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Question 1 10 /^10 pts
CHEM 104 chemistry ii Exam 3 Portage Learning
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Due No due date Points 100 Questions 10 Time Limit 120 Minutes
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Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Table.pdf) This may be helpful throughout the exam. Show the calculation of the molar solubility (mol/L) of Cu(OH) 2 , Ksp of Cu(OH) 2 = 1.6 x 10-^19. Your Answer: Cu (OH) (s) --> Cu ^+2 (aq) + 2OH- (aq) Ksp = [Cu^+2] [OH-] ^ Ksp = (x) (2x) ^2 = 4x^ when x = molar solubility of Cu(OH)2 Ksp = 1.6 x10^-19 = 4x^ solving for x = ((1.6x10^-19)/4) ^(1/3) 3.42 x 10^ (-7) mol/L
Question 2 10 /^10 pts
Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Table.pdf) This may be helpful throughout the exam. Show the calculation of the Ksp of AgCl if the solubility of AgCl is 0.0001921 g/ ml. MW of AgCl = 143. AgCl (s) Ag+^ (aq) + Cl-^ (aq) Your Answer: AgCL(s) --> Ag + (aq) + Cl- (aq) Ksp = [Ag+] [ Cl-] = (x) (x) = x^ where x = molar solubility of AgCl MW of Cu(OH) 2 (s) (s) (s) (2s) 1.6 x 10 -^19 = [s] x [2s]^2 s = 3.42 x 10 -^7 mol/L
Question 3 10 /^10 pts
Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Table.pdf) This may be helpful throughout the exam. Show the calculations and predict if a precipitate of PbI 2 will form if 1.0 x 10-^4 M Pb(NO 3 ) 2 is mixed with 1.0 x 10-^3 M NaI? The Ksp of PbI 2 = 7.1 x 10 -^9. Your Answer: PbI2(s) --> Pb_2 (aq) + 2I - (aq) Q of PbI2 = [Pb^+2][I-] ^ molar solubility of AgCl = ((0.0001921 x 1000)/ (143.35 x 100)) x= 1.34 x 10^(-5) mol/L Ksp of AgCL = x^2 = (1.3x10^(-5)^2) = 1.80 x 10 ^(-10) molar sol of AgCl = (0.0001921 g/100 ml) x (1000 ml / 1 L) x (1 mol / 143.35 g AgCl) molar sol AgCl = 1.34 x 10 -^5 mol / L This means that as 1.34 x 10 -^5 mol / L of AgCl dissolves, 1.34 x 10 - (^5) mol / L of Ag+ (^) and 1.34 x 10- (^5) mol / L of Cl- (^) are formed since one mole of each ion forms from 1 mole of AgCl Ksp = [Ag+] x [Cl-] = (1.34 x 10 -^5 ) x (1.34 x 10 -^5 ) = 1.80 x 10 -^10
Question 4 10 /^10 pts
Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Table.pdf) This may be helpful throughout the exam. State and explain which has greater entropy: CH 3 OH (l) at 65 oC or CH 3 OH (g) at 65 oC. Your Answer: CH3OH (g) at 65C has greater entropy. Since liquids have more orderly structure than gases at same temperature and gases have greater entropy than liquids [Pb ^+2] = 1.0x 10^(-4)M [I-] = 1.0 x10 ^ (-3)M Q of PbI2 = (1x10^-4)(1x10^-3)^ = 1 x 10^- 10 Since Q < Ksp the reaction should go in the forward direction (solution) and PbI2 will not precipitate. PbI 2 (s) Pb+2^ (aq) + 2 I-^1 (aq) Qc = [Pb+2]initial x [I-^1 ]^2 initial =^ [1.0^ x^10 -^4 ]^ x^ [1.0^ x^10 -^3 ]^2 =^ 1.0^ x^10 -^10 Since Qc (1.0 x 10 -^10 ) < Ksp (7.1 x 10 -^9 ) the reaction should go in the FORWARD direction (solution) and PbI 2 WILL NOT precipitate.
Question 6 10 /^10 pts
Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Table.pdf) This may be helpful throughout the exam. 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: = 4 (213.7) + 6(188.7) - (2(229.5) + 7 (205.5)) = 89.5 J/mole entropy for the reaction incerases. there are 2 moles of gase and 7 moles of gas produces 4 moles of gas plus 6 moles of gas so you have a total of 9 moles of gas producing 10 moles of gas, producing more disorder. The entropy is increaed for the reaction which was the prediction. 2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) 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 = [4 (213.7) + 6 (188.7)] – [2 (229.5) + 7 (205.5)] = + 89. J/mole It makes sense that the disorder should increase (ΔS^0 is positive) since 9 moles of gases are forming 10 moles of gases so the system has more disorder (= higher entropy = + ΔS^0 )
Question 8 7.5^ /^10 pts
Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Table.pdf) This may be helpful throughout the exam. Using the values of ΔH^0 and ΔS^0 given, determine the value of ΔG^0 for the reaction below at 25oC. ΔH^0 = 136.4 kJ /mole and ΔS^0 = 120.3 J /mole. Explain whether this reaction is spontaneous as written. C 2 H 6 (g) → C 2 H 4 (g) + H 2 (g) What would be the value of ΔG^0 for the reaction at 900 oC? Explain whether this reaction is spontaneous as written at 900oC. Your Answer: , T = 25+273 = 298 = 136.4 - 298 (0.1203) = 100.55 Kj/mole Since is a large positive number the reaction is nonspontaneous at 25C T = 900 +273 = 1173 = 136.4 - 1173 x (0.1203) = - 4.71 kJ/mole Since for this reaction is negative, the reaction is spontaneous. The reaction is expected to occur at 900C
Question 10 7.5^ /^10 pts
Click this link to access the Periodic Table. (https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Table.pdf) This may be helpful throughout the exam. Using the table and images (A - F) below, predict, and explain by doing a Qc vs Ksp calculation, which tube shows the result when colorless 1 x 10-^7 M AgNO 3 solution was added to a colorless 1 x 10-^7 M NaBr solution. MnS (s) + O 2 (g) → Mn (s) + SO 2 (g) ΔG^0 = - 85.9 kJ S (s) + O 2 (g) → SO 2 (g) ΔG^0 = - 300.1 kJ MnS (s) → Mn (s) + S (s) ΔG^0 = + 214.2 kJ for the overall reaction is = - 85.9 kJ which is a large negative number hence the overall reaction is spontaneous at that temperature 214.2-300.1 = - 85.9 kJ Your Answer: adding the two reactions MnS (s) + O2 (g) -----> Mn(s) + SO2(g) ΔG^0 = - 300.1 kJ (2) S (s) + O 2 (g) → SO 2 (g)
Substance Formula Color K sp Copper (II) carbonate CuCO 3 Blue 1.4 x 10-^10 Lead (II) chromate PbCrO 4 2.8 x 10 -^13 Yellow Silver bromide AgBr White 5.0 x 10 -^13 Your Answer: AgNO3 (aq) + NaBr (aq) --> AgBr (s) + NaNO3 (aq) Q for AgBr = [Ag+] [ Br-] = (1x10^(-7))(1x10(^-7)) = 1 x10^(-14) Ksp for AgBr = 5 x 10 ^(-13) Since Q is < Ksp for AgBr white precipitate will not form (as seen in tube E). the reaction will move in the backward direction.
