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Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Gaseous Chemical Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 12

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Chemistry 1011 Slot 52 12.2 The Equilibrium Constant YOU ARE EXPECTED TO BE ABLE TO: Write an expression for the equilibrium constant, K, for a gaseous reaction Recognize that the expression for K depends on the form of the balanced chemical equation for the reaction. Write an expression for the equilibrium constant, K, for a gaseous reaction that includes a substance in the solid or liquid phase.

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Chemistry 1011 Slot 53 The Equilibrium Constant For a gaseous reaction, the equilibrium constant can be written in terms of the partial pressures (concentrations) of reactants and products For aA (g) + bB (g) cC (g) + dD (g) The equilibrium constant, K p, is K = (P C ) c x (P D ) d (P A ) a x (P B ) b

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Chemistry 1011 Slot 54 The Equilibrium Constant Equilibrium partial pressures of the products are in the numerator (top) Equilibrium partial pressures of the reactants are in the denominator (bottom) Each partial pressure is raises to a power equal to its coefficient in the balanced equation

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Chemistry 1011 Slot 55 Equilibrium Constant Example Ammonia is made industrially by the Haber Process: N 2(g) + 3H 2(g) 2NH 3(g) The equilibrium constant, K, is K p = (P NH 3 ) 2 P N 2 x (P H 2 ) 3

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Chemistry 1011 Slot 56 Equilibrium Constant Example Sulfuric acid is a very important industrial chemical. It is manufactured from sulfur dioxide and oxygen 2SO 2(g) + O 2(g) 2SO 3(g) The equilibrium constant, K, is K p = (P SO 3 ) 2 (P SO 2 ) 2 x P O 2

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Chemistry 1011 Slot 57 Dependence of K on Equation Stoichiometry The expression for K, and its value will depend on how the equation is written For N 2(g) + 3H 2(g) 2NH 3(g) K p = (P NH 3 ) 2 P N 2 x (P H 2 ) 3 For 1 / 2 N 2(g) + 3 / 2 H 2(g) NH 3(g) K p ’ = (P NH 3 ) (P N 2 ) 1/2 x (P H 2 ) 3/2

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Chemistry 1011 Slot 58 Dependence of K on Equation Stoichiometry The Coefficient Rule: –If coefficients in a balanced equation are multiplied by a factor, n, then –The equilibrium constant is raised to the n th power K’ = K n The Reciprocal Rule: –If the equation is written in reverse, then K’’ = 1/K

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Chemistry 1011 Slot 59 Adding Chemical Equations The Rule of Multiple Equilibria If a reaction can be expressed as the sum of two or more reactions, K for the overall reaction is equal to the PRODUCT of the equilibrium constants for the individual reactions –If Reaction 3 = Reaction 1 + Reaction 2 –Then K (Reaction 3) = K (reaction 1) x K (Reaction 2)

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Chemistry 1011 Slot 510 Multiple Equilibria Example Reaction 1: SO 2(g) + 1 / 2 O 2(g) SO 3(g) Kp = 2.2 = (P SO 3 ) (P SO 2 ) x (P O 2 ) 1 / 2 Reaction 2: NO 2(g) NO (g) + 1 / 2 O 2(g) Kp = 4.0 = (P NO ) x (P O 2 ) 1 / 2 (P NO 2 ) Adding the equations: SO 2(g) + NO 2(g) SO 3(g) + NO (g)

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Chemistry 1011 Slot 511 Multiple Equilibria Example The equilibrium constant expression for the total reaction is K p = (P SO 3 ) x (P NO ) (P SO 2 ) x (P NO 2 ) This is obtained by multiplying together the equilibrium constant expressions for the two individual reactions (P SO 3 ) x (P NO ) x (P O 2 ) 1 / 2 (P SO 2 ) x (P O 2 ) 1 / 2 (P NO 2 ) K p = 2.2 x 4.0 = 8.8

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Chemistry 1011 Slot 512 Heterogeneous Equilibria Up to now, all of the reactions considered have been homogeneous gas reactions In some cases, one or more of the substances involved may be a liquid or solid This would be a heterogeneous system Example: I 2(s) I 2(g)

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Chemistry 1011 Slot 513 Heterogeneous Equilibria – the Sublimation of Iodine I 2(s) I 2(g) The rate of the forward process depends only on temperature. Sublimation will occur at constant rate as long as some solid iodine remains The rate of the reverse reaction will depend on the concentration (partial pressure) of iodine gas At equilibrium (in a closed system), this rate will become constant K p = P I 2(g)

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Chemistry 1011 Slot 514 Heterogeneous Equilibria For heterogeneous equilibria, it is found that: –The position of equilibrium is independent of the amount of solid or liquid component, as long as some still remains –Concentrations of solids and liquids are constant –Terms for solid or liquid components do not appear in the expression for K

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Chemistry 1011 Slot 515 Heterogeneous Equilibria – the Decomposition of Calcium Carbonate CaCO 3(s) CaO (s) + CO 2(g) Written in terms of concentrations, the equilibrium constant expression is: K = [CaO (s) ][CO 2(g) ] [CaCO 3(s) ] But the concentrations of the solids are constant, so K = [CO 2(g) ], or K p = P CO 2

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Chemistry 1011 Slot 516 Heterogeneous Equilibria with a Liquid Component Example: CO 2(g) + H 2(g) CO (g) + H 2 O ( l ) In this case, the amount of water vapour in the system is constant, since it will be in equilibrium with the liquid water So, K p = (P CO ) (P CO 3 ) (P H 3 )

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AA + bB cC + dD Equilibrium RegionKinetic Region.

AA + bB cC + dD Equilibrium RegionKinetic Region.

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