CHM152 Exam 2 Key

CHM152 Exam #2 Fall 2012 (Ch. 13, 14) KEY 1. (7 points) T F At equilibrium, the amount of products is always the same as the amount of reactants. T F The forward and reverse reaction rates for a reaction are the same at equilibrium. 2. (7 points) Assume you want to perform the following reaction, 5A(g) → 2B(g), which has a Kp of 1.0x10-5. If you start with 5.0atm of A, what will be the equilibrium pressures? Assume the reaction takes place at 25°C. 5A(g) → 2B(g) I 5.0atm 0atm C -5x +2x E 5.0atm-5x 2x [ B]2 Kp   1.0 x10 5 5 [ A] [2 x] 2  1.0 x10 5 5 [5.0] 4 x 2  0.0313 x  0.0884 Since K is very small the change in x will be very small…so we get rid of it (Ap)eq= 5.0-(5)(0.0884) = 4.6atm 5.0 x100  8.8% (5)(0.0884) OOOPPPSS….since this is >5% we’re supposed to use the quadratic instead. But since it’s an exam, we don’t have time…. 1.0 x10 5  (2.5 points) what would be the Kc for the reaction? K p  K c ( RT ) n  (3.5 points) If you had started with 5.0atm of B and 0.0atm of A, which direction would the reaction proceed to reach equilibrium? forward   reverse Latm (298K )( 25) 0.08206 molK already at equilibrium (3.5 points) If you started with 2.5atm of each reactant, which direction would the reaction proceed to reach equilibrium? forward reverse already at equilibrium (3.5 points) If you would have written the reaction in reverse, 2B(g) → 5A(g), would this change the value of KP? If yes, what would be the value? YES K p '  3. [0.18] 2  1.6 x10 5  K p 5 [4.6] (Bp)eq= 2(0.0884) = 0.18atm 1  1.0 x10 5 Kp (4.5 points) Assume you have two monoprotic weak acid solutions “A” and “B” at the same concentration (1.0M). When you test the two acids, “A” has a pH of 3 and the “B” has a pH of 4. Answer the following questions:  Which acid has the larger Ka? A B SAME  0.14  Which solution has a higher concentration of OH-? A B SAME  Which acid is the stronger acid? A B SAME 4. (4.5 points) Assume you have two monoprotic weak acid solutions “A” (2.0M) and “B” (1.0M). When you test the two acids, they both have a pH of 5. Answer the following questions:  Which acid has the larger Ka? A B SAME  Which solution has a higher concentration of OH-? A B SAME  Which acid is the stronger acid? A B SAME 5. (3 points) Which of the following can act as both an acid and a base (choose all that apply): __x_ H2O ____Al3+ ____H3PO4 __x_H2PO4__x_HPO42____PO436. (7 points) You have a reaction with the balanced equation of A(aq) + B(aq) ⇌ 2C(aq). If you start with 1.0M of A and 1.0M of B, what would be the equilibrium concentrations? The Kc for the reaction is 123. A(aq) + B(aq) ⇌ 2C(aq) I 1.0M 1.0M C -x -x E 1.0M-x 1.0M-x 0M +2x 2x 2 x  11.09  11.09 x Kc  Kc  [C ]2  123 [ A][ B] (2 x) 2  123 (1.0  x) 2 2x  11.09 1.0  x x  0.85 [A]eq = [B]eq = 1.0M-0.85M = 0.15M = 0.2M [A]eq= (2 x) 2  123 (1.0  x) 2 [B]eq= (1.7) 2  128  123 (0.15M )(0.15M ) [C]eq = (2)(0.85) = 1.7M [C]eq= 7. (4 points) What concentration of KOH will give you a pH of 8.00? If you plug-in the rounded value (with only 1 sig fig) for x, your Kc will be 72 !(sometimes you have to plug-in numbers with more sig figs to double check pOH = 14.00 - 8.00 = 6.00 [KOH] = [OH-] = 10-pOH = 10-6.00 = 1.0x10-6M 8. (3 points) In a pH solution of 7.0, what are the following concentrations? [H+]: 1.0x10-7M [OH-]: 1.0x10-7M  (4 points) In a pH solution of 10.19, what are the following concentrations? (show calculations to receive full credit). [H+]: = 10-10.19 = 6.46x10-11M [OH-]: 1.00 x10 14  1.55x10-4M 6.46 x10 11 M 9. (4 points) Put the four acids made with halogens (HCl, HBr, etc. [the hydrohalic acids]) in order of decreasing strength (do not include HAt). HI___ > HBr___ > HCl___ > HF___ 10. (2.5 points) Are all of the acids above strong acids? Explain. No only HI, HBr, HCl are strong. HF is weak. 11. (3 points) Explain the trend in hydrohalic acid strength. The longer the bond between the X-H, the weaker the bond, the stronger the acid. 12. (13.5 points) Complete the following acid/base reactions by predicting the conjugate acid/base pairs. NH3 + HF ⇌ NH4+ + F- reactant favored / product favored HCO3- + S2- ⇌ CO32- + H2S reactant favored / product favored CN- + NH4+ ⇌ HCN + NH3 reactant favored / product favored Made a mistake when I made this problem/table. Should be S2here and H2S on the table. 13. (2 point) Which acid from the table above will give the weakest conjugate base? HF 14. Use the following unbalanced question to answer the following questions: N2(g) + H2(g) → NH3(g) (exothermic)  (2 point) Write the Kp expression: N2(g) + 3H2(g) → 2NH3(g)  (6 point) Determine how to shift the reaction to the right. ____↑_______ the amount of N2 ____↑_______ the partial pressure of H2 ____↓_______ the amount of NH3 so… [ NH 3 ] 2 [ N 2 ][ H 2 ]3 HF NH4+ HCN HCO3H2O H2S Acid Strength ____↓_______ the volume of the container 15. (7 points) Determine the pH of a 0.100M phenolate ion (C6H5O-, a base) solution. Its conjugate acid (C6H5OH) has a Ka of 1.0x10-10. C6H5O- + H2O → OH- + C6H5OH I 0.100M C -x E 0.100M-x 0M +x x 0M +x x Kb  x 2  1.0 x10 5 1.0 x10 14  1.0 x10  4 10 1.0 x10 x  3.2 x10 3 x2  1.0 x10  4 0.100M pOH = -log(3.2x10-3) = 2.49 pH = 14.00-2.49 = 11.51 16. (3 points) Write the K expression for the following reactions/processes: Kp  Pc C or K p  2 (B) PB 2  A(s) + 2B(g) ⇌ C(g)  NaNO3(s) ⇌Na+(aq) + NO3(aq) Kc = [Na+][NO3-]  3OH-(aq) + Fe3+(aq) ⇌ Fe(OH)3(s) Kc = 1 [OH ] [ Fe 3 ]  3