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Chemistry 1011

Chemistry 1011. TOPIC Gaseous Chemical Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 12. 12.2 The Equilibrium Constant. YOU ARE EXPECTED TO BE ABLE TO: Write an expression for the equilibrium constant, K, for a gaseous reaction

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Chemistry 1011

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  1. Chemistry 1011 TOPIC Gaseous Chemical Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 12 Chemistry 1011 Slot 5

  2. 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. Chemistry 1011 Slot 5

  3. 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, Kp , is K = (PC)c x (PD)d (PA)a x (PB)b Chemistry 1011 Slot 5

  4. 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 Chemistry 1011 Slot 5

  5. Equilibrium Constant Example • Ammonia is made industrially by the Haber Process: N2(g) + 3H2(g) 2NH3(g) • The equilibrium constant, K, is Kp =(PNH3)2 PN2 x (PH2)3 Chemistry 1011 Slot 5

  6. Equilibrium Constant Example • Sulfuric acid is a very important industrial chemical. It is manufactured from sulfur dioxide and oxygen 2SO2(g) + O2(g) 2SO3(g) • The equilibrium constant, K, is Kp =(PSO3)2 (PSO2 )2 x PO2 Chemistry 1011 Slot 5

  7. Dependence of K on Equation Stoichiometry • The expression for K, and its value will depend on how the equation is written • For N2(g) + 3H2(g) 2NH3(g) Kp =(PNH3)2 PN2 x (PH2)3 • For 1/2N2(g) + 3/2H2(g) NH3(g) Kp’ =(PNH3) (PN2 )1/2 x (PH2)3/2 Chemistry 1011 Slot 5

  8. 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 nth power K’ = Kn • The Reciprocal Rule: • If the equation is written in reverse, then K’’ = 1/K Chemistry 1011 Slot 5

  9. 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) Chemistry 1011 Slot 5

  10. Multiple Equilibria Example Reaction 1: SO2(g) + 1/2O2(g) SO3(g) Kp = 2.2 = (PSO3) (PSO2 ) x (PO2 )1/2 Reaction 2: NO2(g) NO(g) + 1/2O2(g) Kp = 4.0 = (PNO) x (PO2 )1/2 (PNO2 ) Adding the equations: SO2(g) + NO2(g) SO3(g) + NO(g) Chemistry 1011 Slot 5

  11. Multiple Equilibria Example • The equilibrium constant expression for the total reaction is Kp =(PSO3) x (PNO) (PSO2 ) x (PNO2 ) This is obtained by multiplying together the equilibrium constant expressions for the two individual reactions (PSO3) x (PNO) x (PO2 )1/2 (PSO2 ) x (PO2 )1/2(PNO2 ) Kp = 2.2 x 4.0 = 8.8 Chemistry 1011 Slot 5

  12. 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: I2(s) I2(g) Chemistry 1011 Slot 5

  13. Heterogeneous Equilibria – the Sublimation of Iodine I2(s) I2(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 Kp = PI2(g) Chemistry 1011 Slot 5

  14. 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 Chemistry 1011 Slot 5

  15. Heterogeneous Equilibria – the Decomposition of Calcium Carbonate CaCO3(s) CaO(s) + CO2(g) • Written in terms of concentrations, the equilibrium constant expression is: K = [CaO(s)][CO2(g)] [CaCO3(s)] • But the concentrations of the solids are constant, so K = [CO2(g)], or Kp = PCO2 Chemistry 1011 Slot 5

  16. Heterogeneous Equilibria with a Liquid Component • Example: CO2(g) + H2(g) CO(g) + H2O(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, Kp = (PCO) (PCO3) (PH3) Chemistry 1011 Slot 5

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