Lecture2.13_Class.ppt - UNIT 1 Chemical Chemical...

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UNIT 1 Chemical Equilibrium Chemical Equilibrium Lecture 3
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K Unit conventions K is often referred to as the THERMODYNAMIC K When all concentrations are in Molarity K can be written as Kc When all concentrations are in Pressures K can be written as Kp Because NaCl is a pure solid: [NaCl] = 1 We call this a solubility product It doesn’t matter how much solid we add the degree to which the solid can dissolve is fixed! NaCl (s) + H 2 O (l) Na + (aq) + Cl - (aq) K [ Na ( aq ) ][ Cl ( aq ) ] [ NaCl ( s ) ] [ Na ( aq ) ][ Cl ( aq ) ] Why can we use 1 for for pure solids and liquids? Do these concentration conventions make sense?
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K Unit conventions Why can we use pressures in atmospheres for gases? Do these concentration conventions make sense? K is often referred to as the THERMODYNAMIC K When all concentrations are in Molarity K can be written as Kc When all concentrations are in Pressures K can be written as Kp Ideal Gas Law: PV = nRT n/V = P/RT M = P/RT (n/V = Molarity) Molarity is proportional to Pressure (when T is constant) To convert from P (atm) to M: divide P by RT To convert from M to P(atm): multiply M by RT
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K Unit conventions – An example CaCO 3(s) CaO (s) + CO 2(g) [CaO (s) ]P CO2 ] K = [CaCO 3(s) ] Concentrations of pure solids and liquids are constant and so they are dropped from the expression equilibrium constant (K) expression. K = P CO2
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The equilibrium constant expression for the reaction shown here is: Ag 3 PO 4 (s) 3Ag + (aq) + PO 4 3- (aq) Clicker Q1: L.G. 7 A. A. A. B. K c = [Ag 3 PO 4 ] [Ag ][PO 4 3 ] K c [Ag ][PO 4 3 ] [Ag 3 PO 4 ] K c [Ag ] 3 [PO 4 3 ] K c [Ag 3 PO 4 ] [Ag ] 3 [PO 4 3 ]
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Numerical value of K (Calcs) The numerical value of K depends on how the reaction has been written K = [Products]/[Reactants] to the power of the rxn stoichiometry The numerical value of K for one reaction can be related to the numerical value of K for another reaction without knowing fixed concentrations.
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Rule 1: Relating the value of K When you reverse a rxn the value of K rev is simply the reciprocal value of K for the original rxn. K decomp = 1/ K form K decomp = [CaO (s) ][CO 2 (g) ] [CaCO 3(s) ] K form = [CaCO 3(s) ] [CaO (s) ][CO 2 (g) ] Decomposition reaction CaCO 3(s) CaO (S) + CO 2(g) ) Heat Formation reaction CaO (s) + CO 2(g) ) CaCO 3(s) = [CO 2(g) ] = 1/[CO 2(g) ]
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The equilibrium constant for the reaction NO(g) + ½O 2 (g) NO 2 (g) has a value of K = 1.23 at a certain temperature. What is the value of K for the reaction below? NO 2 (g) NO(g) + ½O 2 (g) A. 2.46 B. 1.51 C. 0.81 D. 0.41 Clicker Q2: L.G. 4
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When you multiply a reaction by N, K is simply raised to the power N K new = K N K = [HI] 4 /[I 2 ] 2 [H 2 ] 2 K = [HI] 2 /[I 2 ][H 2 ] K = [HI] /[I 2 ] 1/2 [H 2 ] 1/2 = 0.71 = 0.25 I 2 (l) + H 2 (g) 2HI (s) K = 0.5 0.5I 2 (l) + 0.5H 2 (g) HI (s) K = 0.71 2I 2 (l) + 2H 2 (g) 4HI (s) K = 0.25 = (0.5) 1/2 = (0.5) 2 Rule 2: Relating the value of K
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The equilibrium constant for the reaction NO(g) + ½ O 2 (g) NO 2 (g) has a value of K = 1.23 at a certain temperature.
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