Unit 4: Equilibrium

1. Unit 4: Equilibrium 7.5 Quantitative Changes in Equilibrium Systems

2. kf kr A + 2B AB2 [AB2] = Kc = kf [A][B]2 kr Chemical Kinetics and Chemical Equilibrium ratef = kf[A][B]2 rater = kr[AB2] Equilibrium ratef = rater kf[A][B]2 = kr[AB2]

3. aA + bB cC + dD [C]c[D]d K = [A]a[B]b Law of Mass Action Equilibrium Will K >> 1 Lie to the right Favor products K << 1 Lie to the left Favor reactants K = 1 [reactants] = [ products]

4. the [ ] of the substances at any closed chemical system: • system is at equilibrium ? • and if not, in which direction the system will shift to reach equilibrium…

5. The reaction quotient (Qc)is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. NOTE: Q is similar to Kc - Kc is calculated using concentrations at equilibrium - Q is calculated using [ ] that may NOT be at equilibrium.

6. IF • Qc > Kcsystem proceeds from right to left to reach equilibrium (REACTANTS) • Qc = Kcthe system is at equilibrium • Qc < Kcsystem proceeds from left to right to reach equilibrium (PRODUCTS)

7. Example • Calculate the value of the equilibrium constant, Kc , for the system shown, if 0.1908 moles of CO2, 0.0908 moles of H2, 0.0092 moles of CO, and 0.0092 moles of H2O vapor were present in a 2.00 L reaction vessel were present at equilibrium.

8. ICE • "I" stands for the initial concentrations (or pressures) for each species in the reaction mixture. • "C" represents the change in the concentrations (or pressures) for each species as the system moves towards equilibrium. • "E" represents the equilibrium concentrations (or pressures) of each species when the system is in a state of equilibrium.

9. How to Make an ICE Chart • Write Balanced Chemical EQ • Set up ICE table. • Write an Equilibrium Expression (remember solids and liq are const) • Express all quantities in terms of MOLARITY (moles per liter) • Use initial quantities (mol/L) when calculating the reaction quotient, Q, to determine the direction the reaction shifts to establish equilibrium. • The change in each quantity must be in agreement with the reaction stoichiometry. • Clearly define the change you choose to be represented by "x." Define all other unknown changes in terms of this change. • Use equilibrium quantities in calculations involving the equilibrium constant, K.

10. A mixture consisting initially of 3.00 moles NH3, 2.00 moles of N2, and 5.00 moles of H2, in a 5.00 L container was heated to 900 K, and allowed to reach equilibrium. Determine the equilibrium concentration for each species present in the equilibrium mixture if Kc=0.0076 @ 900 K 2 NH3(g)  N2(g) + 3 H2(g) REACTANTS

11. Solve.... 

12. At 12800C the equilibrium constant (Kc) for the reaction Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. [Br]2 [Br2] Kc = Br2 (g) 2Br (g) Br2 (g) 2Br (g) = 1.1 x 10-3 Kc = (0.012 + 2x)2 0.063 - x Let x be the change in concentration of Br2 Initial (M) 0.063 0.012 ICE Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x Solve for x

13.  -b ± b2 – 4ac x = 2a Br2 (g) 2Br (g) Initial (M) 0.063 0.012 Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x = 1.1 x 10-3 Kc = (0.012 + 2x)2 0.063 - x 4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x2 + 0.0491x + 0.0000747 = 0 ax2 + bx + c =0 x = -0.0105 x = -0.00178 At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M At equilibrium, [Br2] = 0.062 – x = 0.0648 M

14. PROBLEM: Fuel engineers use the extent of the change from CO and H2O to CO2 and H2 to regulate the proportions of synthetic fuel mixtures. If 0.250 mol of CO and 0.250 mol of H2O are placed in a 125-mL flask at 900K, what is the composition of the equilibrium mixture? At this temperature, Kc is 1.56 for the equation CO(g) + H2O(g) CO2(g) + H2(g) PLAN: We have to find the concentrations of all species at equilibrium and then substitute into a Kc expression. SOLUTION: All concentrations must be recalculated as M, so [CO] = 0.250/0.125L concentration CO(g) + H2O(g) CO2(g) + H2(g) Determining Equilibrium Concentrations from Initial Concentrations and Kc ICE initial 2.00 2.00 0 0 change -x +x +x -x equilibrium 2.00-x 2.00-x x x

15. [CO2][H2] (x) (x) (x)2 Kc = = = [CO][H2O] (2.00-x) (2.00-x) (2.00-x)2 x = 2.00-x Determining Equilibrium Concentrations from Initial Concentrations and Kc continued = +/- 1.25 x = 1.11M [CO2] = [H2] = 1.11M [CO] = [H2O] = 0.89M 2.00 - x = 0.89M

16. 4.00 moles of HI are placed in an evacuated 5.00 L flask and then heated to 800 K. The system is allowed to reach equilibrium. What will be the equilibrium concentration of each species? 2 HI(g) H2(g) + I2(g) Kc = 0.016 @ 800 K 2 HI(g)  H2(g) + I2(g)

17. Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate Keq. Solution Set of a table of concentrations [NOCl] [NO] [Cl2] Before 2.00M 0 0 Change 2x 2x x Equilibrium 0.66M What is X?

18. Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) Solution Set of a table of concentrations [NOCl] [NO] [Cl2] Before 2.00 0 0 2x 2x x Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33

19. Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2] Before 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33

20. The concentrations of an equilibrium mixture of O2 , CO, and CO2 were 0.18 M, 0.35 M, and 0.029 M respectively. Enough CO was added to the flask containing the equilibrium mixture to momentarily raise its concentration to 0.60 M. What will be the concentration of each species in the flask once equilibrium has been re-established after the additional carbon monoxide was added? 2 CO2(g)  2 CO(g) + O2(g)

21. Initially, a mixture of 0.100 M NO, 0.050 M H2, 0.100 M H2O was allowed to reach equilibrium (initially there was no N2 ).  At equilibrium the concentration of NO was found to be 0.062 M.  Determine the value of the equilibrium constant, Kc , for the reaction:

22. A flask is charged with 3.00 moles/L of dinitrogen tetroxide gas and 2.00 moles/L of nitrogen dioxide gas at 25oC and allowed to reach equilibrium.  It was found that the concentration of the nitrogen dioxide decreased by 0.952 moles/L.  Estimate the value of Kc for this system: Kc = (1.048)2/(3.376) = 0.3160

23. PROBLEM: Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It decomposes by the reaction COCl2(g) CO(g) + Cl2(g) Kc = 8.3x10-4 (at 3600C) PLAN: After finding the concentration of starting material, write the expressions for the equilibrium concentrations. When solving for the remaining amount of reactant, see if you can make an assumption about the initial and final concentrations which could simplifying the calculating by ignoring the solution to a quadratic equation. SOLUTION: (a) 5.00 mol/10.0 L = 0.500M (b) 0.100 mol/10.0 L = 0.0100M Calculating Equilibrium Concentration with Simplifying Assumptions Calculate [CO], [Cl2], and [COCl2] when the following amounts of phosgene decompose and reach equilibrium in a 10.0-L flask. (a) 5.00 mol COCl2(b) 0.100 mol COCl2 Let x = [CO]eq = [Cl2]eq and 0.500-x and 0.0100-x = [COCl2]eq, respectively, for (a) and (b).

24. (x) (x) [CO][Cl2] (a) Kc = 8.3x10-4 = Kc = (0.500-x) [COCl2] (x) (x) 8.3x10-4 = (0.500) (x) (x) (b) Kc = 8.3x10-4 = (0.010 - x) continued Calculating Equilibrium Concentration with Simplifying Assumptions assume x is << 0.500 so that we can drop x in the denominator 4.15x10-4 = x2 x ≈ 2 x 10-2 (0.500 - x) = 4.8x10-2 CHECK: 0.020/0.500 = 0.04 or 4% percent error Dropping the -x will give a value for x = 2.9x10-3M. (0.010 - x) ≈ 0.0071M CHECK: 0.0029/0.010 = 0.29 or 29% percent error Using the quadratic formula produces x = 2.5x10-3 and 0.0100-x = 7.5x10-3M