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Chapter 14 - Chemical Equilibrium a system in which the...

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Chemical Equilibrium : a system in which the rates of the forward and reverse reactions are the same. no observable change occurs at equilibrium Reactants Products forward reverse
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H 2 O (g) H 2 O (l) physical equilibrium at equilibrium, liquid water evaporates at the same rate that water vapor condenses N 2 (g) + 3H 2 (g) 2NH 3 (g) chemical equilibrium the reaction reaches a steady state before all the reactants are consumed
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Learn to write the equilibrium constant expression for any chemical reaction. Determine the relationship between the expression for the equilibrium constant and the form of the balanced equation. Learn to convert between equilibrium constants in which the concentrations are expressed in mol/L and those in terms of partial pressures expressed in atm.
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Consider the reaction 2NO 2 (g) N 2 O 4 (g)
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Consider the reaction 2NO 2 (g) N 2 O 4 (g)
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Initial concentration, M Equilibrium concentration, M Concentration ratio at equilibrium [NO 2 ] [N 2 O 4 ] [NO 2 ] [N 2 O 4 ] [N 2 O 4 ] [NO 2 ] [N 2 O 4 ] [NO 2 ] 2 0.0200 0.0000 0.0103 4.86 x 10 -3 0.4718 45.8 0.0000 0.0100 0.0103 4.86 x 10 -3 0.4718 45.8 0.0100 0.0100 0.0134 8.29 x 10 -3 0.6187 46.2 0.0400 0.0000 0.0161 1.19 x 10 -2 0.7391 45.9 The square brackets like [ NO 2 ] indicate molar concentrations
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2 NO 2 (g) N 2 O 4 (g) equilibrium constant K eq = = 45.9 [NO 2 ] 2 [N 2 O 4 ]
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a A + b B c C + d D the equilibrium constant is given by K eq = [C] c [D] d [A] a [B] b [ ] Concentrations are in moles per liter Temperature influences K eq !!!
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2O 3 (g) 3O 2 (g) K eq = Cl 2 (g) 2Cl(g) K eq = CO(g) + H 2 O(g) H 2 (g) + CO 2 (g) K eq = [O 2 ] 3 [O 3 ] 2 [CO] [H 2 O] [H 2 ] [CO 2 ] [Cl 2 ] [Cl] 2
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Learn to write the equilibrium constant expression for any chemical reaction. Determine the relationship between the expression for the equilibrium constant and the form of the balanced equation. Learn to convert between equilibrium constants in which the concentrations are expressed in mol/L and those in terms of partial pressures expressed in atm.
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Lets say we double the coefficients of a chemical reaction - what will happen to the value of K eq ?
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2H 2 (g) + O 2 (g) 2H 2 O(g) K 1 = H 2 (g) + 1/2O 2 (g) H 2 O(g) K 2 = = K 1 [H 2 ] 2 [O 2 ] [H 2 O] 2 [H 2 ] [O 2 ] 1/2 [H 2 O]
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2H 2 O(g) 2H 2 (g) + O 2 (g) K 3 = = 1/K 1 The form of the equilibrium expression depends on the balanced equation For any reaction, K rev = 1/K for [H 2 ] 2 [O 2 ] [H 2 O] 2
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N 2 O 4 (g) 2NO 2 (g) K eq = 4.63 x10 -3 (a) determine K eq for the reaction 2NO 2 (g) N 2 O 4 (g) (b) determine K eq for the reaction NO 2 (g) 1/2N 2 O 4 (g)
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Answer: (a) K = 1/K eq = 1/4.63 x 10 -3 = 2.16 x 10 2 (b) K = 1/ K eq = 14.7
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Learn to write the equilibrium constant expression for any chemical reaction. Determine the relationship between the expression for the equilibrium constant and the form of the balanced equation. Learn to convert between equilibrium constants in which the concentrations are expressed in mol/L and those in terms of partial pressures expressed in atm.
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N 2 O 4 (g) 2NO 2 (g) K c = subscript c indicates molar concentration.
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