General Chemistry (Thall)

1. Run [A] [B] [C] Rate
1 1.00 M 1.00 M 1.00 M 0.150 M/s
2 2.00 2.00 2.00 4.80
3 1.50 1.00 1.00 0.338
4 1.50 3.00 1.00 9.13

(a) Write rate equation and describe how orders determined. (6)
(b) Calculate rate constant. (3)
(c) Determine rate when [A] = [B] = [C] = 3.00M (3)

2. Reaction first-order in A. (15)
(a) Calculate k if 30.0% A remains after 10.0 min.
(b) Calculate time needed for 95.0% of A to react.
(c) Calculate % A remaining after 17.5 min.

3. At 300K the rate constant is 0.100 s^-1 and at 350K the rate constant is 0.800 s^-1. (10)
(a) Determine activation energy.
(b) Determine rate constant at 400 K.

4. A_2(g) + B_2(g) 2 AB_(g) Kc = 16
3.00 mol A₂, 3.00 mol B₂, and 3.00 mol AB added to a 1.00 L container. (8)
Calculate [A₂], [B₂], and [AB] at equilibrium.

5. A_(g) + B_(g) AB_(g) T = 300K
(a) 2.00 mol each A, B, and AB added to a 1.00 L container. Calculate [A], [B], and [AB]
when equilibrium reached (Kc = 1.0). (8)
(b) Calculate Kp. (4)

6. 2 A_(g) + 2 B_(g) A₂B_2(g) + heat (Ea = 10 kJ DH = -90 kJ)
(a) State and account for shift when temperature decreased. (3)
(b) State and account for shift when everything doubled. (3)
(c) State and account for shift when volume increased. (3)
(d) State and account for shift when [B] doubled and [A₂B₂] tripled. (3)
(e) Draw energy profile to scale labeling values for Ea and DH. (3)
(f) Explain why transition state resembles reactant. (3)
(g) Explain why K>1. (3)

7. Provide brief explanation: (16)
(a) A small increase in temperature results in large increase in rate of reaction.
(b) The concentration of solids & liquids are omitted from equilibrium expression.
(c) For a two step reaction, the rate depends on the slow step.
(d) The initial concentration has no influence on the half-life of a first order reaction.

Answers
1. (a) rate = k[A]²[B]³
runs 1&2 establish 5th order overala
runs 3&4 establish 3rd order in B (if triple [B], rate 27x)
runs 1&3 establish 2nd order in A (increase [A] by 1.5x, rate 2.25x)

(b) rate = k[A]²[B]³
0.150 M/s = k [M⁵] k = 0.150 M^-4s^-1

(c) rate = [0.150][3]⁵ = 36.5 M/s

2. (a) ln (A₀/A_t) = kt
ln (100/30) = k(10 min) k = 0.12 min^-1
(b) ln (100/5) = (0.12)t t = 25 min
(c) ln (100/A_t) = (.12)((17.5) A_t = 12.2 or 12.2%

3. (a) E_a = (R)[(T₁T₂)/(T₂-T₁)]ln (k₂/k₁) = (0.00831)[(350)(300)/(50)]ln 8 = 36.3 kJ/mol

(b) ln (k₂/k₁) = (E_a/R)[(T₂- T₁)/(T₁T₂)] = (36.3/.00831)[(400-300)/(300)(400)] = 3.64
(k₂/0.10 s^-1) = 38 k₂ = 3.8 s^-1

4. A_2(g) + B_2(g) 2 AB_(g)
Eq: 3-x 3-x 3+2x

(3+2x)/(3-x) = 4 x = 1.50 [A₂] = [B₂] = 1.50M [AB] = 6.00M

5. A + B AB T = 300K
2-x 2-x 2+x

(a) (2+x)/(2-x)² = 1
x2 - 5x +2 = 0 x = 0.44 [A] = [B] = 1.56M [AB] = 2.44M

(b) Kp = Kc(RT)^Dn
= (1)[(0.0821)(300)]^-1 = 0.041

6. (a) Reaction exothermic, SHIFTS RIGHT when Temperature decreased
(b) K =[A₂B₂]/[A]²[B]² = 2/(2²)(2²) = 1/8 SHIFTS RIGHT
(c) If pressure decreased, favors greater # gas molecules, SHIFTS LEFT
(d) Q = 3/(1²)(2²) = 3/4 SHIFTS RIGHT
(e)

(f) Reactant closer to transition state in terms of energy.
(g) When equilibrium reached, there will be much more product than reactant (see energy profile).

7. (a) Increasing the temperature dramatically increases the number of effective collisions.
(b) These concentrations are constant and can be represented in K.
(c) The slow step has the higher energy barrier and it is assumed the slow step takes about the same
time as the slow and fast steps combined.
(d) Half-life defined as the time required for half the initial concentration to react. Therefore, it is
independent of initial concentration.

1. Construct table and provide CA/CB (if any): CrO₄^-2 HC₂O₄^-1 NH₄⁺¹ HClO₄ PCl₃ (10)
2. List theories (Arrenhius, Bronsted-Lowry, or Lewis) that are applicable: (10)
(a) Br + Br Br₂ (b) HClO₃ + H₂O H₃O⁺¹ + ClO₃^-1
(c) Cl⁺¹ + Cl^-1 Cl₂ (d) H + Cl HCl (e) HNO₃ + H₂S H₃S⁺¹ + NO₃^-1

3. (a) For HAc/Ac^-1 Prove K_w = K_a^. K_b (12)
(b) Explain how HAc/NaAc works as buffer.
(c) Using Ka/Kb, explain why NH₄CN is basic in water.

4. Calculate pH resulting from mixing: (16)
(a) 40 mL of 1.00 M HCl and 60 mL of 1.00 M NaOH.
(b) 50 mL of 1.00 M HCN and 30 mL of 1.00 M NaOH.

5. Calculate pH: (20)
(a) 3.00 M H₃PO₄ (b) 10.0 M Na₃PO₄ (c) 0.100 M NaClO (d) 0.50M NaHCO₃

6. When 2.95 g of C₃H₉N diluted to exactly 1 liter, pH of 11.40 results. Calculate K_b (8)
7. Calculate pH, [HF] & [F^-1] for 0.200 M HF. (8)

8. Calculate Ka if 1.0M weak acid ionizes 3.0%. (8)

9. 1.00 L of buffer contains 1.00 mol HAc and 0.50 mol NaAc: (12)
(a) Calculate pH after adding 0.25 mol OH^- (assume volume remains 1.00L)
(b) Calculate pH after adding 0.25 mol H⁺ (assume volume remains 1.00L)

Answers
1. CrO₄^-2 HC₂O₄^-1 NH₄⁺¹ HClO₄ PCl₃
CA HCrO₄^-1 H₂C₂O₄ none none HPCl₃⁺¹
CB none C₂O₄^-2 NH₃ ClO₄^-1 none

2. (a) none (b) Arrenhius/BL/Lewis (c) Lewis (d) none (e) BL/Lewis

3. (a) HAc H⁺¹ + Ac^-1 Ac- HAc + OH^-1
K_a = [H⁺¹][Ac^-1]/[HAc]K_b = [HAc][OH^-1]/[Ac^-1]K_a^. K_b = [H⁺¹][OH^-1] = K_w
(b) If additional base added, H⁺¹ combines with OH^-1 and shift takes place to the RIGHT
If additional acid added, Ac^-1 combines with H⁺¹ and shift takes place to the LEFT

(c) NH₄⁺¹ H⁺¹ + NH₃ K_a = 5.6x10^-10
CN^-1 HCN + OH^-1 K_b = 1.6x10^-5 Solution basic because K_b > K_a

4. (a) In: 0.040 mol 0.060 mol
HCl + NaOH ®NaCl + H₂O
Eq: 0 0.020 mol
NaOH in excess by 0.020 mol [OH^-1] = 0.020mol/0.10L = 0.20M pOH = 0.70 pH = 13.30

(b) In: 0.050 mol 0.030 mol
HCN + NaOH NaCN + H₂O
Eq: 0.020 mol 0 0.030 mol

HCN/NaCN buffer now exists
Ka = [H⁺][CN^-]/[HCN] = [H⁺][0.030mol/0.080L]/[0.020mol/0.80L] = 6.2x10^-10
[H⁺] = 4.1x10^-10 M pH = 9.38

5. (a) H₃PO₄ H⁺¹ + H₂PO₄^-1 K_a = 7.1x10^-3 = x²/3 x = [H⁺¹] = 0.15 M pH = 0.84
3-x x x
(b) PO₄^-3HPO₄^-2 + OH^-1 K_b = 2.22x10^-2 = x²/10 x = [OH^-1] = 0.47 M pOH = 0.33
10-x x x pH = 13.67

(c) ClO^-1 HClO + OH^-1 K_b = 3.3x10^-7 = x²/0.1 x = [OH^-1] = 1.8x10^-4 pOH = 3.74
0.1-x x x pH = 10.26

(d) HCO₃^-1 H⁺¹ + CO₃^-2 Ka = 4.7x10^-11
HCO₃^-1 H₂CO₃ + OH^-1 Kb = 2.2x10^-8 Since Kb > Ka, solution BASIC
0.5-x ~ 0.5 x x
x²/0.5 = 2.2x10^-8 x = [OH^-1] = 1.1x10^-4
pOH = 3.96 pH = 10.04

6. mol B = 2.95/59.0 = 0.0500M [B] = 0.0500 M when pH = 11.40, [OH^-1] = 0.0025 M
B = BH⁺¹ + OH^-1
0.0475M 0.0025M 0.0025M
K_b = [BH⁺¹][OH^-1]/[B] = (0.0025)²/0.0475 = 1.3x10^-4

7. HF H⁺¹ + F^-1 Ka = 6.8x10^-4
0.2-x x x

x²/(0.2-x) = 6.8x10^-4 = x²/0.2
x = 0.012 pH = 1.92 [F-1] = 0.012M [HF] = 0.19M

8. HA H⁺¹ + A^-1
0.97 0.03 0.03 Ka = [H⁺¹][A^-1]/[HA] = (0.03)2/0.97 = 9.3x10^-4

9. (a) Addition of 0.25M OH^-1 causes 0.25M HAc to shift RIGHT
at equilibrium: [HAc] = 0.75M [Ac^-1] = 0.75M
K_a = [H⁺¹][Ac^-1]/[HAc] = [H⁺¹][0.75]/[0.75] = 1.8x10^-5
[H⁺¹] = 1.8x10^-5 pH = 4.74
(b) Addition of 0.25M H⁺¹ causes 0.25M Ac^-1 to shift LEFT
at equilibrium: [HAc] = 1.25M [Ac^-1] = 0.25M
K_a = [H⁺¹][Ac^-1]/[HAc] = [H⁺¹][0.25]/[1.25] = 1.8x10^-5
[H⁺¹] = 9.0x10^-5 pH = 4.05

1. Calculate solubility for PbI₂ in g/L (K_sp = 8.7x10^-9) (10)
2. Calculate molar solubility for PbSO₄ in 0.50 M Na₂SO₄ (10)
3. Calculate K_sp if 0.062 g Ca₃(PO₄)₂soluble in 1.00 L of H₂O (10)
4. Calculate Ksp for M(OH)₃ if saturated solution gives pH of 9.00 (10)
5. Calculate pH range for separating 1.0M Mg⁺² from 1.0M Zn⁺². (10)
6. If 50.0mL 0.020M Ca⁺² mixed with 50.0mL 0.030M CrO₄^-2, (10)
will precipitate form? Explain with calculation!
7. 4.00 M Na₃PO₄ solution has density of 1.20 g/mL. (12 )
Calculate: (a) weight % Na₃PO₄ (b) molality (c) mole fraction Na₃PO₄
8. The vapor pressure of pure water is exactly 600 torr. (10)
Solution consisting of 108 g H₂O and 54.0 g of nonvolatile solute displays
vapor pressure of 500 torr. Calculate FW of solute.
9. Solution prepared using 33.3 g CaCl₂ and exactly 400 g H₂O. (10)
Determine boiling & freezing points of this solution.
10. Determine F.P. if 3.00m CaBr₂ solution ionizes 80.0%. (10)

Answers
1. PbI₂ Pb⁺² + 2 I^-1 K = 8.7x10^-9 = [Pb⁺²][I^-1]² =(x)(2x)² = 4x³
x x 2x x = 0.0013
(0.0013 mol/L)(461 g/mol) = 0.60 g PbI₂

2. PbSO₄ Pb⁺² + SO₄^-2 Ksp = 6.3x10^-7 = [Pb⁺²][SO₄^-2] = (x)(x+0.50) @ 0.50x
x x x+0.50 x = 1.3x10^-6 M

3. Solubilty Ca₃(PO₄)₂ = (0.062g/L)/(310 g/mol) = 2.0x10^-4 M
Ca₃(PO₄)₂ 3 Ca⁺² + 2 PO₄^-3
2.0x10^-4 M 6.0x10^-4 M 4.0x10^-4 M K_sp = [Ca⁺²]³[PO₄^-3]² = [6.0x10^-4]³[4.0x10^-4]² =3.5x10^-17

4. When pH=9.00, [OH^-1] = 1.0x10^-5
M(OH)₃ M⁺³ + 3 OH^-1
3.3x10^-6 M 1.0x10^-5 M Ksp = [M⁺³][OH^-1]³ = [3.3x10^-6][1.0x10^-5]³ = 3.3x10^-21

5. Mg(OH)_2(s) Mg⁺² + 2 OH^{-1
1.0M x M}
Mg(OH)₂: Ksp = 7.1x10^-12 = [Mg⁺²][OH^-1]²
7.1x10^-12 = [1.0][OH^-1]²
[OH^-1] = 2.7x10^-6 M pOH = 5.58 pH =8.42 pH range equal or less than 8.42

6. After mixing: [Ca⁺²] = 0.010M [CrO₄^-2] = 0.015M Ksp for CaCrO₄ = 7.1x10^-4
Ion Product = [Ca⁺²] [CrO₄^-2] = [0.010] [0.015] = 1.5x10^-4
Since IP < Ksp, NO PPT forms!

7. Assume 1.00 L or 1200 g total mass
contains 4.00 mol Na₃PO₄ or 656 g Na₃PO₄ and 544g H₂O (1200 - 656)
(a) % Na₃PO₄ = (656/1200)x100 = 54.7
(b) m = 4.00mol/0.544kg = 7.35
(c) mol H₂O = 544/18 = 30.2 mol fraction Na₃PO₄ = 4.00/34.2 = 0.117

8. P_sol = P_w X_w
500 torr = (600 torr)X_w X_w = 0.833 n_w = 108/18 = 6.00 n_s = 54.0/FW

X_w = n_w/(n_w + n_s) = 6.00/(6.00 + 54.0/FW) = 0.833 FW = 45.0 g/mol

9. mol CaCl₂ = 33.3/111 = 0.300 m = 0.300/0.400kg = 0.750
DT_b = imK_b = 3(0.750)(0.512) = 1.15 BP = 101.2°C
DT_f = imK_f = 3(0.750)(-1.86) = -4.19 FP = -4.2° C

10. Assume 100 particle CaBr₂ CaBr₂ Ca⁺² + 2Br^-1
20 80 160 = 260 i = 260/100 = 2.60

DT_f= imK_f = (2.60)(3.00)(1.86) = 14.5 F.P. = normal F.P. - DT_f = -14.5°C

1. Use DG/DH/DS to explain: (12)
(a) Water does not solidify at 10°C.
(b) Salts dissolve more readily in hot water than in cold water.
(c) Endothermic crystal formation.

2. Determine temperatures range needed to become spontaneous. (9)
Provide brief explanation!
(a) DH = 0 kJ (b) DH = -0.500 kJ (c) DH = +300 kJ
DS = -20 J/K DS = -500 J/K DS = +300 J/K

3. CaS_(s) + 2 O_2(g) ® CaSO_4(s) (16)
CaS_(s): DH_f = -482 kJ/mol DG_f = -477 kJ/mol
CaSO_4(s): DH_f = -1434 kJ/mol DG_f = -1321 kJ/mol
Determine: (a) DH_rxn (b) DG_rxn (c) DS_rxn (d) Is reaction favored at high/low temp? Explain!

4. NH₄Cl_(s) ® NH_3(g) + HCl_(g) (16)
NH₃: S° = 192 J/K/mol DH_f° = -46 kJ/mol
HCl: S° = 187 J/K/mol DH_f° = -92 kJ/mol
NH₄Cl: S° = 95 J/K/mol DH_f° = -314 kJ/mol
Determine: (a) DS°_rxn (b) DH°_rxn (c) DG°_rxn (d) K

5. Using Tables 20.3/20.4, estimate enthalpy of formation for N₂H₄. (10)

6. Write balanced ox/red equations and determine E° net: (10)
(a) AuBr₃ ® Au + Br₂ (b) SnCl₂ + Zn ® ZnCl₂ + Sn

7. Write balanced anode/cathode/overall equations for electrolysis: (10)
(a) LiNO_3(aq) (produce H₂/O₂) (b) KBr_(aq) (produce H₂)

8. (a) Determine mass Ni obtained if Ni⁺² subjected to 3.50 A for 35.0 minutes. (10)
(b) Determine minutes required for 15.0 g Cu⁺² to deposit if current 4.00 A.

9. Design galvanic cell using Al/Ag electrodes. Indicate anode, cathode, direction of (8)
current, and cell solutions. Provide balanced anode/cathode equations and calculate potential.

Answers
1. (a) when water solidifies: DH negative (favorable) DS negative (not favorable)
water does not solidify at 10°C because TDS>DH
(b) DH ? DS positive (favorable)
because DS favorable, want TDS as large as possible
(c) DH positive DS negative for endothermic crystal formation
endothermic crystal formation not possible!

2. (a) Never (b) T<1K (c)T>1000K

3. (a)DH = -DH_CaS + DH_CaSO4 = -(-482kJ) + (-1434kJ) = -952 kJ
(b) DG = -(-477kJ) + (-1321kJ) = -844kJ
(c) DS = [DH - DG]/T = [-952kJ - (-844kJ)]/298K = -0.362kJ/K = -362J/K
(d) Since enthalpy driven, favored at low temperature

4. (a) DS = -95J/K + 192J/K + 187J/K = +284J/K
(b) DH = +314kJ + (-464kJ) + (-92kJ) = +176kJ
(c) DG = DH - TDS = 176kJ - (298K)(0.284kJ/K) = +91.4kJ
(d) ln K = -DG/RT = (-91.4)/[(0.00831)(298)] = -36.9 K = 9.4x10^-17

5. DH_f(N) = 2(473 kJ) form 4 N-H bonds 4(-388 kJ)
DH_f(H) = 4(218 kJ) form 1 N-N bond -240kJ
+1818 kJ -1792kJ
DH = +1818kJ - 1792kJ = 26kJ

6. (a) Ox: 6 Br^-1® 3Br₂ + 6e^-1 -1.08 V (b) Ox: Zn°® Zn⁺² + 2e^-1 +0.76 V
Red: 2 Au⁺³ + 6e^-1 ® 2Au° +1.50 V Red: Sn⁺² + 2e^-1 ® Sn° -0.14 V
+0.42 V +0.62 V

7. (a) anode: 2H₂O ® 4H⁺¹ + O₂ + 4e^-1 (b) anode: 2Br-1 ® Br₂ + 2e^-1
cathode: 4H₂O + 4e^-1 ® 4OH^-1 + 2H₂ cathode: 2H₂O + 2e^-1® 2OH^-1 + H₂
overall: 2H₂O ® 2H₂ + O₂ overall: 2H₂O + 2Br^-1 ® H₂ + Br₂+ 2OH^-1

8. (a) Ni⁺² + 2e^-1 ® Ni° Q = I x t = (3.50)(2100) = 7350 coul = 0.0762F
0.0762F 0.0381 mol (2.24g Ni)

(b) Cu⁺² + 2e^-1 ® Cu° t = Q/I = 45500/4.00 = 11380 s = 190 min
0.472F 0.236 mol
or 45500 coul

9.

1. Reaction first-order in A. (12)
Calculate: (a) Rate constant k if 65.0% A remains after 6.00 min.
(b) Time needed for 75.0% of A to react. (c) % A remaining after 25.0 min. (d) t_½

2. A_2(g) + B_2(g) 2 AB_(g)
(a) 10.0 mol A₂ and 10.0 mol B₂, added to a 1.00 L container. At equilibrium 5.00 mol A₂ (4)
and 5.00 mol B₂ remain. Calculate equilibrium constant K.
(b) If 5.00 mole AB added to the system already at equilibrium in part (a), (8)
determine [A₂], [B2], and [AB] when new equilibrium reached.

3. Calculate pH: (a) 0.100 M NaClO (b) 1.00 M NH₄CN (12)

4. If 1.00 L of a buffer contains 1.00 mol HAc and 0.500 mol NaAc, calculate: (15)
(a) pH of buffer.
(b) pH after adding 0.30 mol NaOH to the buffer (assume volume remains 1.00L).
(c) pH after adding 0.30 mol HCl to the buffer (assume volume remains 1.00L).

5. Calculate K_sp for Pb(OH)₂ if 0.00100 g soluble in 1.00 L of water. (10)

6. Solution prepared from 95.3g MgCl₂ and 450g H₂O. Calculate: (12)
(a) MgCl₂ molality (b) solution freezing point (c) solution vapor pressure (pure water 600 torr)

7. SnCl_4(g) + 2 H₂O_(g) ® SnO_2(s) + 4 HCl_(g) (15)

DH_f (-472) (-286) (-581) (-92) ( kJ/mol)
DG_f (-432) (-237) (-520) (-95) ( kJ/mol)

Determine: (a) DH°_rxn (b) DG°_rxn (c) DS°_rxn (d) K (e) account for DS°

8. Design galvanic cell using Zn/Ni electrodes. Indicate anode, cathode, direction of current, and cell (12)
solutions. Provide balanced anode/cathode equations and calculate potential.

9. (a) Write balanced ox/red equations and determine overall E° for CuCl₂ ® Cu + Cl₂
(b) If CuCl₂ subjected to 8.0 amp for 5 hours, calculate grams of Cu obtained.

Answers
1. (a) ln(A₀/A) = kt
ln 1.54 = 0.43 = k(6.00 min) k = 0.072 min^-1
(b) ln(A₀/A) = kt
ln (100/25) = ln 4 = 1.39 = (0.072 min^-1)t t = 19.3 min
(c) ln(A₀/A) = kt
ln(100/A) = (0.072 min^-1)(25.0 min) = 1.80
100/A = 6.05 A = 16.5 or 16.5%
(d) t½ = 0.693/.072min^-1 = 9.63 min

2. (a) A₂ + B₂2AB
Eq: 5M 5M 10M K = 10²/5² = 4.0

(b) A₂ + B₂2AB
Eq: 5+x 5+x 15-2x K = (15-2x)²/(5+x)² = 4 or (15-2x)/(5+x) = 2
15-2x = 10+ 2x x = 1.25
[A₂] = [B₂] = 6.15M [AB] = 12.5M

3. (a) ClO^-1HClO + OH^-1
Eq: 0.1-x x x K_b = 4.0x10^-8 = x²/0.1 x = [OH^-1] = 2.0x10^-4
pOH = 3.40 pH = 10.60

(b) CN^-1HCN + OH^-1 K_b = 2.0x10^-5 soluton basic (K_b > K_a)
NH₄⁺¹NH₃ + H⁺¹ K_a = 5.6x10^-10
CN^-1HCN + OH^-1
Eq: 1-x x x K_b = 2.0x10^-5 = x²/1 x = [OH^-1] = 4.5x10^-3
pOH = 11.650 pH = 2.35

4. (a) [H⁺¹] = (K_a)([HAc])/[AC^-1] = (1.8x10^-5)(1.00M)/(0.500M) = 3.6x10^-5 pH =4.44

In: 1.00M 0.50M
(b) HAc H⁺¹ + Ac^-1 [H⁺¹] = (K_a)([HAc])/[AC^-1] = (1.8x10^-5)(0.70M)/(0.80M) = 1.6x10^-5
Eq: 0.70M 0.80M pH =4.80

(c) [H⁺¹] = (K_a)([HAc])/[AC^-1] = (1.8x10^-5)(1.30M)/(0.20M) = 1.2x10^-4 pH = 3.93

5. [Pb(OH)₂] = 0.00100/241 = 4.1x10^-6 M Pb(OH)₂ Pb⁺² + 2OH^-1
4.1x10^-6 M 4.1x10^-6 M 8.2x10^-6 M
Ksp = [Pb⁺²][OH^-1]² = (4.1x10^-6 M)(8.2x10^-6 M)² = 2.8x10^-16

6. (a) mol MgCl₂ = 1.00 m = 1.00mol/0.450kg = 2.22m
(b) DT_f = imk_f = (3)(2.22)(-1.86) = -12.4°C
(c) mol ions = 3.00 mol H₂O = 25.0 X_H2O = 25/28 = 0.89
P =(X_H2O)(P_H2O)= (0.89)(600 torr) = 534 torr

7. (a) DH° = +85 kJ (b) DG° = +6 kJ (c) DS° = +265 J/K (d) K = 0.089 (e) 3 gas particles ® 4 gas particles

8.

9. (a) RED: Cu⁺² + 2e^-1 ® Cu° +0.34 V
OX: 2Cl^-1® Cl₂ + 2e^-1 -1.36 V E° = -1.02 V

(b) Q = I x t = (8.0A)(5)(3600s) = 144,000 C = 1.5 F Cu⁺² + 2e^-1® Cu°
1.5F 0.75 mol 48g Cu°