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1 Chemical Equilibria Chapter 16

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2 Chemical reactions Can reverse (most of the time) Can reverse (most of the time) Though might require a good deal of energy Though might require a good deal of energy or double arrows = reversible rxn or double arrows = reversible rxn It “swings both ways” It “swings both ways” Once fwd/rev rxns occur at equal rates Once fwd/rev rxns occur at equal rates = equilibrium (no net change seen) = equilibrium (no net change seen)

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3 Equilibrium constant aA + bB cC + dD aA + bB cC + dD Thus, Thus,

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4 Equilibrium constant in action Give the equilibrium constant for: Give the equilibrium constant for: H 2(g) + I 2(s) 2HI (g) Initially, 0.0175 M of reactants Initially, 0.0175 M of reactants Decrease to 0.0037 M reactants Decrease to 0.0037 M reactants What is the concentration of HI formed? What is the concentration of HI formed?

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5 Solution

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6 ICE Table Initial Change Equilibrium Table Initial Change Equilibrium Table Great way to explain and show concentration changes Great way to explain and show concentration changes Let’s make an ICE table for the previous reaction on the board Let’s make an ICE table for the previous reaction on the board

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7 Calculate K eq for the reaction

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8 Problems Express K eq for: Express K eq for: CH 3 OH (g) CO (g) + 2H 2(g) Express K eq for: Express K eq for: C 3 H 8(g) + 5O 2(g) 3CO 2(g) + 4H 2 O (g) Solve K eq for the following: Solve K eq for the following: A (g) 2B (g) –Given [A] i = 1.00M, [B] i = 0.00M, and [A] eq =0.75M

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10 Caveats to K eq : solids Solids in reversible rxns are excluded from expression since concentration derived from constant densities: Solids in reversible rxns are excluded from expression since concentration derived from constant densities: S (s) + O 2(g) SO 2(g) S (s) + O 2(g) SO 2(g) K eq = ? K eq = ?

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11 Caveats to K eq : aqueous solns Same rule for pure liquids Same rule for pure liquids Ex: NH 3(aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Ex: NH 3(aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) K eq = ? K eq = ?

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12 Caveats to K eq : gases & K p For gases we can use partial pressures For gases we can use partial pressures –Why? Hint: Relationship between pressure and concentration Hint: Relationship between pressure and concentration –See ideal gas law Given: Given: aA + bB cC + dD K p not necessarily equal to K eq K p not necessarily equal to K eq

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13 The relationship between K eq and K p

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14 Problem Given Given 2NO (g) + O 2(g) 2NO 2(g) K p = 2.2 x 10 12 @ 25°C K p = 2.2 x 10 12 @ 25°C Find K eq Find K eq

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15 Solution

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16 Determining equilibrium constant Given 2SO 2(g) + O 2(g) 2SO 3(g) Given 2SO 2(g) + O 2(g) 2SO 3(g) And: [SO 2 ] i = 1.00 M, [O 2 ] i = 1.00 M & [SO 3 ] f = 0.925 M And: [SO 2 ] i = 1.00 M, [O 2 ] i = 1.00 M & [SO 3 ] f = 0.925 M What is K eq ? What is K eq ?

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17 Solution

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18 More problem solving Sulfuryl chloride (SO 2 Cl 2 ) dissociates into sulfur dioxide and chlorine in the gas phase: SO 2 Cl 2(g) SO 2(g) + Cl 2(g) In an experiment, 3.174 g of SO 2 Cl 2 (MW = 134.96g/mol) is placed in a 1.000 L flask and is at 100.0°C. At equilibrium, the total pressure in the flask is 1.30 atm. Calculate: a) The partial pressures of each gas at equilibrium. b) The K p at 100.0°C for the reaction.

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19 Solution

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20 Yet another Isopropanol can dissociate into acetone and hydrogen: (CH 3 ) 2 CHOH (g) (CH 3 ) 2 CO (g) + H 2(g) At 179°C, the equilibrium constant is 0.444. Calculate the equilibrium partial pressures of all three gases if 10.00 g (MW = 60.10g/mol) of isopropanol are initially placed in a 10.00 L vessel.

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21 Solution

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22 Another but with a twist At 25°C, the equilibrium constant for the reaction below is 5.9 x 10 -13. 2NO 2(g) 2NO (g) + O 2(g) Suppose a container is filled with 0.89 atm of NO 2 Calculate the equilibrium partial pressures of each gas OK to approximate if “x [A] 0 < 5%“ – –“5% rule”

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23 Solution

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24 More on K eq Is the rxn product or reactant favored? Is the rxn product or reactant favored? –I.e., Will it form more product or reactant? If K 1, then prod concentration higher than reactant concentration prod favored If K 1, then prod concentration higher than reactant concentration prod favored –Makes mathematical sense See right See right If K < 1, then reactant concentration higher than prod concentration reactant favored If K < 1, then reactant concentration higher than prod concentration reactant favored If K = 1, neither favored; both equal concentration If K = 1, neither favored; both equal concentration

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25 Reaction quotient, Q When rxn not at equilibrium, use Q When rxn not at equilibrium, use Q Where Where aA + bB cC + dD –Then Remember, Q used for system when system NOT in equilibrium Remember, Q used for system when system NOT in equilibrium

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26 Q – its benefits Answers the question: Answers the question: Is the system at equilibrium (i.e., does Q = K)? If not, we can predict which way the reaction will continue to proceed If not, we can predict which way the reaction will continue to proceed If Q < K, rxn still needs to go to prod side to achieve equilibrium (i.e., where Q = K) If Q < K, rxn still needs to go to prod side to achieve equilibrium (i.e., where Q = K) –In other words, insufficient product formed for equilibrium conditions If Q > K, rxn has “overshot” K and needs to go to reactant side to achieve equilibrium (i.e., where Q = K) If Q > K, rxn has “overshot” K and needs to go to reactant side to achieve equilibrium (i.e., where Q = K) –In other words, an excess of product is formed for equilibrium conditions

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27 Work on this… N 2(g) + 3H 2(g) 2NH 3(g) The equilibrium constant at 400°C is K eq =0.5. Suppose we make a mixture with the following concentrations: [NH 3 ] = 5.0M, [N 2 ] = 3.5M, [H 2 ] = 1.9M In which direction will the reaction go? a) products b) reactants

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29 Manipulating K eq C (s) + ½ O 2(g) CO (g) C (s) + ½ O 2(g) CO (g) K p ’ = ? K p ’ = ? Changing it: 2C (s) + O 2(g) 2CO (g) Changing it: 2C (s) + O 2(g) 2CO (g) K p ” = ? K p ” = ? How are K p ’ and K p ” related mathematically? How are K p ’ and K p ” related mathematically?

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30 More on K eq Given 2NO 2(g) N 2 O 4(g) Given 2NO 2(g) N 2 O 4(g) K eq = ? K eq = ? What is K eq when rxn is reversed? What is K eq when rxn is reversed? Therefore, what can we say about K eq fwd and K eq rev ? Therefore, what can we say about K eq fwd and K eq rev ?

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31 Even more on K eq Remember Hess’s Law? Remember Hess’s Law? 1) A+B C 2) B+C D What is the net rxn and it’s K net using Hess’s Law? Can one obtain the same values as above not using Hess’s Law? What can we say about K net using each rxn’s K eq ?

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32 Disturbing chemical equilibria Le Châtelier’s Principle Le Châtelier’s Principle Change one component of the rxn & the rxn will attempt to rectify it Change one component of the rxn & the rxn will attempt to rectify it Think of it this way: Think of it this way: If something is changed, how can it be undone or controlled so that equilibrium is achieved once again? If something is changed, how can it be undone or controlled so that equilibrium is achieved once again?

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33 Temperature variation on equilibrium 2NO 2(g) N 2 O 4(g) + heat 2NO 2(g) N 2 O 4(g) + heat K p = ? K p = ? H° = -57.1 kJ H° = -57.1 kJ @ 273 K, K p = 1300, and @ 298 K, K p = 170 @ 273 K, K p = 1300, and @ 298 K, K p = 170 Hence, if one raises the temp to 298 K, which way will it swing? Hence, if one raises the temp to 298 K, which way will it swing?

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34 Pressure & volume change on equilibrium If volume decreased (pressure increased) favors smaller # of molecules If volume decreased (pressure increased) favors smaller # of molecules If volume increased (pressure decreased) favors larger # of molecules If volume increased (pressure decreased) favors larger # of molecules If reversible rxn has = # of molecules on each side, a volume/pressure change will do nothing If reversible rxn has = # of molecules on each side, a volume/pressure change will do nothing

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35 Question What change in equilibrium will be seen when one adds a solid and why? What change in equilibrium will be seen when one adds a solid and why?

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36 How about this? 2H 2 S (g) + O 2(g) 2S (s) + 2H 2 O (g) ; ΔH = -221.19 kJ/mol If O 2(g) is added to the reaction vessel, what happens to the amount of S (s) ? – –a) It increases b) It decreases c) Nothing If the volume of the vessel is cut in half, what happens to the ratio of P H 2 O /P H 2 S ? – –a) It increases b) It decreases c) Nothing If the temperature is increased, what happens to the equilibrium constant K? – –a) It increases b) It decreases c) Nothing If S (s) is added to the reaction, what happens to P H 2 O ? – –a) It increases b) It decreases c) Nothing

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