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# 1 more nh 3 will be formed correct 2 nothing the

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1. More NH 3 will be formed. correct 2. Nothing; the system is at equilibrium. 3. More N 2 and H 2 will be formed. Explanation: K = 4 . 0 × 10 8 [NH 3 ] = 44 . 0 mol 10 L [N 2 ] = 0 . 452 mol 10 L [H 2 ] = 0 . 108 mol 10 L Q = [NH 3 ] 2 [N 2 ] [H 2 ] 3 = (4 . 40 M) 2 (0 . 0452 M) (0 . 0108 M) 3 = 3 . 4 × 10 8 Since Q < K equilibrium will shift to the right, forming more NH 3 . 019 3.3points The reaction Ag + ( aq ) + Cl ( aq ) AgCl(s) has an equilibrium of 10 20 (K eq = 10 20 ). If you have a beaker containing 1 liter of wa- ter with 0.1 mol of AgCl(s) in it along with 10 6 M Ag + ( aq ) and 10 15 M Cl ( aq ), how does the reaction shift? 1. not enough information 2. not at all 3. left correct 4. right Explanation: Q = 1 [Ag + ] · [Cl ] = 1 10 6 · 10 15 = 10 21 Because Q is greater than K, the reaction will shift to the left. 020 3.3points Which reaction would be the most useful, in terms of thermodynamics, if you were trying to make “C?” 1. A + B ←→ C K = 10 40 correct 2. F + E ←→ C + D K = 10 12 3. 2 C ←→ F K = 10 5 4. 2 D ←→ C K = 10 20 5. C + A ←→ D K = 10 10 Explanation:

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casey (rmc2555) – Homework 6 – holcombe – (51395) 6 The reaction whose equilibrium constant lies farthest toward the side with “C” on it is A + B ←→ C , K = 10 40 . 021 3.3points For the reaction POCl 3 (g) POCl(g) + Cl 2 (g) K c = 0 . 30. An initial 0 . 31 moles of POCl 3 are placed in a 3 . 9 L container with initial concentrations of POCl and Cl 2 equal to zero. What is the final concentration of POCl 3 ? ( Note: You must solve a quadratic equation.) 1. final concentration = 0 . 269487 M 2. final concentration = 0 . 0142056 M cor- rect 3. final concentration = 0 . 0284112 M 4. final concentration = 0 . 19 M 5. final concentration = 0 . 065 M Explanation: K c = 0.30 V container = 3 . 9 L [POCl 3 ] initial = 0 . 31 mol 3 . 9 L = 0 . 0794872 M POCl 3 (g) POCl(g) + Cl 2 (g) ini, M 0 . 0794872 Δ, M x x x eq, M 0 . 0794872 x x x K c = [Cl 2 ] [POCl] [POCl 3 ] = x 2 0 . 0794872 x = 0 . 3 x 2 = 0 . 0238462 0 . 3 x x 2 + 0 . 3 x 0 . 0238462 = 0 x = 0 . 3 ± radicalbig (0 . 3) 2 4 (1) ( 0 . 0238462) 2 (1) = 0 . 0652816 or 0 . 365282 Reject 0 . 365282 as x because it leads to negative concentrations for POCl and Cl 2 and a concentration larger that the orig- inal concentration for POCl 3 . Therefore x = 0 . 0652816 M and [POCl 3 ] = 0 . 0794872 M 0 . 0652816 M = 0 . 0142056 M 022 3.3points Consider the following decomposition reac- tion at 700 K. 2 CaSO 4 (s) 2 CaO(s) + 2 SO 2 (g) + O 2 (g) If K p = 0 . 032 at this temperature, what will be the equilibrium overall pressure start- ing from pure CaSO 4 (s)? 1. 0.20 bar 2. 0.60 bar correct 3. 0.011 bar 4. 0.22 bar 5. 0.40 bar Explanation: K p = P 2 SO 2 P O 2 P SO 2 = 2 x P O 2 = x 0 . 032 = (2 x ) 2 x = 4 x 3 x = (0 . 032 / 4) 1 / 3 x = 0 . 2 bar P total = P SO 2 + P O 2 P total = 0 . 4 + 0 . 2 = 0 . 6 bar 023 3.3points The equilibrium constant (dimensionless) for A B is 2. If the initial concentrations of A and B are each 1.0 M, what are the final concentra- tions of A and B respectively? 1. [A] = 0.67 M; [B] = 1.33 M correct 2. [A] = 0.50 M; [B] = 1.00 M 3. [A] = 0.67 M; [B] = 0.33 M
casey (rmc2555) – Homework 6 – holcombe – (51395) 7 4. [A] = 1.00 M; [B] = 1.33 M 5. [A] = 1.00 M; [B] = 2.00 M Explanation: K = 2 [A] ini = 1 . 0 M [B] ini = 1 . 0 M Q = [B] [A] = 1 . 0

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