Chm 152 Final Exam Review Key

Final Exam Review 1 (Ch. 14, 15, 16) KEY 1. a. Rate =  1 [HBr] 1 [H2 ] 1 [Br2 ]   2 t 1 t 1 t b. Rate of HBr = 0.455M - 0.500M  3.0x10-3 M/s 15.0s  0.0s Rate of Br 2 =  c. 1 [HBr] 1 [Br2 ]  2 t 1 t 1 [Br2 ] 1 [HBr]  1 t 2 t = Since the word “disappear” is already in the question, we sometimes omit the “-“ sign since it’s understood that the amount is decreasing, 1  0.455 M - 0.500M  1 [Br2 ]     2  15.0s - 0.0s  1 t 1 1 Rate of HBr =  (1.2 x10 3 M s 1.5x10-3 M/s 1  [HBr]  -3 )   =-2.4x10 M/s 2  t  2. a. s-1 or 1/s b. M-1s-1 or 1/M s c. M·s-1 or M/s Rate = k[NO2]m [F2]n 3. m [NO2](M) 0.100 0.200 0.200 0.400 [F2](M) 0.100 0.100 0.200 0.400  0.051M/s   0.200M   0.100M        0.026M/s   0.100M  0.100M Initial Rate (M/s) 0.026 0.051 0.103 0.411 0.026 M/s = k(0.100M)1 (0.100M)1 2  2m m1 m k = 2.6 M-1s-1 n  0.103M/s   0.200M   0.100M        0.051M/s   0.200M  0.200M 2  2n n  1 n Rate = 2.6 M-1s-1[NO2][F2] 4. a. t 1/2  d. 4.88x103 s .  25%  - 4 -1 ln    (1.42x10 s ) t  100%  b. c. ln2  1.42x10-4 t 1/2  t = 9.8x103 s t = 1.8x103 s  0.78M  - 4 -1 ln    (1.42x10 s ) t  1.00M  .  At  - 4 -1 2 ln    (1.42x10 s )(2.00x10 s)  0.150M  . e   At    ln      0.150M   e .    (1.42x10 - 4 s -1 )(2.00x10 2 s)      ln2 k [SO2Cl2] 200s = 0.146M [SO2Cl2] 500s = 0.140M 1.8x103 s A ln  t  A0    kt  k ln  2  k1 5. a. 121000 J/mol  Ea  1 1      R  T1 T2  0.0117 s -1  Ea 1   1  ln      122000 J/mol -1   0.689 s  8.314 J/K mol  450K 400K  b. ~0.100 1/s  Depends on which k/temp you use. ln  k2 1   121000J/mol  1     -1  0.0117 s  8.314 J/K mol  400K 425K  6. a. Cl2(g) ⇌ 2Cl(g) Cl(g) + CHCl3(g) → HCl(g) + CCl3(g) Cl(g) + CCl3(g) → CCl4(g) Cl2(g) + CHCl3(g) → HCl(g) + CCl4(g) b. Cl(g), CCl3(g) c. There isn’t one, DOH! Remember, a catalyst would be used in an earlier step and reappear in a later step. d. Rate = k[Cl2] 7. The reaction below has an equilibrium constant of Kp=2.26x104 at 298K. CO(g) + 2H2(g) ⇌ CH3OH(g)  Kp   Kp   Kp ' (PCH 3OH ) (PCO )(PH 2 ) Kp = 2.26x104 = product favored 2 (PCH 3OH ) (0.10atm)(0.50atm) (PCO )(PH 2 ) 2 (PCH 3OH ) 2  2.26x10 4 Pressure of CH2OH = 570atm So Kp’ = 1/Kp = 2.26x104 = 4.42x10-5, reactants favored 8. Calculate Kc for each of the following reactions (assume 298K): a. I2(g) ⇌ 2I(g) Kp = 6.26x10-22 Kp = Kc(RT)Δn Kc  6.26x10-22 = 2.56x10-23 Kc  1 L atm [0.08206 molK (298.15K)] Kp RT n    b. CH4(g) + H2O(g) ⇌ CO + 3H2(g) Kp = 7.7x10 24 7.7x1024 = 1.3x1022 Kc  2 L atm [0.08206 molK (298.15K)] 9. For the following reaction, Kc = 1.52 2A(g) ⇌B(g) + C(g) If a reaction vessel initially contains [A] = 0.500M at 500K, what are the equilibrium concentrations of A, B, & C at equilibrium? 2A(g) ⇌ B(g) + C(g) I 0.500M 0M 0M C -2x +x +x E 0.500-2x x x [A] = 0.500M-(2)(0.178) = 0.144M [B][C] K  1.52 [A]2 K (x)(x)  1.52 (0.500 - 2x) 2 (x)(x)  1.52 (0.500 - 2x) 2 x  1.23 0.500 - 2x x  0.615  2.46x 3.46x  0.615 x  0.178 [B] = [C] = 0.178M 10. Consider the following reaction: NiO(s) + CO(g) ⇌ Ni(s) + CO2(g) Kc = 4.0x103 at 1000K If a mixture of solid nickel(II) oxide and 0.10M carbon monoxide were allowed to come to equilibrium at 1000K, what would be the equilibrium concentration of CO2 NiO(s) + CO(g) ⇌ Ni(s) + CO2(g) 0.10M 0M -x +x 0.10M-x x x = 0.10 [CO] = 0.10 – 0.10 = 0M Kc  [CO 2 ]  4.0x10 3 [CO ] Solids do not appear in the equilibrium expression (pure liquids also do not appear) Kc  x  4.0x103 0.10  x x  4.0x10 2  4.0x103 x 4001x  4.0 x10 2 [CO2] = x = 0.10M Since Kc >>1, it makes sense that the reaction will go to completion (near 100%). 11. Consider the following reaction at equilibrium: K c  [O 2 ] 3 2KClO3(s) ⇌2KCl(s) + 3O2(g) Solids do not appear in the equilibrium expression (pure liquids also do not appear) Predict whether the reaction will shift left, shift right, or remain unchanged upon each of the following disturbances a. O2 is removed from the reaction mixture. Shift to the right (removing a product) b. KCl is added to the reaction mixture. No change. Addition of more solid does not affect eq. c. KClO3 is added to the reaction mixture. No change. Addition of more solid does not affect eq. d. O2 is added to the reaction mixture. Shift to the right (adding a product) e. Volume of the container is decreased. Same as increasing pressure. Since gas is only on products side, it will shift toward reactants. Determine the pH and pOH of the following 4 solutions: 12. For the following reaction, Kc = 0.513 at 500K. N2O4(g) ⇌2NO2  A 4.5 x 10-3 M HBr solution. Strong acid: pH = -log (H3O+) = 2.35, pOH = 14 – 2.35 = 11.65  A 3.67 x 10-5 M KOH solution. Strong base: pOH = -log (OH-) = 4.435; pH = 14 – 4.435 = 9.565 13.  A solution made by diluting 25 mL of 6.0 M HCl until the final volume of the solution is 1.75 L. M1V1 = M2V2 (6.0M)(0.025 L) = (M2)(1.75 L) M2 = [H3O+] =0.086M [OH-] = 1x10-14/0.0857 = 1.2x10-13M pH = -log(0.086) = 1.07  *5.0 L of an aqueous solution that contains 1.0 grams of HBr and 1.0 grams of nitric acid. 1.0 g HBr (1 mol / 79.908 g) = 0.0125 mol; 1.0 g HNO3 (1 mol / 63.008) = 0.0159 mol total moles = 0.0284 mol / 5.0 L = [H3O+] = 0.0057 M; since both acids are strong. [OH-] = 1.x10-14/0.00568 = 1.76x10-12M pH = -log(0.0057) = 2.24 14. Identify the following as strong or weak acids, strong or weak bases, neutral salts, basic salts or acidic salts. LiCl __ neutral salt _______, NH3_______ weak base____, NH4ClO4___ acidic salts _____, HClO4__ strong acid__, Ba(OH)2__ strong base ___, Ba(CH3COO)2__basic salts __, HF__weak acid ______, NaF_____ basic salt_____, AlBr3____ acidic salt __, KOH_____ strong base ____, K2CO3___ basic salt______, Ca(NO3)2___ neutral salts ____. 15. Write a balanced chemical equation to show that sodium cyanide (NaCN) is basic in water solution. NaCN  Na+ + CN— CN- + H2O  HCN + OH- 16. What is the Kb expression and value for the equilibrium above? Kb = [HCN][OH-] / [CN-] Kb = Kw / Ka = 1.0 x 10-14 / 4.9 x 10-10 = 2.0 x 10-5 17. Which one of the following salts, when dissolved in water, produces the solution with the highest pH? a. b. c. d. KI KBr KF KCl KF is a basic salt (The product of a weak acid and strong base). The others are neutral salts (The product of a strong acid and a strong base. 18. Calculate the pH of a 0.100 M CH3COOH solution. Ka for acetic acid is 1.8 x 10-5. (Use HA as CH3COOH): HA + H2O  A- + H3O+ 0.100 0 0 -x +x +x 0.100 – x x x Ka = x2 / (0.100 – x) = 1.8x10-5 Assume x << 0.100 (and check it!) Ka = x2 / 0.100 = 1.8x10-5 x2 = 1.8x10-6; x = 1.3 x 10-3; pH = -log(1.3x10-3) = 2.89 19. Calculate the % ionization for the solution above. 1.3x10-3 / 0.100 x 100 = 1.3% 20. Would the pH of the solution above be different if the concentration of CH3COOH was 0.200M? Explain. Yes, at a higher concentration, the pH should be lower since there will be more [H+] ions in solution. 21. Would the percent ionization of the solution above be different if the concentration of the CH3COOH was 0.200M? Explain. Yes, the higher the concentration of a WA or WB, the lower the % ionization (although the [H+] overall is higher at the higher concentration, the [H+] concentration is lower relative to the 0.200M acid). 22. Assume you have two acids, one with a Ka of 1.0x10-5 and another with a Ka of 1.0x10-10. Asuming the y have the same concentration, which acid should have a higher % ionization? Assuming both acids are the same concentration, the one with the higher Ka will have a higher % ionization. 23. Calculate the pH of a 0.100 M NaCH3COOH. (Use A- as CH3COO-): A- + H2O  HA + OH0.100 0 0 -x +x +x 0.100 – x x x Kb x Ka = K w Kb = 1.0x10-14 / 1.8x10-5 = 5.6x10-10 Kb = x2 / (0.100 – x) = 5.6 x 10-10 Assume x << 0.100 (and check it!) Kb = x2 / 0.100 = 5.6 x 10-10 x2 = 5.6 x 10-11; x = 7.48 x 10-6; pOH = 5.13, pH = 8.87