CHEM 104 Module 3 Exam: Solubility, Entropy, and Gibbs Free Energy, Exams of Nursing

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CHEM 104 Module 3 Exam Portage Learning

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Portage Learning CHEM 104 Module 3 Exam

Question 1 Show the calculation of the molar solubility (mol/L) of BaF 2 , Ksp of BaF 2 = 1.0 x 10 -^6. Your Answer: Ksp = [Ba+2] [F-]^2 = 1.0x10-^6 1.0x10-^6 = [s] [2s]^2 1.0x10-^6 = 4s^3 s = 6.3x10-^3 mol/L Question 2 Show the calculation of the Ksp of MnS if the solubility of MnS is 0.0001375 g/100 ml. MW of MnS = 87. MnS (s) Mn+2^ (aq) + S-^2 (aq) Your Answer: molar solubility of MnS = (0.0001375 g/100 ml) x (1000 mL) x (1 mol/87.01 g MnS) molar solubility of MnS = 1.580x10-^5 Ksp = [Mn+2] [S-^2 ] Ksp = (1.580x10-^5 ) (1.580x10-^5 ) = 2.496x10-^10

The reaction is from solid H 2 O to liquid H 2 O and liquid has greater entropy because it has more

available motions than solid. Thus, reaction from solid to liquid increases the entropy. Question 6 Determine ΔS^0 for the following reaction using the data given. Explain whether this value agrees with prediction of the ΔS^0 value. ΔS^0 C 2 H 6 = 229.5 J/mole, ΔS^0 O 2 = 205.5 J/mole, ΔS^0 CO 2 = 213.7 J/mole, ΔS^0 H 2 O = 188.7 J/mole 2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) Your Answer: ΔS^0 = ΔS^0 products - ΔS^0 reactants ΔS^0 = [4 ΔS^0 CO 2 + 6 ΔS^0 H 2 O] - [2 ΔS^0 C 2 H 6 + 7 ΔS^0 O 2 ] ΔS^0 = [(4x213.7) + (6x188.7)] - [(2x229.5) + (7x205.5)] ΔS^0 = 89.5 J/mol Reaction has a higher entropy and system is more disorder because ΔS^0 is positive. Question 7 Determine ΔG^0 for the following reaction using the data given and use it to predict the spontaneity of the reaction. ΔGf^0 C 2 H 6 = - 32.9 kJ/mole, ΔGf^0 CO 2 = - 394.4 kJ/mole, ΔGf^0 H 2 O = - 228.6 kJ/mole 2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) Your Answer:

Question 9 The formation of Mn from MnS is nonspontaneous as written in reaction 1 below but can be accomplished by coupling it with the oxidation of S to SO 2 (shown in reaction 2 below). Show the overall spontaneous reaction resulting from the coupling of these two reactions and calculate the ΔG^0 of the overall reaction. (1) MnS (s) → Mn (s) + S (g) ΔG^0 = + 214.2 kJ(2) S (s) + O 2 (g) → SO 2 (g) ΔG^0 =

• 300.1 kJ Your Answer: MnS (s) + O 2 (g) → Mn (s) + SO 2 (g) ΔG^0 = - 85.9 kJ Question 10 Using the table and images (A - F) below, predict, and explain by doing a Qc vs Ksp calculation, whichtube shows the result when colorless 1 x 10 -^6 M AgNO 3 solution was added to colorless 1 x 10-^6 M NaBr solution. A B C D E F

Substance Formula Color K sp Copper (II) carbonate CuCO 3 Blue (^) 1.4 x 10 -^10 Lead (II) chromate PbCrO 4 Yellow 2.8 x 10 -^13 Silver bromide Your Answer: AgBr White 5.0 x 10 -^13 Ksp = [Ag+] [Br-] = 5 .0 x 10-^6 Qc = [Ag+]initial [Br-]initial = [1.0 x 10-^6 ] [ 1.0 x 10 -^6 ]Qc = 1.0x10-^12 Qc (1.0x10-^12 ) < Ksp (5 .0 x 10 -^6 ) thus the reaction should go in the forward direction (solution) andAgBr will not precipitate. Image B shows the result of when colorless AgNO 3 solution was added to colorless NaBr solutionbecause it does not precipitate, its color is still colorless. Qc = (1 x 10-^6 ) x (1.0 x 10 -^6 ) = 1 x 10 -^12 Ksp for AgBr = 5.0 x 10 -^13 Since Qc > Ksp a white precipitate will form = E