LG 3.3 Kc & Kp for Gaseous Systems PDF

Chemical Equilibrium LG 3.3 Kc & Kp FOR GASEOUS SYSTEM  DESIRED LEARNING COMPETENCIES 1. By the end of this learning guide, the students should be able to relate Kc and Kp for homogeneous gaseous equilibria.  THE GAS CONCEPT RECALL!! 1. What causes the fizzing sound wh...

Chemical Equilibrium LG 3.3 Kc & Kp FOR GASEOUS SYSTEM  DESIRED LEARNING COMPETENCIES 1. By the end of this learning guide, the students should be able to relate Kc and Kp for homogeneous gaseous equilibria.  THE GAS CONCEPT RECALL!! 1. What causes the fizzing sound when you open a can of soda? 2. Why does squeezing the can cause the pressure of gas in the solution to increase? https://tenor.com/search/coke-exploding-gifs  IDEAL GAS LAW It show the relationship between the pressure, volume, number of moles, and temperature of gas assuming ideal conditions PV = nRT 𝑛 P= RT 𝑉 Where: P = pressure (atm) V = volume (liters) Where: R = gas constant (0.08206 n/V = molar concentration(M) L.atm/K.mol) n = number of moles  EQUILIBRIUM CONSTANT, Kp For the general equation: aA + bB ⇋ cC + dD (𝑃𝐷)𝑑 (𝑃𝐸)𝑒 Kp = (𝑃𝐴)𝑎 (𝑃𝐵)𝑏 Since both the Kc and Kp expressions use the stoichiometric coefficients of the balanced equation, it is also ideal to examine first if the reactions given are balanced.  LEARNING CHECK Write the Kp expression for the following reactions. 1. N2O4(g) ⇌ 2NO2(g) 2. Ni(CO)4(g) ⇌ Ni(s) + 4CO(g) 3. 4HCl(aq) + O2(g) ⇌ 2H2O(l) + 2Cl2(g)  LEARNING CHECK ANSWER!! 1. N2O4(g) ⇌ 2NO2(g) 2. Ni(CO)4(g) ⇌ Ni(s) + 4CO(g) 3. 4HCl(aq) + O2(g) ⇌ 2H2O(l) + 2Cl2(g) 𝑃NO 2 𝑃CO 4 𝑃Cl 2 Kp= 2 Kp= Kp= 2 𝑃N O 𝑃Ni(CO) 𝑃O 2 4 4 2  INTEGRATING Kc in Kp 𝑛 PA = RT 𝑉 PA = [A]RT The partial pressure of reactant A is the product of its concentration temperature and gas constant. We can integrate this equation to obtain: Kp = Kc(RT)∆n Where ∆n is the difference between the moles of the gaseous product minus the moles of gaseous reactant.  WORKED EXAMPLES In the synthesis of ammonia from nitrogen and hydrogen, N2(g) + 3H2(g) ⇌ 2NH3(g) Kc = 9.60 at 300°C. Calculate the Kp for this reaction at this temperature. T = 273.15 + 300oC = 573.15 K R = 0.08206 L.atm/K.mol. Kc = 9.60 Kp = Kc(RT)∆n Kp = 9.60(0.08206 L.atm/K.mol x 573.15 K)2-4 Kp = 4.34 x 10-3  WORKED EXAMPLES The equilibrium constant Kc equals 0.0952 for the following reaction at 227°C CH3OH(g) ⇌ CO(g) + 2H2(g) What is the value of Kp at this temperature? T = 273.15 + 227oC = 500.15 K R = 0.08206 L.atm/K.mol. Kc = 0.0952 Kp = Kc(RT)∆n Kp = 0.0952(0.08206 L.atm/K.mol x 500.15 K)3-1 Kp = 1.603 x 103  REFERENCES Albarico, J.M. (2013). THINK Framework. Based on Ramos, E.G. and N. Apolinario. (n.d.) Science LINKS. Quezon City: Rex Bookstore Inc. 2. Brown, T. L., Bursten, B., Lemay, H. E., Murphy, C. J., & Woodward, P. M. (2012). Chemistry: The Central Science (12th ed.). Glenview, Illinois: Pearson Education, Inc. 3. Burdge, J., & Overby, J. (2012). Chemistry: Atoms First. New York, New York: McGraw-Hill Companies, Inc. 4. Ebbing, D. & Gammon, S. (2009). General Chemistry (9th ed.). Boston, New York: Houghton Mifflin Company. 5. Silberberg, M. S. (2010). Chemistry: The Molecular Nature of Matter and Change (5th ed.). New York, New York: McGraw-Hill Companies, Inc.