_Gr.11_Chem_PPT_L.14.2_ writing equilibrium constant expressions.pdf

14.2: writing equilibrium constant expressions Objectives 1. Differentiate between heterogeneous and homogeneous equilibria 2. Convert between Kp and Kc for reactions involving gaseous species 3. Determine the value of equilibrium constant 2 ...

14.2: writing equilibrium constant expressions Objectives 1. Differentiate between heterogeneous and homogeneous equilibria 2. Convert between Kp and Kc for reactions involving gaseous species 3. Determine the value of equilibrium constant 2  Heterogeneous equilibrium For example, the thermal decomposition of calcium carbonate in the commercial preparation of lime occurs by a reaction involving both solid and gas phases However, experimental results show that the position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. REASON is that the concentrations of pure solids and liquids cannot change.  Homogeneous Heterogeneous equilibria equilibria Reactions in which all Reactions in which reactants and products are in reactants and products same phase are in different phases  Equilibrium Expressions for Heterogeneous Equilibria We can generalize from this result as follows: If pure solids or pure liquids are involved in a chemical reaction, their concentrations are not included in the equilibrium expression for the reaction. This simplification occurs only with pure solids or liquids, not with solutions or gases, since in these last two cases the concentrations can vary.  Equilibrium Expressions for Heterogeneous Equilibria For example, in the decomposition of liquid water to gaseous hydrogen and oxygen, Introducing Kp Equilibrium is described involving gases and solutions in terms of concentrations. Equilibrium involving gases also can be described in terms of pressures. 7  Equilibrium including GASES K or Kc Kp Equilibrium in terms Equilibrium in terms of concentration of Partial Pressures 8  Example 1: Equilibrium partial pressures of the gases Kp Equilibrium concentrations K Equilibrium partial pressures of the gases Kp 9 10  Write expressions for KP for the following reactions. REACTION KP expression a) 2NH3(g) + CO2(g) ⇌ N2CH4O(s) + H2O(g) b) 2NBr3(s) ⇌ N2(g) + 3Br2(g) c) 2KClO3(s) ⇌ 2KCl(s) + 3O2(g) d) CuO(s) + H2(g) ⇌ Cu(l) + H2O(g) 11 Example 2:  N2(g) + 3H2(g) → 2NH3(g) Find Kp given the equilibrium pressures at a certain temperature: 13 Practice The following equilibrium pressures at a certain temperature were observed for the reaction: 2NO2(g) ⇌ 2NO(g) + O2(g) P(NO2) = 0.55 atm; P(NO) = 6.5 x 10-5 atm; P(O2) = 4.5 x 10-5 atm Calculate the value for the equilibrium constant Kp at this temperature. 14 Practice The following equilibrium pressures were observed at a certain temperature for the reaction: N2 (g) + 3H2(g) ⇌ 2NH3(g) P(NH3) = 3.1 x 10-2 atm, P(N2) = 8.5 x 10-1 atm; P(H2) = 3.1 x 10-3 atm Calculate the value for the equilibrium constant Kp at this temperature. If P(N2) = 0.525 atm, P(NH3) = 0.0167 atm, and P(H2) = 0.00761 atm, does this represent a system at equilibrium? 15 Kp vs. Kc 16 For the general reaction The relationship Δn between K and Kp is Kp = K (RT) Δn = sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. Δn = (l + m) - (j + y) R = 0.08206 L·atm/mol·K T = temperature (in Kelvin)  Example: Calculate the value of K at 25℃ for the reaction: Example At 1100 K, Kp = 0.25 for the reaction: 2SO2 (g) + O2(g) ⇌ 2SO3(g) What is the value of K at this temperature? 19 When is K=Kp 20  For the general reaction The relationship Δn between K and Kp is Kp = K (RT) When Δn = 0, Kp = K (RT)0 so Kp = Kc K is equal to Kp when the sum of the coefficients on either side of the balanced equation is identical, so the terms in RT cancel out. Practice For which reactions is Kp equal to Kc? a. 2Fe(s) + 3/2 O2(g) ⇌ Fe2O3(s) b. CO2(g) + MgO(s) ⇌ MgCO3(s) c. C(s) + H2O(g)⇌ CO(g) + H2(g) d. 4KO2(s) + 2H2O(g) ⇌ 4KOH(s) + 3O2(g) 22 23 24 Thank you 25