1. Entry Task: Dec 16 th Monday Self-Check Ch. 15 sec. 1-4 Pre-lab Determine Keq Lab MAYHAN

2. I can… Write equilibrium expression for a reaction and explain calculate how concentrations of reactant and product affect the reactions equilibrium. Write the equilibrium expression for gases and calculate the how the changes in gas pressure/concentration can affect the equilibrium. Determine the direction of the reaction by the magnitude of the Kc. Use the ICE tables to predict changes concentration and its effect on the Kc value. MAYHAN

4. 1)Write the equilibrium expression (K eq ) for the following reactions. BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq) 2 Hg(s) + O 2 (g)  2 HgO(s) MAYHAN

5. 1) Write the equilibrium expression (K eq ) for the following reactions. 2 NaHCO 3 (s)  Na 2 CO 3 (s) + CO 2 (g) + H 2 O(g) CaCO 3 (s) + 2 HCl(aq)  CaCl 2 (aq) + CO 2 (g) + H 2 O(l) MAYHAN

6. For each of the following, you MUST write the K eq expression, substitute in the values, then solve. 2) SO 3 (g) + H 2 O(g)  H 2 SO 4 (l) At equilibrium, the [SO 3 ] = 0.400 M and the [H 2 O] = 0.480 M. Calculate the value of the equilibrium constant. Kc = 5.21 MAYHAN

7. For each of the following, you MUST write the K eq expression, substitute in the values, then solve. 3) PCl 5 (s) + H 2 O(g)  2 HCl(g) + POCl 3 (g) A 2.0 L flask at equilibrium at 100 o C contains 0.075 mol of PCl 5, 0.050 mol of H 2 O, 0.750 mol of HCl, and 0.500 mol of POCl 3. Calculate the K eq for the reaction. Change into molarity 0.050mol/2.0L = 0.025 M H2O 0.75mol/2.0L = 0.375 M HCl 0.500mol/2.0L = 0.25 M POCl 3 Kc= 1.41 MAYHAN

8. For each of the following, you MUST write the K eq expression, substitute in the values, then solve. 4) 2 NO 2 (g)  N 2 O 4 (g). If 2.00 moles of NO 2 and 1.60 moles of N 2 O 4 are present in a 4.00 L flask at equilibrium, what is the equilibrium constant? Change into molarity 2.00mol/4.0L = 0.5 M Kc= 1.6 1.60mol/4.0L = 0.4 M MAYHAN

9. For each of the following, you MUST write the K eq expression, substitute in the values, then solve. 5) H 2 (g) + I 2 (g)  2 HI(g). If, at equilibrium, [H 2 ] = 0.200 M and [I 2 ] = 0.200 M, what is the concentration of HI in the 5.0 L flask? K eq = 55.6 at 250 o C. Kc= 1.49 MAYHAN

10. For each of the following, you MUST include an ICE table and the K eq expression. A reaction vessel had 1.95 M CO and 1.25 M H 2 O introduced into it. After an hour, equilibrium was reached according to the equation: CO 2 (g) + H 2 (g)  CO(g) + H 2 O(g) Analysis showed that 0.85 M of CO 2 was present at equilibrium. What is the equilibrium constant for this reaction? INITIAL CHANGE EQUILIBRIUM CO 2  H2H2 CO 1.95M 0 M 1.25M H2OH2O 0.85M 0.85 0.85M -0.85 1.1M0.4M MAYHAN

11. Use the equilibrium concentrations to find Kc- plug n chug Kc= 1.64 if you divide 1/1.64 you’ll get the forward reaction kc= 0.61 INITIAL CHANGE EQUILIBRIUM CO 2  H2H2 CO 1.95M 0 M 1.25M H2OH2O 0.85M 0.85 0.85M -0.85 1.1M0.4M MAYHAN

12. For each of the following, you MUST include an ICE table and the K eq expression. 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) Into a 2.00 L container is placed 1.00 mol of SO 2 (g) and 1.00 mol of O 2 (g). At equilibrium, [SO 3 ] = 0.150 M. Calculate the equilibrium constant for this reaction. Change into molarity 1.00mol/2.0L = 0.5 M SO 2 1.00mol/2.0L = 0.5 M O 2 MAYHAN

13. For each of the following, you MUST include an ICE table and the K eq expression. 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) Into a 2.00 L container is placed 1.00 mol of SO 2 (g) and 1.00 mol of O 2 (g). At equilibrium, [SO 3 ] = 0.150 M. Calculate the equilibrium constant for this reaction. INITIAL CHANGE EQUILIBRIUM 2SO 2  O2O2 2SO 3 0.150M 0.5 M 0.0M 0.150 0.150 mol SO 3 L 1 mol O 2 2 mol of SO 3 = 0.075M Same goes for SO 2 -0.075-0.150 0.350M0.425M MAYHAN

14. Use the equilibrium concentrations to find Kc- plug n chug Kc= 0.432 INITIAL CHANGE EQUILIBRIUM 2SO 2  O2O2 2SO 3 0.150M 0.5 M 0.0M 0.150-0.075-0.150 0.350M0.425M MAYHAN

15. For each of the following, you MUST include an ICE table and the K eq expression. 2 NOCl(g)  2 NO(g) + Cl 2 (g) When 0.50 mol of NOCl was put into a 1.0 L flask and allowed to reach equilibrium, 0.10 mol of Cl 2 was found. What is K eq for this reaction? Change into molarity 0.50mol/1.0L = 0.5 M NOCl 0.10mol/1.0L = 0.1 M Cl 2 MAYHAN

16. For each of the following, you MUST include an ICE table and the K eq expression. 2 NOCl(g)  2 NO(g) + Cl 2 (g) When 0.50 mol of NOCl was put into a 1.0 L flask and allowed to reach equilibrium, 0.10 mol of Cl 2 was found. What is K eq for this reaction? INITIAL CHANGE EQUILIBRIUM 2NOCl  2NO Cl 2 0.1M 0.5 M 0.0 M 0.1 0.1 mol NO L 2 mol NOCl 1 mol of NO = 0.2 0.2 -0.2 0.30M0.2M MAYHAN

17. Use the equilibrium concentrations to find Kc- plug n chug Kc= 0.044 INITIAL CHANGE EQUILIBRIUM 2NOCl  2NO Cl 2 0.1M 0.5 M 0.0 M 0.10.2 -0.2 0.30M0.2M MAYHAN

18. HW: Determine Keq Lab READ IT!!!! MAYHAN