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TOPIC A: EQUILIBRIUM Equilibrium, Le Chatelier’s Principle, Acid- Base Equilibrium, Ksp, pH

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Dynamic Equilibrium Many chemical reactions and processes are reversible From Biology: as oxygen binding and then being released from hemoglobin the continual evaporation and condensation of water in the atmosphere From Chemistry: a solid dissolving and then crystallizing (solid aqueous) electrons being lost and gained in REDOX reactions (electrochemical cells, ΔG = 0) H+ ions being exchanged in acid-base reactions. (acid base)

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A dynamic equilibrium exists when the rate of the forward reaction is equal to the rate of the backward reaction. At the beginning of the reaction: As the reaction proceeds: ReactantsProducts reactant concentrations are high rate of the forward reaction is high product concentrations are low rate of the backward reaction is low. ReactantsProducts the reactant concentrations will decrease rate of the forward reaction slows the product concentrations increase rate of the backward reaction will increase.

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Equilibrium: The point at which the rates of forward and backward reaction are the same

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Reaction Rate Time Forward Reaction Reverse reaction Equilibrium

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If more reactants are introduced to the equilibrium: the forward reaction will increase a new, constant forward reaction will be established. The same effect can be observed in the reverse reaction rate when a product is introduced. When equilibrium has been achieved: it appears that the reaction has stopped on the molecular scale, it is still occurring

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What is equal at Equilibrium? Rates are equal Concentrations are not. Rates are determined by concentrations and activation energy. The concentrations do not change at equilibrium. or if the reaction is very slow.

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If the products and reactants have observable differences (colors for example), the reaction mixture will often appear as a mixture of the two. [Co(H 2 O) 6 ] Cl- [CoCl 4 ] H 2 O Red Blue For example, the mixture above will often appear purple if reactants and products are both present. For example

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Law of Mass Action For any reaction jA + kB lC + mD K = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power K is called the equilibrium constant. is how we indicate a reversible reaction Where [ ] represents equilibrium concentrations (M)

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Examples: (1) Write the equilibrium expression for the following reaction: 4NH 3 (g) + 7O 2 (g) 4NO 2 (g) + 6H 2 O (g)

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Calculating the values of K If we write the reaction in reverse. lC + mD jA + kB Then the new equilibrium constant is K ’ = [A] j [B] k = 1/K [C] l [D] m

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Calculating the values of K If we multiply the equation by a constant njA + nkB nlC + nmD Then the equilibrium constant is K’ = [A] nj [B] nk = ([A] j [B] k ) n = K n [C] nl [D] nm ( [C] l [D] m ) n

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Example N 2 + 3H 2 2NH 3 For the following concentrations: [NH 3 ] = 3.1 x mol/L [N 2 ] = 8.5 x mol/L [H 2 ] = 3.1 x mol/L 1. Calculate the K at 127 o C 2. Calculate the K for: 2NH 3 N 2 + 3H 2 3. Calculate K for: 1/2N 2 + 3/2H 2 NH 3

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The units for K Are determined by the various powers and units of concentrations. They depend on the reaction.

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K is CONSTANT At any temperature. Temperature affects rate. The equilibrium concentrations don’t have to be the same only K. Equilibrium position is a set of concentrations at equilibrium. There are an unlimited number.

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Calculate K N 2 + 3H 2 2NH 3 Initial At Equilibrium [N 2 ] 0 =1.000 M [N 2 ] = 0.921M [H 2 ] 0 =1.000 M [H 2 ] = 0.763M [NH 3 ] 0 =0 M [NH 3 ] = 0.157M

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Calculate K N 2 + 3H 2 2NH 3 Initial At Equilibrium [N 2 ] 0 = 0 M [N 2 ] = M [H 2 ] 0 = 0 M [H 2 ] = M [NH 3 ] 0 = M [NH 3 ] = 0.203M K is the same no matter what the amount of starting materials

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(2) Equilibrium Practice Problems (1) – Handout The relative size of K can give a simple, quantitative guide to the amounts of products and reactants present in any given equilibrium mixture. Large K values - relatively large amount of products Small K values - relatively large amount of reactants

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Equilibrium in gaseous reactions, Kp Equilibrium constants for gaseous reactions are usually found in terms of the partial pressures of the components. Partial Pressure of A = (mole fraction of A) (Total Pressure) Mole Fraction of A = mols of A total mols

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Consider the Equilibrium: 2SO 2 (g) + O 2 (g) 2SO 3 (g) K p = (P SO3 ) 2 (P SO2 ) 2 (P O2 ) K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] Equilibrium and Pressure

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The relationship between Kc and Kp An equilibrium constant ( K ), can be expressed two ways Kc (concentration) or Kp (partial pressure) The two terms are interchangeable using: Kp = Kc (RT) Δn Δn = (mols of products – mols of reactants) T = Temperature in K R = atm L / K mol (Kp is in atm)

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If K p = 1.9 x 10 3, calculate the value of K at 25 o C for the reaction 2NO (g) + Cl 2 (g) 2NOCl (g) Example Calculating K from K p

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Tells you the direction the reaction will go to reach equilibrium Calculated the same as the equilibrium constant, but for a system not at equilibrium Use initial concentrations Q = [Products] coefficient [Reactants] coefficient Compare value to equilibrium constant The Reaction Quotient

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If Q

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for the reaction 2NOCl(g) 2NO(g) + Cl 2 (g) K = 1.55 x M at 35ºC In an experiment: 0.10 mol NOCl, mol NO(g) and mol Cl 2 are mixed in 2.0 L flask. Which direction will the reaction proceed to reach equilibrium? Example

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Given the starting concentrations and one equilibrium concentration. Use stoichiometry to figure out other concentrations and K. Learn to create a table of initial and final conditions. Solving Equilibrium Problems

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Consider the following reaction at 600ºC 2SO 2 (g) + O 2 (g) 2SO 3 (g) In a certain experiment 2.00 mol of SO 2, 1.50 mol of O 2 and 3.00 mol of SO 3 were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO 3 were found to be present. Calculate the equilibrium concentrations of O 2 and SO 2, K and K P

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Consider the same reaction at 600ºC In a different experiment.500 mol SO 2 and.350 mol SO 3 were placed in a L container. When the system reaches equilibrium mol of O 2 are present. Calculate the final concentrations of SO 2 and SO 3 and K

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1. Write the balanced equation for the reaction 2. Write the equilibrium expression for the reaction 3. Set up an ICE table 4. List the initial concentrations 5. Calculate Q and determine the direction of the shift 6. Define the change needed to reach equilibrium and define the equilibrium concentrations 7. Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown Procedure to Solve Equilibrium Problems

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Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K the equilibrium constant is Calculate the equilibrium concentrations of all species if mol of each component is mixed in a L flask. Sample Ex

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Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 x 10 2 at a certain temperature. In a particular experiment, mol of each component was added to a L flask. Calculate the equilibrium concentrations of all species. Sample Ex

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H 2(g) + F 2(g) 2HF (g) If 3.0 mol of H 2 and 6.0 mol of F 2 are combined in a 3.0-L flask and allowed to reach equilibrium. Find the concentrations of each component at equilibrium. K=1.15 x 10 2 (Quadratic) Example: Large K

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Gaseous NOCl decomposes to form the gases NO and Cl 2. At 35 o C the equilibrium constant is 1.6 x In an experiment in which 1.0 mol NOCl is placed in a 2.0-L flask, what are the equilibrium concentrations? Example: Small K

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The rule of thumb is that if the value of the shift (x or 2x etc.) is less than 5% of all the initial concentration, our assumption was valid. If not we would have had to use the quadratic equation More on this later. Our assumption was valid. Checking the assumption

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