H (g) + I (g) HI (g) [I] [C] [E] 1mole/.0 L = 0.50 1mole/.0 L = 0.50 0 Given Information 139 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type II: 4. Consider the equilibrium H (g) + I (g) HI (g) The K eq = 50. at a certain temperature. What are the equilibrium concentrations of each substance if a.0 L flask that initially contains only 1.0 mole H and 1.0 mole I. Assume the same temperature is maintained. [I] H (g) + I (g) HI (g) 1mole/.0 L = 0.50 1mole/.0 L = 0.50 [C] x x +x [E] 0 Stoichiometric ratio 140 70

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type II: 4. Consider the equilibrium H (g) + I (g) HI (g) The K eq = 50. at a certain temperature. What are the equilibrium concentrations of each substance if a.0 L flask that initially contains only 1.0 mole H and 1.0 mole I. Assume the same temperature is maintained. [I] H (g) + I (g) HI (g) 1mole/.0 L = 0.50 1mole/.0 L = 0.50 [C] x x +x [E] 0.50 x 0.50 x x 0 Equilibrium Concentrations 141 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type II: 4. Consider the equilibrium H (g) + I (g) HI (g) The K eq = 50. at a certain temperature. What are the equilibrium concentrations of each substance if a.0 L flask that initially contains only 1.0 mole H and 1.0 mole I. Assume the same temperature is maintained. K eq [HI] (x) [H ][I ] (0.50 x) 50 (x) (0.50) 50 50 or x 0.39 M, 0.70 M (0.50 x) 50 Since x [ HI], x 0, x 0.39 M 14 71

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type II: 4. Consider the equilibrium H (g) + I (g) HI (g) The K eq = 50. at a certain temperature. What are the equilibrium concentrations of each substance if a.0 L flask that initially contains only 1.0 mole H and 1.0 mole I. Assume the same temperature is maintained. [I] H (g) + I (g) HI (g) 1mole/.0 L = 0.50 1mole/.0 L = 0.50 [C] x x +x [E] 0.50 0.39 0.50 0.39 (0.39) 0 Calculate equilibrium concentrations 143 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type II: 4. Consider the equilibrium H (g) + I (g) HI (g) The K eq = 50. at a certain temperature. What are the equilibrium concentrations of each substance if a.0 L flask that initially contains only 1.0 mole H and 1.0 mole I. Assume the same temperature is maintained. Substitute these [H ] = 0.50 0.39 = 0.11 M equilibrium concentration into [I ] = 0.50 0.39 = 0.11 M the equilibrium constant expression to confirm that [HI] = (0.39) = 0.78 M K eq = 50. 144 7

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type III: 5. Consider the equilibrium. H (g) + I (g) HI (g) What are the final equilibrium concentrations of each substance if a 0.50 L flask initially contains.0 moles H, 1.0 mole I and 10. mole HI, then 1.0 mole I is added. H (g) + I (g) HI (g) [I].0 mole/0.50 L = 4.0 M [A] [C] [E] 1.0 mole/0.50 L =.0 M 1.0 mole/0.50 L =.0 M 10 mole/0.50 L = 0. M Given information 145 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type III: 5. Consider the equilibrium. H (g) + I (g) HI (g) What are the final equilibrium concentrations of each substance if a 0.50 L flask initially contains.0 moles H, 1.0 mole I and 10. mole HI, then 1.0 mole I is added. H (g) + I (g) HI (g) [I].0 mole/0.50 L = 4.0 M 1.0 mole/0.50 L =.0 M 10 mole/0.50 L = 0. M [A] 1.0 mole/0.50 L =.0 M [C] x x +x [E] 4.0 x 4.0 x 0. +x Equilibrium concentrations 146 73

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type III: 5. Consider the equilibrium. H (g) + I (g) HI (g) What are the final equilibrium concentrations of each substance if a 0.50 L flask initially contains.0 moles H, 1.0 mole I and 10. mole HI, then 1.0 mole I is added. K eq [HI] (0. x) [H ][I ] (4 x) 50 (0 x) 4 50 0 50 or x 0.91 M, 9.5 M (4 x) 50 Since 4 x [H ], x 4, x 0.91 M 147 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type III: 5. Consider the equilibrium. H (g) + I (g) HI (g) What are the final equilibrium concentrations of each substance if a 0.50 L flask initially contains.0 moles H, 1.0 mole I and 10. mole HI, then 1.0 mole I is added. [H ] = 4.0 0.91 = 3.09 M [I ] = 4.0 0.91 = 3.09 M [HI] = 0. + (0.91) = 1.8 M The addition of I resulted in a shift to the right. Substitute these equilibrium concentration into the equilibrium constant expression to confirm that K eq = 50. 148 74

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type II: 6. Consider the following equilibrium at 5 o C. NH 3 (g) 3 H (g) + N (g) Initially a 5.0 L flask contains only 1.0 mole NH 3 and 0.40 mole N. After equilibrium is established, 0.78 mole NH 3 remain. Find the new equilibrium concentrations of H and the value of K eq. NH 3 (g) 3 H (g) + N (g) [I] 1.0 mole/5.0 L = 0.0 M 0 M [C] [E] 0.78/5 = 0.156 M The addition of I resulted in a shift to the right. 0.40 mole/5.0 L = 0.08 M Given information 149 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type II: 6. Consider the following equilibrium at 5 o C. NH 3 (g) 3 H (g) + N (g) Initially a 5.0 L flask contains only 1.0 mole NH 3 and 0.40 mole N. After equilibrium is established, 0.78 mole NH 3 remain. Find the new equilibrium concentrations of H and the value of K eq. NH 3 (g) 3 H (g) + N (g) [I] 1.0 mole/5.0 L = 0.0 M 0 M [C] 0.044 M 3/(0.044)= 0.066 M [E] The addition 0.78/5 of I = resulted 0.156 M 0.066 M 0.10 M in a shift to the right. 0.40 mole/5.0 L = 0.08 M 0.044/= 0.0 M Equilibrium Conditions 150 75

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type I: Calculating the Equilibrium Constant Type II: Calculating the Equilibrium Concentrations Type III: Calculating the Equilibrium Concentrations when a reagent is added Type IV: Determining the shift and/or calculating more than one variable (K eq and Concentration) 153 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Consider the magnitude of K eq and the position of equilibrium. Reactants Products When K eq is very large - Reaction proceeds towards completion. The position of (K eq >> 1) equilibrium lies far to the right, toward the products. When K eq 1 - The concentrations of reactants and products are nearly the same at equilibrium. The position of equilibrium lies approximately midway between reactants and products. When K eq is very small - Extremely small amounts of products are formed. The (K eq << 1) position of equilibrium lies far to the left, toward the reactants. 154 77

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) For non-equilibrium conditions, K is replaced by the reaction quotient, Q. aa (aq) + bb (aq) cc (aq) + dd (aq) c [C] [D] Q a [A] [B] What is Q and what does Q have to do with K eq? d b 155 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Depending on the relationship between Q and K, one can predict the direction the reaction will shift until equilibrium is achieved. There are three conditions, which should be considered. Q = K eq, then the reaction is at equilibrium. Q > K eq, then the reaction is not at equilibrium and some products will be converted to reactants. Q < K eq, then the reaction is not at equilibrium and some reactants will be converted to products. c [C] [D] Q a [A] [B] d b 156 78

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type IV: 7. Consider the following equilibrium at 5 o C. NH 3 (g) 3 H (g) + N (g).0 L flask initially contains.5 M NH 3 (g), 0.40 M H (g), 0.50 M N (g). Show which way will the system shift to reach equilibrium. From example 6, K eq = 0.0011 NH 3 (g) 3 H (g) + N (g) [I].5 M 0.40 M 0.50 M [H ] [N Q [NH ] 3 3 3 ] (0.40) (0.50) (.5) 0.0051 157 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Compare Q to K eq, Q = 0.0051 K eq = 0.0011 * Q = K eq, then the reaction is at equilibrium. Q > K eq, then the reaction is not at equilibrium and some products will be converted to reactants. Q < K eq, then the reaction is not at equilibrium and some reactants will be converted to products. Equilibrium will adjust to convert products to reactant, shift to the left. * Calculated in example 6. 158 79

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Compare Q to K eq, 159 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Predicting the Direction of Net Change Example Consider the equilibrium among N (g), H (g) and NH 3 (g) in 1 liter flask. N (g) + 3 H (g) NH 3 (g) K = 9.5 at 300 C n N n H n NH3 Q Shift mass NH 3 or 1 1 1 1.7 0.3 1 0.3 1.7 1 Place the last mixture in a 10 liter flask (conc. by a factor of 10) 0.3 1.7 1 160 80

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Predicting the Direction of Net Change Example Consider the equilibrium among N (g), H (g) and NH 3 (g) in 1 liter flask. N (g) + 3 H (g) NH 3 (g) K = 9.5 at 300 C n N n H n NH3 Q Shift mass NH 3 or 1 1 1 1 1.7 0.3 1 0 0.3 1.7 1 0.7 Place the last mixture in a 10 liter flask (conc. by a factor of 10) 0.3 1.7 1 70 161 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) We can make assumptions to simplify the math. Equilibrium calculations can usually be simplified when K eq is very small (ie position of the equilibrium lies far to the left), toward the reactants. Usually this permits us to simplify the calculations considerably. example: 0.100 x 0.100 where 0.100 is the initial reactant concentration and the amount of dissociation, x, is small. This is valid if the initial reactant concentration of 0.100 is at least 1000 x K eq. But always check to see if the assumption is valid. 16 81

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) [ initial] K eq 1000 Sometimes assumptions can be made to simplify the math of the problems. Any time that an assumption is made, the validity of the assumption must be checked. If it is not valid, the quadratic equation should be used. The quadratic equation will be given in an exam. 163 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) The quadratic equation is used to determine the value of x when x is in a mathematical equation in this form: ax + bx +c = 0 where a, b and c are numbers (a 0). The quadratic equation is then set up to solve for x: x b b a 4ac 164 8

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) The quadratic equation will give two possible values of x, x b b a 4ac one from the + sign up and one from the sign in the numerator. Only one of these values will be correct. The incorrect value is usually obvious because it is usually results in a negative concentration. 165 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Type IV: 8. The equilibrium constant for the dissociation of iodine, I (g), is 3.76 x 10-3 at 1000 K. I (g) I (g) Suppose 1.00 mole of iodine is placed in a.00 L flask at 1000 K. What are the concentrations of iodine and the iodine atom when the system comes to equilibrium? [I] 1mole/.0 L = 0.50 M I (g) I(g) [C] x +x [E] 0.50 x x 0 Stoichiometric ratio 166 83

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) I (g) I (g) K eq [ -3 I] (x) 3.76x10 [I ] (0.500 x) Test to see if we can simplify. We can simplify if 0.500 3.76x10 [ initial] 1000 3 K eq 13.98 ( < 1000) We cannot simplify! We need to use the quadratic equation to solve for x. 167 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) I (g) I (g) K eq [ -3 I] (x) 3.76x10 [I ] (0.500 x) Rearrange, 3.76x10-3 (0.500 x) = 4x 0.00188 (3.76x10-3 )x = 4x 4x + (3.76x10-3 ) x - 0.00188 = 0 a = 4, b = 3.76x10-3, c = -0.00188 x b b a 4ac 168 84

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) I (g) I (g) K eq [ -3 I] (x) 3.76x10 [I ] (0.500 x) Use the quadratic formula. a = 4, b = 3.76 x 10-3, c = -0.00188 Compare: If we had made the assumption that 0.500 x 0.500 then x = 0.018 x = 0.01, -0.015 [I ] = 0.500 0.01 = 0.479 M [I] = 0.044 M K eq b x Check: 0.044) (0.479) b a 4ac ( -3 3.76x10 169 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) We have covered the following type of K eq calculation in this presentation. Type IV: Determining the shift and/or calculating more than one variable (K eq and Concentration) 170 85

Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Synthesis of NH 3 N (g) + 3 H (g) NH 3 (g) H = -9 kj/mol The optimum conditions are only for the equilibrium position and do not take into account the rate at which equilibrium is attained. 171 Equilibrium Constant, K eq, Calculations Hebden Unit (page 57 63) Schematic Diagram of the Haber Process Synthesis of NH 3 N (g) + 3 H (g) NH 3 (g) H = -9 kj/mol 17 86