Equilibrium

1. Equilibrium

2. Question One • What does it mean if the Kc is greater than one? • The reaction is product favored

3. Question Two • Write the expression for the equilibrium constant for the following reaction: carbon monoxide reacts with oxygen to form carbon dioxide • 2 CO (g) + O2 (g)  2 CO2 (g) Keq = [CO2]2 [CO]2 [O2]

4. Question Three • Write the expression for the equilibrium constant for the production of HI gas from hydrogen gas and iodine gas. H2 (g) + I2 (g)  2 HI (g) K = [HI]2 [H2] [I2]

5. Question Four • When the following reaction is at equilibrium at a certain temperature, sulfur dioxide + oxygen <-> sulfur trioxide, [SO2] = 0.20 M, [O2] = 0.20 M, and [SO3] = 0.40 M. Calculate Keq. • 2 SO2 (g) + O2 (g)  2 SO3 (g) [0.20] [0.20] [0.40] K = [SO3]2 = (0.40)2 = 20. [SO2]2[O2] (0.20)2 (0.20)

6. Question Five • 2 NF2 (g)  N2F4 (g) 0.500M N2F4 is mixed with no NF2 initially present. The equilibrium concentration of N2F4 is 0.200M. Calculate the Keqvalue. • 2 NF2 (g)  N2F4 (g) initial [0] [0.500] change +0.600 -0.300 equilibrium [0.600] [0.200] Keq = [0.200] = 0.556 [0.600]2

7. Question Six • Consider the following reaction: 2 HCl (g)  H2 (g) + Cl2 (g) The initial concentration of HCl is 2.0M and there is no H2 or Cl2present.After equilibrium conditions have been established, the concentration of Cl2 is 0.10M. What are the equilibrium concentrations for HCl and H2? Calculate Kc 2 HCl (g)  H2 (g) Cl2 (g) [2.0] 0 0 -0.20 +0.10 +0.10 [1.80] [0.10] [0.10] Keq = [0.10] [0.10] = 0.0031 [1.80]2

8. Question Seven • Calculate the enthalpy of reaction for the combustion of C3H6. • 2C3H6 + 9O2 6CO2 + 6H2O • -3809 kJ

9. Question Eight • What three things are necessary for a successful reaction to occur? • Reactant molecules must collide • The collision must occur with sufficient energy to get over the activation energy hump • The collision must occur with the proper orientation

10. Question Nine • What are four ways to speed up a reaction? • Increase concentration of reactants • Increase temperature • Increase pressure of a gaseous system • Add a catalyst • Increase Surface Area

11. Question Ten • The reaction, A + B  C + D has a Keq of 6.57 x 10-3. Determine the final equilibrium concentrations of all substances if 0.200M C and 0.200M D are mixed. A + B  C + D Initial 0 0 [0.200] [0.200] Change +x +x -x -x Eq x x 0.2 - x 0.2 - x [0.185M] [0.185M] [0.015M] [0.015M] 6.57 x 10-3 = [C] [D] = (0.200 -x)2 x = 0.185 [A] [B] x2 *simplify by taking the square root immediately. So: 0.0811 = 0.2 – x or 0.0811x = 0.20 – x x 1.0811x = 0.20 x = .185

12. Question Eleven • How does increasing the temperature increase reaction rate? • Molecules have more energy and therefore there will be more collisions and the increased energy means a greater fraction of the molecules will have enough energy to get over the activation energy hump

13. Question Twelve • What is entropy? • A measure of the disorder in a system

14. Question Twelve.Five • How many grams is required to make a 0.250 M solution of plumbic dichromate if I want to make a 450. mL solution? • Pb(Cr2O7)2 • 0.25 x 0.45 = 0.1125 moles x 639 = 71.9 g

15. Question Thirteen • For the reaction, N2 (g) + O2 (g)  2 NO (g), the equilibrium constant is found to be 1.0 at room temperature. If 1.00M N2 is mixed with 1.00M O2, find the eventual equilibrium concentration of all substances. N2+ O2 2 NO • initial 1.0 1.0 0 change -x -x +2x eq. 1 – x 1 – x 2x 1 – 0.33 1 – 0.33 2(0.33) [0.67] [0.67] [0.67] • 1.0 = (2x)2= x = 0.33 (1-x)(1-x)

16. Question Fourteen • How does adding a catalyst affect the reaction rate? • It speeds up a reaction by lowering the activation energy so that more molecules have sufficient energy to get over the activation energy.

17. Question Fifteen • Kc is 0.50 for synthesis of nitrogen monoxide gas from its elements. Find the eventual equilibrium concentration of all substances if 0.25 moles of NO are placed in a 4.0L container. N2 + O2 2 NO initial 0 0 [0.0625] change +x +x -2x equilibrium x x 0.0625 – 2x [0.023] [0.023] [0.017] • 0.50 = (0.0625-2x)2 x2

18. Question Sixteen • Which has more entropy, solid water or water vapor? Why? • Water vapor has more entropy because there is more disorder in the gaseous state than in the solid state due to the increased molecular motion of gases

19. Question Seventeen • What is the definition of rate of reaction? • Change in concentration of reactant or product per unit time

20. Question Eighteen • In the reaction of hydrogen gas and chlorine gas to form hydrogen monochloride gas, the Keq = 1.00 x 102. The initial concentrations of both the hydrogen gas and chlorine gas are 1.0M. What are the equilibrium concentrations of all three species? H2+ Cl2 2 HCl Initial [1.0] [1.0]0 Change - x - x + 2x Equilibrium 1 – x 1 – x 2x [0.17] [0.17] [1.7] 1.00 x 102 = (2x)2 take the square root so: ( 1 – x)(1 – x) 10 = 2x x = 0.83 (1 – x)

21. Question Nineteen • In the reaction CO (g) + H2O (g)  CO2 (g) + H2 (g) (the Keq is 5.0), if the initial concentrations for all species is [1.00] find the equilibrium concentrations of each substance. CO + H2O  CO2 + H2 Initial [1.0] [1.0] [1.0] [1.0] Change - x - x + x + x Equilibrium 1 – x 1 – x 1 + x 1 + x [0.62] [0.62] [1.4] [1.4] *because Keq is greater than 1, the reaction proceeds toward products. 5.0 = (1 + x)(1 + x) *take the square root so: (1 – x)(1 – x) 2.24 = 1 + x 1 – x x = 0.38

22. Question Twenty • For an exothermic reaction, which is higher – the enthalpy of the bonds in the reactants or the enthalpy of the bonds in the products? • The enthalpy of the bonds in the products

23. Question Twenty One N2 (g) + 3H2 (g)  2NH3 (g) DH = -230 kJ 1. Write the equilibrium constant expression Keq= [NH3]2 [N2][H2]3 2. Calculate the value of Keq if at equilibrium nitrogen is 2.0M, hydrogen is 1.0M, and ammonia is 0.50M. Keq = [0.50]2 = 0.13 [2.0][1.0]3 3. Will there be more reactant or product? Favors the reactant

24. Question Twenty Two N2 (g) + 3H2 (g)  2NH3 (g) DH = -230 kJ Which direction will the reaction proceed if: a. the pressure is decreased. reactant b. the reaction is cooled. product c. a catalyst is added. neither d. NH3 is added. reactant e. H2 is added. product f. some nitrogen is removed. reactant

25. Question Twenty Three • Initially 4.0M H2 and 2.0M N2 are placed in a container at a certain temperature. After the equilibrium has been established 1.0M NH3 has been produced. What are the equilibrium concentrations of H2 and N2? What is Kc? 3 H2 + N2 2 NH3 initial [4.0] [2.0] 0 change - 1.5 - 0.50 +1.0 equilibrium 2.5 M 1.5 M 1.0M Keq = [1.0]2 = 0.043 [2.5]3[1.5]

26. Question Twenty Four For the following reaction: N2O4 (g) + 58.9 kJ  2 NO2 (g) A 1.0L flask at 55oC is found to contain 3.6 moles of N2O4 and 1.75 moles of NO2 at equilibrium. What is the value of Keq? Keq = [1.75]2 = 0.85 [3.6] • If the reaction is heated, in what direction would the equilibrium shift? • The reaction would shift toward the products. • Would the Keq change? If so, would it increase or decrease? • Yes, the Keq would increase (products are in the numerator). • If NO2 is removed from the flask, in what direction would the equilibrium shift? • The reaction would shift toward the products. • Would the Keq change? If so, would it increase or decrease? • No, the Keq would not change. • What effect would increasing the pressure have on this equilibrium? • The reaction would shift toward the reactants.

27. Question Twenty Five • Write the equilibrium expression for the following reactions: • CaCl2 (s)D Ca+2(aq) + 2Cl-(aq) • Kc = [Ca+2] [Cl-1] 2 • ZnO(s) + CO(g))D Zn(s) + CO2(g) • Kc = [CO2] / [CO]

28. Question Twenty Seven • The colourless gas dinitrogen tetroxide decomposes to the brown coloured air pollutant nitrogen dioxide and exists in equilibrium. A 0.125 mol sample of dinitrogen tetroxide is introduced into a 1.00 L container and allowed to decompose at a given temperature. When equilibrium is reached, the concentration of the dinitrogen tetroxide is 0.0750 mol/L. What is the value of Keq for this reaction? • N2O4D 2NO2 Keq = .12/0.075 = 0.133 0.125 0 - 0.05 +2(0.05) 0.075 0.1

29. Question Twenty Seven • A 0.921 mol sample of dinitrogen tetroxide is placed in a 1.00 L vessel and heated to 100°C. At equilibrium it is found that 20.7 % of the dinitrogen tetroxide has decomposed to nitrogen dioxide. Calculate the Keq for this reaction. • N2O4D 2NO2 .207(.921) = .191 0.921 0 - 0.191 +2(.191) Kc = .3822/.73 = 0.200 0.73 0.382

30. Question Twenty Nine • The following reaction has Keq value of 85.0 at 460°C: SO2(g) + NO2(g) ⇌ NO(g) + SO3(g) If a mixture of sulfur dioxide and nitrogen dioxide is prepared, each with an initial concentration of 0.100 mol/L, calculate the equilibrium concentrations of nitrogen dioxide and nitrogen monoxide at this temperature. • SO2(g) + NO2(g) ⇌ NO(g) + SO3(g) 85 = x2/(.1-x)2 0.10 0.10 0 0 9.22 = x/(.1-x) -x -x +x +x .922 – 9.22x = x 0.10 –x 0.10 –x xx .922 = 10.22x x = 0.0902 M [SO2] = [NO2] = 0.0098 M [NO] = [SO3] = 0.0902M

31. Question Thirty • Hydrogen and iodine gases react to form hydrogen iodide gas. If 6.00 mol of H2 and 3.00 mol of I2 are placed in a 3.00 L vessel and allowed to come to equilibrium at 250°C calculate the equilibrium concentrations of all species. The Keq for the reaction is 4.00 at 250 °C. • H2 + I2D 2HI Keq = (2x)2/(2-x)(1-x) 2 1 0 4.00 = 4x2 / (2 -3x + x2) -x -x +2x 8 – 12 x + 4x2 = 4x2 2-x 1-x 2x 8 – 12x = 0 8 = 12x x = 0.67 M [H2] = 1.33 M [I2] = 0.33M [HI] = 1.34M

32. Question Thirty One • The equilibrium constant for the reaction below is 0.11. Calculate all equilibrium concentrations if 0.33 mol of iodine chloride gas is placed in a 1.00 L vessel and allowed to come to equilibrium. 2 ICl(g) ⇌ I2(g) + Cl2(g) 0.11 = x2 / (0.33 – 2x)2 0.33 0 0 0.33 = x / (.33 – 2x) -2x +x +x 0.1089 – 0.66x = x 0.33 -2x x x 0.1089 = 1.66x x = 0.0656 [ICl] = 0.20 M [I2] = [Cl2] = 0.066M

33. Question Thirty Two • The dissociation of nitrogen monoxide gas to nitrogen and oxygen gases has a Kc of 2.63 x 10-2 at 27°C. If 2.00 mol of nitrogen monoxide is placed in a 2.00 L vessel and allowed to reach equilibrium, what is the concentration of oxygen and nitrogen? • 2NO D N2 + O2 2.63 x 10-2 = x2 / (1-2x)2 1.00 0 0 0.162 = x / (1-2x) -2x +x +x 0.162 – 0.324 x = x 1-2x x x 0.162 = 1.324x x = 0.122 M [N2] = [O2] = 0.122M

34. Question Thirty Three For the reaction below, which change would cause the equilibrium to shift to the right? • CH4(g) + 2H2S(g) ↔ CS2(g) + 4H2(g) DH = -43.0 kJ (a) Decrease the concentration of dihydrogen sulfide. Shift to the Left (b) Increase the pressure on the system. Shift to the Left (c) Decrease the temperature of the system. Shift to the Right (d) Increase the concentration of carbon disulfide. Shift to the Left (e) Decrease the concentration of hydrogen. Shift to the Right

35. Question Thirty Four • Explain WHY a reaction will shift towards products if a reactant is added • More reactant molecules present will result in more effective collisions causing the forward reaction to occur where reactants will collide and form products

36. Question Thirty Five • How many grams of calcium phosphate are needed to make 250. mL of a 3.5 M solution? • 0.35 x 0.25 = 0.0875 mol Ca3(PO4)2 x 310 = 27g

37. Question Thirty Six • What is the volume of an ammonium acetate solution that has a mass of ammonium acetate of 45.0 grams and a concentration of 1.2 M? • 45 g NH4C2H3O2 / 77 = 0.584 mol • 1.2 = 0.584 / v • V = 0.49 L