AP Chemistry Name……………………………………...…….. Equilibrium Test Review Hour……. Directions: Complete the following problems. 1. Consider the reaction system, CoO(s) + H2(g) ↔ Co(s) + H2O(g) The equilibrium constant expression is Note that solids are not included! [CoO][H 2 ]
[Co][H 2 ]
a.
c. [Co][H 2O]
[CoO][H 2 ]
b. [H 2O]
[H 2 ]
d. [H 2 ]
[H 2O]
2. Given the equilibrium, 2SO2(g) + O2(g) ↔ 2SO3(g) if this equilibrium is established by beginning with equal number of moles of SO2 and O2 in 1.0 L bulb, then the following must be true at equilibrium: Watch the balanced equation! Once started, [SO2] changes by twice that of [O2]. So [SO2] < [O2] a. [SO2] = [SO3] c. [SO2] < [O2] b. 2[SO2] = 2[SO3] d. [SO2] > [O2] 3. At a given temperature, 0.300 mol NO, 0.200 mol Cl2 and 0.500 mol ClNO were placed in a 25.0 L container. The following equilibrium is established: 2ClNO(g) ↔ 2NO(g) + Cl2(g) At equilibrium 0.600 mol of ClNO was present. The number of moles of Cl2 present at equilibrium is 2ClNO(g) ↔ 2NO(g) + Cl2(g) I 0.0200 0.012 0.0080 C +0.0040 -­‐0.0040 -­‐0.0020 E 0.0240 0.0080 0.0060 0.0060 M Cl2 × 25 L = 0.150 moles Cl2 a. 0.050 c. 0.150 b. 0.100 d. 0.200 4. The equilibrium constant, Kc for #3 is a. 4.45×10-­‐4 c. 0.167×10-­‐4 b. 6.67×10-­‐4 d. 0.111 Kc =
[Cl 2 ][NO] 2 [0.0060][0.0080] 2
=
=06.67 × 10 34 2
2
[ClNO]
[0.0240]
Page 1 of 5 5. At 985°C, the equilibrium constant for the reaction, H2(g) + CO2(g) ↔ H2O(g) + CO(g) is 1.63. What is the equilibrium constant for the reverse reaction? Reverse reaction has K value equal to inverse of original K. a. 0.815 c. 0.613 b. 2.66 d. 1.00 6. What is the relationship between Kp and Kc for the reaction, 2ICl(g) ↔ I2(g) + Cl2(g)? Note that ∆n = 0 a. Kp = Kc c. Kp = Kc(RT) -­‐1
b. Kp = Kc(RT) d. Kp = Kc(RT)2 7. For the reaction 2NO2(g) ↔ N2O4(g), Kp at 298 K is 7.3 when all partial pressures are expressed in atmospheres. What is Kc for this reaction? Kp = Kc(RT)∆n; 7.3 = Kc(0.0821×298)-­‐1 a. 4270 c. 179 b. 0.291 d. 2.06 8. 0.200 mol NO is placed in a one liter flask at 2273 K. After equilibrium is attained, 0.0863 mol N2 and 0.0863 mol O2 are present. What is Kc for this reaction? 2NO(g) ↔ N2(g) + O2(g) I 0.200 0 0 C -­‐0.1726 0.0863 0.0863 E 0.0274 0.0863 0.0863 [N ][O ] [0.0863][0.0863]
K c = 2 22 =
=09.92 [NO]
[0.0274] 2
a. 9.92 c. 0.0372 b. 3.15 d. 39.7 9. For the reaction system, N2(g) + 3H2(g) ↔ 2NH3(g) + heat the conditions that would favor maximum conversion of reactants to products would be a. high temp & high pressure b. high temp & low pressure c. low temp & low pressure d. low temp & high pressure (LeChatelier!) Page 2 of 5 10. Solid HgO, liquid Hg and gaseous O2 are placed in a glass bulb and are allowed to reach equilibrium at a given temperature. 2HgO(s)↔2Hg(l)+O2(g) ∆H=+43.4 kcal The mass of HgO in the bulb could be increased by a. Adding more Hg (no change) b. Removing O2 (shift right…less HgO!) c. Reducing the volume of the bulb (fewer moles of gas on reactant side!) d. Increasing the temperature (shift right...less HgO!) e. Removing some Hg (no change) 11. For the equilibrium system H2O(g) + CO(g) ↔ H2(g) + CO2(g) ∆H=-­‐42 kJ/mol Kc equals 0.62 at 1260 K. If 0.10 mole each of H2O, CO, H2 and CO2 (each at 1260 K) were placed in a 1.0 L flask at 1260 K, when the system came to equilibrium the mass of H2 ___ and mass of CO would ___. Calculate Q, compare to K. [H ][CO2 ] [0.10][0.10]
Q"=" 2
=
="1 Q > K Reaction will shift to the left [H 2O][CO] [0.10][0.10]
a.
b.
c.
d.
e.
decreases, increases decreases, decreases remains constant, increases increases, decreases increases, increases 12. Which one of the following is the solubility product constant for Mn(OH)2? a. Ksp = [Mn2+][OH-­‐]2 b. Ksp = [Mn2+]2[OH-­‐]2 c. Ksp = [Mn2+]2[OH-­‐] d. Ksp = [Mn2+][2OH-­‐]2 13. The solubility of HgS is 5.5×10-­‐27 mol/L. What is Ksp for HgS? Remember, solubility is x! Ksp = [Hg2+][S2-­‐] = [x][x] = [x]2 = [5.5×10-­‐27]2 a. 4.0×10-­‐3 c. 7.4×10-­‐14 b. 1.3×10-­‐13 d. 3.0×10-­‐53 14. Which expression best describes the relationship between solubility product, Ksp and the solubility s of MgF2? Remember, solubility is x (or s in this case)! Ksp = [s][2s]2 a. Ksp = 4s3 c. Ksp = s2 2
b. Ksp = 4s d. Ksp = 2s3 Page 3 of 5 15. Calculate the molar solubility of Fe2S3, Ksp = 1.4×10-­‐88 In other words…calculate x! Ksp = [2x]2[3x]3 = 108x5 = 1.4×10-­‐88 a. 5.5×10-­‐62 M c. 1.1×10-­‐18 M b. 1.2×10-­‐44 M d. 4.8×10-­‐24 M 16. For BaSO4, Ksp = 1.1×10-­‐10. If you mix 200. mL of 1.0×10-­‐4 M Ba(NO3)2 and 500. mL of 8.0×10-­‐2 M H2SO4 what will be observed? Compare Q with K. Precipitation occurs when Q > K. Don’t forget total volume! '4
0.200$L 1× 10 mole$Ba(NO3 )2
1$mole$Ba 2+
=$2 × 10 '5 mole$Ba 2+
1$L
1$mole$Ba(NO3 )2
'2
2'
0.500$L 8.0 × 10 mole$H 2 SO4 1$mole$SO4
=$0.0400$mole$SO42'
1$L
1$mole$H 2 SO4
2'
2 × 10 '5 mole$Ba 2+ 0.0400$mole$SO4
Q$=
=$1.63 × 10 '6
0.700$L
0.700$L
Q > Ksp…precipitate! a. Q > Ksp; a precipitate will form b. Q = Ksp; no precipitation c. Q < Ksp; no precipitation d. Q < Ksp; a precipitate will form 17. Calculate the molar solubility of AgCl in a 0.10 M solution of NaCl (Ksp of AgCl is 1.8×10-­‐10) Common ion! AgCl(s) ↔ Ag+(aq) + Cl-­‐(aq) I 0 0.10 M C -­‐x +x +x E +x 0.10+x Ksp = [x][0.10+x] = 1.8×10-­‐10…assume [0.10+x] = 0.10…[x][0.10] = 1.8×10-­‐10…[x] = 1.8×10-­‐9 M a. 1.3×10-­‐5 M c. 5.5×108 M b. 1.8×10-­‐9 M d. 4.2×10-­‐5 M 18. For the equilibrium system Co(H2O)62+(aq) + 4Cl-­‐(aq) ↔ CoCl42-­‐(aq) + 6H2O(l) (pink) (blue) the ∆H is positive. Which of the following would result in a system that appears blue at equilibrium? a. Addition of H2O b. Removal of Cl-­‐ by addition of AgNO3(aq) c. Place system in an ice bath d. Addition of KCl(aq) Page 4 of 5 19. Gaseous NOCl decomposes to form the gases NO and Cl2. 2NOCl(g) ↔ 2NO(g) + Cl2(g) At a given temperature, Kc is 1.6×10-­‐5. 1 mol NOCl is placed in a 2.0 L flask. What is [NO] at equilibrium? I 2NOCl(g) ↔ 2NO(g) + Cl2(g) 0.50 0 0 C -­‐2x +2x +x E 0.50-­‐2x +2x +x [NO] 2 [Cl 2 ] [2x] 2 [x]
Kc =
=
=01.6 × 10 /5
2
2
[NOCl]
[0.50/2x]
assume0x0is0small0as0K0is0small
4x 3
=01.6 × 10 /5
[0.50] 2
x0=01.0 × 10 /2
[NO]0=02x0=02.0 × 10 /2
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