1. Chapter 15: Chemical Equilibrium N2O4(g)⇋ 2 NO2(g) A reaction is in equilibrium when the rate of the forward reaction: ratefor = kfor[N2O4] equals the rate of the reverse reaction: raterev = krev[NO2]2 ratefor = raterev kfor[N2O4] = krev[NO2]2 Kc = the equilibrium constant for the reaction when Molar concentrations are used to calculate K.

2. Note that it is only possible to establish an equilibrium for a reaction that proceeds via a single step mechanism. Hence the concentration of each reactant and product is raised to its corresponding coefficient. It is important to remember that even though the [reactants] and [products] are constant at equilibrium the reaction hasn’t stopped Chemical equilbrium is a dynamic process. ⇋ N2O4 2 NO2

3. What is Kc for the following reactions? N2(g) + 3 H2(g)⇋ 2 NH3(g) H2O(l)⇋ H2O(g) CO2(g) + H2O(l)⇋ H+(aq) + HCO3(aq) 2 Ag+(aq) + CO32-(aq) ⇋ Ag2CO3(s) Note that reactants and products in the solid or liquid state do not appear in the equilibrium constant expression because their molarity does not change as a result of the position of the equilibrium.

4. Sometimes, it is more convenient to determine the equilibrium constant for gas phase reactions in terms of the partial pressures of the gasses. N2O4(g)⇋ 2 NO2(g) Kp and Kc are related through the ideal gas law as follows: PNO2V = nNO2RT, so PNO2 = (nNO2/V)RT = [NO2]RT Similarly: PN2O4 = [N2O4]RT In general: Kp = KcRTn

5. 2 NOBr(g)⇋ 2 NO(g) + Br2(g) (A) What is Kc for 2 NO(g) + Br2(g) ⇋ 2 NOBr(g) (B) Which reaction, A or B, favors product formation? Rxn B because Kc > 1 Q: What is the relative value of Kc if reactants are favored? Kc < 1 for reactant favored rxns Q: If neither reactants nor products are favored? Kc = 1

6. Combining equilibria a la Hess’s Law: If you know that: 2 NOBr(g)⇋ 2 NO(g) + Br2(g) And that Br2(g) + Cl2(g)⇋ 2 BrCl(g) What is Kc for the combined reaction? 2 NOBr(g) + Cl2(g) ⇋ 2 NO(g) + 2BrCl(g) Kc = (Kc1)(Kc2) = 0.10

7. Calculations with Equilibrium Constants: ICE Tables 15.24 A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00 L vessel at 700K. These substances react as follows: At equilibrium, the vessel is found to contain 0.566 g of H2. What are the equilibrium partial pressures of each species? H2(g) + Br2(g) ⇋ 2 HBr(g) Initial mol H2 = 1.374 g/(2.01588 g/mol) = 0.68159 mol Initial mol Br2 = 70.31 g/(159.808 g/mol) = 0.43997 mol Eq mol H2 = 0.566 g/(2.01588 g/mol) = 0.2808 mol H2(g) + Br2(g) ⇋ 2 HBr(g) Initial 0.68159 mol 0.43997 mol 0 mol Change -0.4008 mol -0.4008 mol +0.8016 mol Equilibrium 0.2808 mol 0.0392 mol 0.8016 mol

8. Use the number of moles of each gas present at equilibrium to determine their partial pressures. PH2 = nRT/V = (0.2808 mol)(0.0821Latm)(700K)/(2.00 L) PH2 = 8.07 atm PBr2 = nRT/V = (0.0392 mol)(0.0821Latm)(700K)/(2.00 L) PBr2 = 1.1 atm PHBr = nRT/V = (0.8016 mol)(0.0821Latm)(700K)/(2.00 L) PHBr = 23.0 atm d) What is Kp? = 58 e) What is Kc? Kp = KcRTn n = 0, so Kp = Kc

9. Stoichiometry and Equilibrium ICE Tables Ex: A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium constant Kc for the reaction shown below at 448°C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter? H2(g) + I2(g) ⇋ 2 HI(g) I 1.000 M 2.000 M 0 C - x +2x - x E 1.000 - x 2x 2.000 - x E 0.444 M 1.11 M 1.444 M Solve quadratic and get that x = 0.556 M

10. Using the reaction quotient to predict equilibrium shifts H2(g) + I2(g) ⇋ 2 HI(g) Kc = 50.5 Ex: Predict in which direction the reaction will proceed to reach equilibrium if we start with 2.0 × 10−2 mol of HI, 1.0 × 10−2 of H2, and 3.0 × 10−2 of I2 in a 2.00-L container. Initial concentrations: [HI] = (0.020 mol HI)/2.00 L = 0.010 M [H2] = (0.010 mol H2)/2.00 L = 5.0 x 10−3 M [I2] = (0.030 mol H2)/2.00 L = 0.015 M Qc < Kc so not enough products Right shift

11. Le Chatelier’s Principle: When a stress is applied to an equilibrium, the equilibrium will shift to alleviate the stress. Fe+3(aq) + SCN-1(aq) ⇋ FeSCN+2(aq) Colorless ⇋ Dark red Reactants Products Left shift = lighter color Right shift = darker color Initial color

12. Other ways to cause a Le Châtelier Shift: N2(g) + 2 O2(g)⇋ 2 NO2(g) H = 67.7 kJ What kind of shift would you see if: Pressure increased? Right shift  Volume increased?  Left shift Right shift  Heating temperature increased?

13. CH4(g) + 2 Cl2(g)⇋ CCl4(g) + 2 H2(g) H = -32 kJ What kind of shift would you see if: Pressure increased? No Change Heating temperature increased?  Left shift H2 removed? Right shift  CH4 removed?  Left shift Catalyst added? No Change