Review Unit #11 1. H 2 O (g) + CO (g) H 2(g) + CO 2(g) H>1 K c = 1.6 What effect would these changes have on the equilibrium position? a. Cool the mixture b. Increase the volume of the flask c. Add H 2(g) Review Unit #11 1. H 2 O (g) + CO (g) H 2(g) + CO 2(g) H>1 K c = 1.6 What effect would these changes have on the equilibrium position? a. Cool the mixture b. Increase the volume of the flask c. Add H 2(g) NO effect = equal moles of gases Review #11, continued 2. 2 NOCl (g) 2 NO (g) + Cl 2(g) What effect would these have on the equilibrium position? a) Add a catalyst No effect b) Decrease volume of flask c) Remove NO (g) 1
3) 2 NOCl (g) 2 NO (g) + Cl 2(g) If the initial concentration of each species was 0.10 M, what is the correct representation of K c? K c = (0.10 + 2x) 2 (0.10 +x) (0.10 2x) 2 4.) If Q = 1.9 x 10-5 and K c = 4.2 x 10-5, which direction does the reaction proceed to reach equilibrium? 5. PCl 5(g) PCl 3(g) + Cl 2(g) at 600K the equilibrium pressure constant is 11.5. Suppose that 2.45 g of PCl 5 is placed in an evacuated 500. ml bulb, which is then heated to 600 K. a. What would the pressure of PCl 5 be if it did not dissociate? b. What is the partial pressure of PCl 5 at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the degree of dissociation of PCl 5 at equilibrium? 5.Suppose that 2.45 g of PCl 5 is placed in an evacuated 500. ml bulb, which is then heated to 600 K. a. What would the pressure of PCl 5 be if it did not dissociate? 2.45 g PCl 5 1 mol = 0.0118 mol P = (n/v)rt 208.5 g P = (0.0118 mol/.500l)(.0821) (600 K) P = 1.16 atm 2
b. What is the partial pressure of PCl 5 at equilibrium? [PCl 5 ] [PCl 3 ] [Cl 2 ] I 1.16 0 0 C -x +x +x E 1.16 x x x K = 11.5 = (x) 2 (1.16 x) use the quadratic formula to find x X= 1.06 atm [PCl 5 ] = 1.16 1.06 = 0.10 atm c. What is the total pressure in the bulb at equilibrium? P t = (1.06 atm) + (1.06 atm) + (0.10 atm) P t = 2.22 atm d. What is the degree of dissociation of PCl 5 at equilibrium? (1.16 0.10) x 100 = 91.4% (1.16) Second Review for Unit #11 AP Chemistry 6. MgCl 2(s) Mg 2+ (aq) + 2 Cl - (aq) H < 0 When the substances in the above reaction are at equilibrium, which of the following can shift the equilibrium to favor the products Choose all answers that apply. a. Heat the reaction b. Cool the reaction c. Adding a catalyst d. Adding more chloride ion e. Adding more solid 3
7. 2 F 2(g) + 2 H 2 O (g) 4 HF (g) + O 2(g) Temperature = 25 o C When 0.200 mol of F 2 and.200 mol H 2 O are put into a 1.00 L flask, what must be true at equilibrium? Choose all that are true. a. [F 2 ] must be greater than [H 2 O] b. [F 2 ] must be less than [HF] c. [O 2 ] must be equal to [H 2 O] d. [HF] must be greater than [O 2 ] e. [F 2 ] must be equal to [H 2 O] 8. Calculate the K c and K p for the above reaction, if at equilibrium 0.056 M of oxygen are present. K c = [HF] 4 [O 2 ] [F 2 ] 2 [H 2 O] 2 [F 2 ] [H 2 O] [HF] [O 2 ] I.200.200 0 0 C - 2x -2x +4x +x C -.112 -.112 +.224 +.056 E.088.088.224.056 K c = [.224] 4 [.056] = 2.35 [.088] 2 [.088] 2 K p = K c (RT) n = K p =2.35 [(.0821)(298)] 1 K p = 57.5 9. Complete the chart below for this reaction 2 C 2 H 2(g) + 5 O 2(g) 4CO 2(g) + 2 H 2 O (g) H = -258 kj Stress Shift [C 2 H 2 ] [O 2 ] [CO 2 ] [H 2 O] Heat the reaction Decrease the volume of the container Add a catalyst NO SHIFT Remove water vapor 4
10. At elevated temperatures, SbCl 5 gas decomposes into SbCl 3 gas and Cl 2 gas. a. Write the balanced equation. SbCl 5(g) SbCl 3(g) + Cl 2(g) b. An 89.7 g sample of SbCl 5 is placed into an evacuated 15.0 L container at 182 o C. i. What is the concentration of SbCl 5 in the container before any decomposition occurs? 89.7 g SbCl 5 1 mol =.300 mol/15.0l =.0200 M SbCl 5 299.05 g ii. What is the pressure of SbCl 5 in the container before any decomposition occurs? P= nrt/ V = (.300 mol) (.0821)(455 K) =.747 atm SbCl 5 (15.0 L) 10 C. If the SbCl 5 is 29.2 % decomposed when equilibrium is established at 182 o C, calculate the value for K c and K p..292 x.0200 M =.00584 M [SbCl 5 ] [SbCl 3 ] [Cl 2 ] K c = [.00584][.00584] [.0142] 2 K c =.00240 I.0200 0 0 C -x +x +x C -.00584 +.00584 +.00584 E.0142.00584.00584 K p =K c (RT) n K p =.00240 (.0821 455) 1 K p = 0.0897 11. 6. At 200. o C, the equilibrium constant for the reaction below is 1.21 x 10-2 2 HI (g) H 2(g) + I 2(g) a. Is this reaction product or reactant favored? Reactant favored b. At this temperature what are the equilibrium concentrations of each species if 122 g of HI are put into a 1.50 L container? 122 g HI 1 mol =.954 mol / 1.50L = 0.636 M HI [HI] [H 127.9 g 2 ] [I 2 ] I.636 0 0 C -2x +x +x E.636-2x x x 5
11b continued K c = 1.21 x 10-2 = [x 2 ] [.636 2x] 2 Take the square root of both sides 0.110 = x (.636 x) X = 0.573 [HI] =.521M [H 2 ] =.0573 M [I 2 ] =.0573 M 11C. At this temperature what are the equilibrium concentrations of each species if 1.27 M HI is put into a flask containing 1.21 M hydrogen gas? [HI] [H 2 ] [I 2 ] I 1.27 1.21 0 C - 2x + x + x E 1.27 2x 1.21 + x x K c = 1.21 x 10-2 = [H 2 ][I 2 ] = (1.21 +x)(x) [HI] 2 (1.27 2x) 2 11 c continued If you neglect + x and -2x x= 0.0161 [HI] = 1.27 2x = 1.24 M [H 2 ] = 1.21 + x = 1.23 M [I 2 ] = x = 0.0161 M If you use the quadratic x = 0.0157 [HI] = 1.27 2x = 1.24 M [H 2 ] = 1.21 + x = 1.23 M [I 2 ] = x = 0.0157 M 6
11 d. At this temperature what is the concentration of HI and H 2 if 0.200 M of all species are in the flask to begin with and there is 0.125 M I 2 at equilibrium? [HI] [H 2 ] [I 2 ] I 0.200 0.200 0.200 C + 2x -x -x 0.200-x = 0.125 x = 0.075 E 0.200+ 2x 0.200 - x 0.125 Since [I 2 ] has decreased, the reverse reaction must have taken place. Complete the chart. [ HI] = 0.350 M [ H 2 ] =.125 M 7