CHEM 130 Exam 4 Review Solutions.pdf - CHEM 130 Exam 4 Review Buehler 555 Monday April 17th 4:30-6:30 pm Useful Equations and Constants = 1 14 = = = A

# CHEM 130 Exam 4 Review Solutions.pdf - CHEM 130 Exam 4...

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Unformatted text preview: CHEM 130 Exam 4 Review Buehler 555 Monday, April 17th, 4:30-6:30 pm Useful Equations and Constants: = 1. + 14 = + = ∙ = A 1.00 L buffer solution is 0.150 M in HC7H5O2 and 0.250 M in LiC7H5O2. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HC7H5O2 is 6.5 × 10-5. . = ∙ . = = . + I 0.25 mol C -0.1 mol E 0.15mol . . = . + = . ∙ = . ∙ = . ⇌ + 0.1 mol 0.15 mol -0.1 mol +0.1 mol +0.1mol 0.25 mol 0.2 mol 0 . = = . 0.1 mol - = . = . = − log(6.5 10 0.136 0.227 )+ = 3.96 2. A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.0500 moles of solid NaOH. Assume no volume change upon the addition of base. The Ka for HF is 3.5 × 10-4. = = . . + ∙ . = . ∙ . = . ⇌ + CHEM 130 Exam 4 Review Buehler 555 Monday, April 17th, 4:30-6:30 pm I 0.375 mol C -0.05 mol E 0.325mol = = = + 0.05 mol 0.375 mol - -0.05 mol +0.05 mol - 0.425 mol - 0 . . . . = . = . = − log(3.5 10 0.283 0.217 )+ = 3.57 3. A titration of a weak acid gives an equivalence point of 9.18. What is the pK a? pKa is half of the equivalence point pKa = 4.59 4. A 100.0 mL sample of 0.18 M HClO4 is titrated with 0.27 M LiOH. Determine the pH of the solution after the addition of 30.0 mL of LiOH. . = ∙ . = . = . ∙ + . = . ⇌ I 0.018 mol 0.0081 mol C -0.0081 mol -0.0081 mol E 0.0099 mol 0 + - - +0.0081 mol - 0.0081 mol - CHEM 130 Exam 4 Review Buehler 555 Monday, April 17th, 4:30-6:30 pm . = = . . = ( = − 5. = . )= − ( . )= . A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 200 mL of KOH. The Ka of HF is 3.5 × 10-4 . = ∙ . = . = . ∙ + . ⇌ I 0.02 mol 0.02 mol C -0.02 mol -0.02 mol E = . 0 + +0.02 mol - 0.02 mol - 0 . = - = . . + ⇌ + I 0.067 M - - - C -x - +x +x E 0.067 - x - x x = = ∙ = => . = = − 6. ( = = . . . . . = )= − ( . + = = − . = . What are the Three Laws of Thermodynamics? = )= . CHEM 130 Exam 4 Review Buehler 555 Monday, April 17th, 4:30-6:30 pm First Law - Also known as Law of Conservation of Energy, states that energy cannot be created or destroyed in an isolated system. Second Law – The entropy of any isolated system always increases. Third Law -The entropy of a system approaches a constant value as the temperature approaches absolute zero. 7. Place the following in order of increasing entropy at 298 K. C2H6(g), Pb(s), Mg(s), CH4(g) Gases have greater entropy than solids. Compounds with higher molecular weights have higher entropy. Mg (s) < Pb (s) < CH4 (g)< C2H6 (g) 8. Calculate ΔS°rxn for the following reaction. The S° for each species is shown below the reaction. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) S°(J/mol∙K) = (4 − (4 192.8 ) 210.8 + (6 ∙ ) 192.8 ∆ 9. 205.2 ∙ ° + (5 210.8 188.8 ) 188.8 ) 205.2 ∙ ∙ = 178.8 / For the following example, identify the sign of ΔH and ΔS H2O(l) → H2O(g) Since you have to add energy (it’s a reactant) to go from a liquid to a gas, the reaction is endothermic, making ΔH positive. Since gases have higher entropy than liquids, ΔS is positive. 10. Consider a reaction that has a positive ΔH and a positive ΔS. Which of the following statements is true? a. This reaction will be nonspontaneous at all temperatures. b. This reaction will be nonspontaneous only at high temperatures. c. This reaction will be spontaneous only at high temperatures. d. This reaction will be spontaneous at all temperatures. e. It is not possible to determine without more information. CHEM 130 Exam 4 Review Buehler 555 Monday, April 17th, 4:30-6:30 pm 11. Above what temperature does the following reaction become nonspontaneous? FeO(s) + CO(g) → CO2(g) + Fe(s) ΔH = -11.0 kJ; ΔS = -17.4 J/K At ∆ = 0, ∆ = ∆ = −11000 = 632.18 −17.4 / 12. Determine ΔG°rxn using the following information. H2(g) + CO(g) → CH2O(g) H°= +1.9 kJ; ΔS°= -109.6 J/K T=25oC ∆ =∆ =( . )−( − ∆ ° + = . . ~ ) − . . 13. Given the following equation, N2O(g) + NO2(g) → 3 NO(g) ΔG°rxn = -23.0 kJ Calculate ΔG°rxn for the following reaction. 3 NO(g) → N2O(g) + NO2(g) Flip the reaction, flip the sign of ΔG°rxn so: ΔG°rxn = 23.0 kJ 14. Calculate the ΔG°rxn at 298K using the following information. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔH°f (kJ/mol) S°(J/mol∙K = ( − ( ) -207.0 91.3 33.2 -285.8 146.0 210.8 240.1 70.0 +( . ) − ∆ +( . ° = . / ) − . ) . ΔG°rxn = ? CHEM 130 Exam 4 Review Buehler 555 Monday, April 17th, 4:30-6:30 pm = ( ) . − ( ) . +( ∙ +( ∙ ° ∆ = = ∙ . ∙ / − ∆ )−( . . ) . ∆ =∆ =( ) . ) ~ . . 15. Use Hess's law to calculate ΔG°rxn using the following information. NO(g) + O(g) → NO2(g) 2 O3(g) → 3 O2(g) ΔG°rxn = ? ΔG°rxn = +489.6 kJ O2(g) → 2 O(g) NO(g) + O3(g) → NO2(g) + O2(g) ΔG°rxn = +463.4 kJ ΔG°rxn = - 199.5 kJ 2 O(g) → O2(g) (2 O3(g) → 3 O2(g) ΔG°rxn = -463.4 kJ ΔG°rxn = +489.6 kJ)(1/2) (NO(g) + O3(g) → NO2(g) + O2(g) ΔG°rxn = - 199.5 kJ)(1/2) _________________________________________________________________ NO(g) + O(g) → NO2(g) ΔG°rxn = -676 kJ 16. Use the free energies of formation given below to calculate the equilibrium constant (K) for the following reaction at 298 K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) K=? ΔG°f (kJ/mol) -110.9 = ( ) − ( 87.6 +( . ) − 51.3 +( . -237.1 ) − . ) . =51 kJ ( )= − = . ∙ =− = . . ( ) CHEM 130 Exam 4 Review Buehler 555 Monday, April 17th, 4:30-6:30 pm 17. Calculate ΔGrxn at 298 K under the conditions shown below for the following reaction. SO3(g) + H2O(g) → H2SO4(l) P(SO3) = 0.20 atm, ∆ = (− . )+ P(H2O) = 0.88 atm . =− . ΔG°= -90.5 kJ ∗ ∙ ( ~ ) . ( . )( . ) 18. In the Haber process, ammonia is synthesized from nitrogen and hydrogen: N2(g) + 3H2(g) → 2NH3(g) ΔG° at 298 K for this reaction is -33.3 kJ/mol. The value of ΔG at 298 K for a reaction mixture that consists of 1.5 atm N2, 1.6 atm H2, and 0.65 atm NH3 is ∆ = (− . )+ . =− . ∙ ∗ ( ~ ( . ) ( . ) ) ( . ) . 19. What element is being reduced in the following redox reaction? MnO4⁻(aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g) MnO4⁻ H2C2O4 O = -2 Mn => (x) + (4)(-2)=-1 Mn = +7 H = +1 Mn = +2 O = -2 C => (2)(+1)+(2)(x)+(4)(-2)=0 C = +3 CO2 Mn+2 O = -2 C => (x)+(2)(-2)=0 C = +4 Mn gained 5 electrons, so it is being reduced 20. What element is being oxidized in the following redox reaction? Cr(OH)4⁻(aq) + ClO⁻(aq) → CrO42-(aq) + Cl⁻(aq) Cr(OH)4⁻ ClO⁻ OH- = -1 Cr => (x) + (4)(-1)=-1 Cr = +3 O = -2 Cl => (x) + (-2) = -1 Cl = +1 ClCl = -1 CrO42O = -2 Cr => (x)+(4)(-2)=2Cr = +6 CHEM 130 Exam 4 Review Buehler 555 Monday, April 17th, 4:30-6:30 pm Cr lost 3 electrons, so it is being oxidized 21. Balance the following redox reaction if it occurs in acidic solution. Sn2+(aq) + NH4+(aq) → Sn(s) + NO3⁻(aq) (2 e- + Sn2+(aq) → Sn(s))4 3H2O (l) + NH4+(aq) → NO3⁻(aq) +10H+(aq) + 8e_______________________________________________________________________________________________________________________________________ 4Sn2+(aq) + 3H2O (l) + NH4+(aq) → 4Sn(s) + NO3⁻(aq) + 10H+(aq) 22. Balance the following redox reaction if it occurs in a basic solution. MnO4⁻(aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g) ...
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