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Chemical Equilibrium

Chemical Equilibrium. Chapter 14. Equilibrium: the extent of a ________. In stoichiometry we talk about ______yields, and the many reasons actual yields may be _________.

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Chemical Equilibrium

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  1. Chemical Equilibrium Chapter 14

  2. Equilibrium: the extent of a ________ In stoichiometry we talk about ______yields, and the many reasons actual yields may be _________. Another critical reason actual yields may be lower is the _______ of chemical reactions: some reactions may produce only 70% of the product you may calculate they ought to produce. Equilibrium looks at the ______ of a chemical reaction.

  3. H2O (l) H2O (g) N2O4(g) 2NO2(g) Equilibrium is a state in which there are no observable _______ as time goes by. • _______equilibrium is achieved when: • the rates of the ______ and _____ reactions are ______ and • the _______ of the reactants and products remain _______ Physical equilibrium Chemical equilibrium 14.1

  4. Rate of ______ of cookies =Rate of ________ cookies

  5. The Concept of Equilibrium • Consider _______ frozen N2O4. At room temperature, it _________ to brown NO2: • N2O4(g)  2NO2(g). • At some time, the _______ stops changing and we have a mixture of N2O4 and NO2. • ________equilibrium is the point at which the rate of the ______ reaction is equal to the rate of the _______ reaction. At that point, the concentrations of all species are ________. • Using the collision model: • as the amount of NO2 builds up, there is a chance that two NO2 molecules will _____ to form ______. • At the beginning of the reaction, there is no ____ so the ______ reaction (2NO2(g)  N2O4(g)) does not occur.

  6. The Concept of Equilibrium • As the substance warms it begins to ________: • N2O4(g)  2NO2(g) • When enough NO2 is formed, it can react to form N2O4: • 2NO2(g)  N2O4(g). • At _________, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4 • The _____ _______ implies the process is dynamic.

  7. The Concept of Equilibrium • As the reaction progresses • [A] __________ to a constant, • [B] __________ from zero to a constant. • When [A] and [B] are constant, ________ is achieved.

  8. _________ __________ __________ N2O4(g) 2NO2(g) Start with NO2 Start with N2O4 Start with NO2 & N2O4 14.1

  9. ________ 14.1

  10. The Equilibrium Constant • No matter the starting _________ of reactants and products, the ______ ______ of ________ is achieved at equilibrium. • For a general reaction • the equilibrium constant expression is • where ____ is the equilibrium constant.

  11. N2O4(g) 2NO2(g) K = aA + bB cC + dD [NO2]2 [C]c[D]d [N2O4] K = [A]a[B]b = 4.63 x 10-3 Equilibrium Will K >> 1 Lie to the ______ ______ products K << 1 Lie to the ______ ______reactants 14.1

  12. N2O4(g) 2NO2(g) aA (g) + bB (g)cC (g) + dD (g) _________________applies to reactions in which all reacting species are in the _________ ______. In most cases Dn = moles of _________ products – moles of ______ reactants = (c + d) – (a + b) 14.2

  13. CH3COOH (aq) + H2O (l) CH3COO-(aq) + H3O+ (aq) ‘ ‘ Kc = Kc [H2O] General practice _____ to include units for the equilibrium constant. = _______________ Equilibrium [H2O] = constant Kc = 14.2

  14. The Equilibrium Expression • Write the equilibrium expression for the following reaction:

  15. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2(g) COCl2(g) Kc= 14.2

  16. The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm? 2NO2 (g) 2NO (g) + O2 (g) 2 14.2

  17. CaCO3(s) CaO (s) + CO2(g) _________equilibrium applies to reactions in which reactants and products are in ________ phases. The concentration of ______ and _______are not included in the expression for the equilibrium constant. 14.2

  18. CaCO3(s) CaO (s) + CO2(g) PCO PCO 2 2 does not depend on the amount of CaCO3 or CaO = Kp 14.2

  19. Writing Equilibrium Constant Expressions • The concentrations of the reacting species in the condensed phase are expressed in ____. In the gaseous phase, the concentrations can be expressed in ___ or in ___. • The concentrations of pure ____, pure ____ and solvents do ____ appear in the equilibrium constant expressions. • The equilibrium constant is a ________ quantity. • In quoting a value for the equilibrium constant, you must specify the ______________ and the __________. • If a reaction can be expressed as a _________ of two or more reactions, the equilibrium constant for the overall reaction is given by the _____ of the equilibrium _______ of the individual __________. 14.2

  20. Calculating Equilibrium Concentrations • Express the equilibrium concentrations of all species in terms of the ______ concentrations and a single unknown ___, which represents the ______ in concentration. • Write the equilibrium constant expression in terms of the equilibrium ______. Knowing the value of the equilibrium constant, solve for x. • Having solved for x, calculate the equilibrium concentrations of all species. 14.4

  21. At 12800C the equilibrium constant (Kc) for the reaction Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br2 (g) 2Br (g) 14.4

  22. 14.4

  23. Example Problem: Calculate Concentration Note the moles into a 10.32 L vessel stuff ... calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M 2 HI H2 + I2 0.242 M 0 0 Initial: Change: Equil:

  24. And yes, it’s a quadratic equation. Doing a bit of rearranging: Example Problem: Calculate Concentration x = 0.00802 or –0.00925 Since we are using this to model a real, physical system, we reject the negative root. The [H2] at equil. is 0.00802 M.

  25. Example Problem: Calculate Keq This type of problem is typically tackled using the “three line” approach: 2 NO + O2 2 NO2 Initial: Change: Equilibrium:

  26. Approximating If Keq is really small the reaction will not proceed to the ______ very far, meaning the equilibrium concentrations will be nearly the _____ as the _____ concentrations of your ______. 0.20 – x is just about 0.20 is x is really dinky. If the difference between Keq and initial concentrations is around _____ orders of magnitude or more, go for it. Otherwise, you have to use the ________.

  27. Example Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I2 2 I More than ___ orders of mag. between these numbers. The simplification will work here. 0.20 0 -x +2x 0.20-x 2x Initial change equil: With an equilibrium constant that small, whatever x is, it’s near dink, and 0.20 minus dink is 0.20 (like a million dollars minus a nickel is still a million dollars). 0.20 – x is the same as 0.20 x = 3.83 x 10-6 M

  28. Example Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I2 2 I These are too close to each other ... 0.20-x will not be trivially close to 0.20 here. 0.20 0 -x +2x 0.20-x 2x Initial change equil: Looks like this one has to proceed through the quadratic ...

  29. A + B C + D C + D E + F A + B E + F [C][D] [E][F] [E][F] [C][D] [A][B] [A][B] Kc = ‘ ‘ ‘ Kc Kc Kc = ‘ ‘ Kc = x Kc = ‘ ‘ ‘ ‘ Kc Kc Kc 14.2

  30. N2O4(g) 2NO2(g) 2NO2(g) N2O4(g) ‘ K = K = = 4.63 x 10-3 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. [NO2]2 [N2O4] [N2O4] [NO2]2 1 K = = 216 14.2

  31. The ____________is calculated by substituting the ______ concentrations of the reactants and products into the equilibrium constant (Kc) expression. • IF • Qc __ Kcsystem proceeds from right to left to reach equilibrium • Qc __Kc the system is at equilibrium • Qc __Kcsystem proceeds from left to right to reach equilibrium 14.4

  32. N2(g) + 3H2(g) 2NH3(g) Equilibrium shifts left to offset stress Add NH3 __________’s Principle If an _____ stress is applied to a system at equilibrium, the system _____ in such a way that the stress is partially _____as the system reaches a new _______ position. • Changes in _______________ 14.5

  33. aA + bB cC + dD Le Châtelier’s Principle • Changes in Concentration continued Change Shifts the Equilibrium Increase concentration of product(s) Decrease concentration of product(s) Increase concentration of reactant(s) Decrease concentration of reactant(s) 14.5

  34. A (g) + B (g) C (g) Le Châtelier’s Principle • Changes in Volume and Pressure Change Shifts the Equilibrium _______ pressure Side with ______ moles of gas ________ pressure Side with _____ moles of gas ________ volume Side with _____ moles of gas ________ volume Side with _____ moles of gas 14.5

  35. Le Châtelier’s Principle • Changes in Temperature Change Exothermic Rx Endothermic Rx _______ temperature K _______ K _______ _______temperature K ________ K ________ hotter colder 14.5

  36. uncatalyzed catalyzed Le Châtelier’s Principle • Adding a _______ • does not change ___ • does not shift the _______of an equilibrium system • system will ______ equilibrium sooner Catalyst lowers Ea for ______ forward and reverse reactions. Catalyst ________ change equilibrium constant or shift equilibrium. 14.5

  37. Example

  38. Change Equilibrium Constant Le Châtelier’s Principle Change Shift Equilibrium Concentration Pressure Volume Temperature Catalyst 14.5

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