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Chem 113 Final Exam Study Guide

by: Caroline Hurlbut

Chem 113 Final Exam Study Guide Chem 113

Marketplace > Colorado State University > Chemistry > Chem 113 > Chem 113 Final Exam Study Guide
Caroline Hurlbut
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This 60-question study guide covers everything we've learned this semester, from thermodynamics and microstates to electrochemical cells.
General Chemistry II
Ingrid Marie Laughman
Study Guide
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This 10 page Study Guide was uploaded by Caroline Hurlbut on Thursday May 5, 2016. The Study Guide belongs to Chem 113 at Colorado State University taught by Ingrid Marie Laughman in Spring 2016. Since its upload, it has received 131 views. For similar materials see General Chemistry II in Chemistry at Colorado State University.


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Date Created: 05/05/16
1. What are the first 3 laws of thermodynamics? 2. Define spontaneity and name characteristics of a spontaneous system. 3. Describe how the number of microstates in a system varies with entropy. 4. List the ways that entropy can be increased. 5. Define elementary reaction. 6. What are the 3 components needed for a reaction to occur? 7. Explain why termolecular reactions are not possible. 8. Define rate determining step. 9. How many transition states (also called activated complexes) are in this energy profile? 10. How is the order of a reaction determined? 11. Given the reaction Cu2S(s)—>2Cu(s) + S(s) and the following ∆S values, calculate the entropy of the reaction. ∆S S(s)=321 J/K/mol ∆S Cu(s)=33.2 J/K/mol ∆S Cu2S(s)=120.9 J/K/mol 12. Explain why the rate for reactants is negative and the rate for products is positive. 13. Write the Arrhenius equation and rearrange it to y-intercept form. 14. What are the units of the rate constant k for zero, first, and second order reactions? 15. The reaction 2N2O5(g)→2N2O4(g) + O2(g) has a first order rate constant of 0.00693 s^-1 at a certain temperature and [N2O5] = 0.0250 M. How long will it take N2O5 to reach 1% of its original concentration? 16. Explain what is meant by ∆Gsys = 0. 17. Write the Gibbs free energy formula. 18. Name some ways that the rate of a reaction can be increased. 19. Why does increasing temperature increase reaction rate? 20. Explain the effect of a catalyst on a reaction and how this effect happens. If a reaction is spontaneous only at high temperatures, what does this tell you about 21. the signs of ∆H and ∆S? 22. What is the purpose of a driving reaction? 23. What happens to the extra energy left over from coupled reactions? 23. Given the coupled reaction and the values for ∆S, calculate the net change in enthalpy for the formation of one mole of lead(II) sulfate: H2SO4(l)—>SO3(g) +H2O(l) ∆H=113. kJ Pb(s) + PbO2(s) + 2SO3(g)—>2PbSO4(s) ∆H=-775. kJ 24. Define equilibrium. 25. Explain the difference between K and Q. 26. Given the following equation and ∆G˚ values, calculate K at 298K: N2H4(g)⁶N2(g) + 2H2(g) ∆G˚ N2H4(g) = 159.4 kJ/mol ∆G˚ N2(g) = ∆G˚H2(g) = 0.0 kJ/mol 27. For CO(g) + H2O(g)⁶CO2(g) + H2(g), K = 0.64. You start with 1.0 mol CO and 1.0 mol CO2 in a 1 L flask. You then add 2.0 mol H2O. Calculate the concentration of H2 at equilibrium. 28. At 1000K, these reactions have the given equilibrium constants: N2O4(g)⁶2NO2(g) K = 1.5 x 10^6 (1/2)N2(g) + O2(g)⁶NO2(g) K = 1.2 x 10^-5 Calculate K for the following reaction at 1000K: N2(g) + 2O2(g)⁶N2O4(g) 29. What is Le Chatelier’s principle? 30. Can the K value of a reaction be changed so that there is a new equilibrium? If so how? 31. Why are only aqueous and gaseous components included in an equilibrium constant expression? 32. This reaction take place at 25˚C: N2O4(g)⁶2NO2(g) and ∆G˚ = 5.4 kJ/mol If the partial pressure of NO2 is 1.0 atm, what partial pressure of N2O4 is required for the reaction to be spontaneous at 25˚C? (Note: R = 0.008314 kJ/K/mol) 33. When is it appropriate to neglect x in equilibrium calculations and why can it be neglected in these situations? 34. Ka HNO2 = 4.0 x 10^-4. You mix HNO2 in water to make a 1.0 M HNO2 solution. Calculate the pH at equilibrium. 35. You have a 0.1 M solution of HClO at a pH of 4.27. Calculate the Ka. 36. The autoionization constant for water is Kw. What is Kw at 25˚C and relate it to conjugate acid-base pairs. 37. Given that Ka C2H2O4 = 5.6 x 10^-2 and Kb NH3 = 1.8 x 10^-5, is a solution of NH4HC2O4 at 25˚C acidic, basic, or neutral? 38. At 50˚C, Kw = 5.48 x 10^-14. At this temperature, what is neutral pH? 39. Write the formulas for conversions between [H3O+]/[OH-] and pH/pOH. 40. What are the Bronsted-Lowry and Lewis definitions for acids and bases? 41. List the important strong acids and bases. 42. Would the salt NH4Cl produce an acidic, basic, or neutral solution? Explain. 43. On a titration curve, the inflection point is the of the reaction. 44. At what point in a titration does pH = pKa? 45. Draw the titration curves for a strong and weak acid titrated with a strong base. 46. What is the common ion effect? 47. When can the Henderson-Hasselbalch equation be used? 48. Define buffer. 49. You have a solution that is 0.20 M HC2H3O2 and 0.50 M C2H3O2- with Ka = 1.8 x 10^-5. Calculate the pH. 50. Calculate the solubility of Ca(OH)2. 51. Calculate the molar solubility of PbF2 in water buffered at 0.05 M H3O+. Ksp PbF2= 3.2 x 10^-8 and Ka HF = 6.8 x 10^-4. 52. List the rules for oxidation states. 53. Identify the oxidizing and reducing agents in the following redox reaction: Cu(s) + 2AgNO3(aq)→2Ag(s) + Cu(NO3)2(aq). 54. Balance the half reactions of following redox reaction using the half reaction method: Cr2O7 2- + I-→Cr3+ + I2. 55. Completely balance the following redox reaction in basic solution using the half reaction method: MnO4- + CN-→CNO- + MnO2. 56. Describe the flow of electrons in an electrochemical cell. 57. What is the purpose of a salt bridge? 58. For a reaction to occur in a cell, what must the signs of n and E˚cell be? 59. Using the standard reduction potentials given, calculate ∆G˚ in kJ for the following half reactions: Br2(l) + 2e-→2Br-(aq) E˚red = 1.065 Cu(s)→Cu2+(aq) + 2e- E˚red = +0.339 60. Calculate the ∆G˚ in kJ for the following cell: Zn2+(aq) + 2e-→Zn(s) E˚ = -0.762 V Cu2+(aq) + 2e-→Cu(s) E˚ = +0.342 V Answers 1. First law: energy is conserved and cannot be created or destroyed. Second law: entropy of the universe (or an isolated system) always increases during spontaneous processes. Third law: as temperature decreases toward absolute zero (0K), the number of microstates moves toward 1 and entropy moves toward 0. 2. Spontaneity is defined as a reaction, once started, will proceed as long as there are reactants. In spontaneous systems, ∆S>0 and ∆G<0. Since microstates refer to the ways to arrange the available energy in a system, the 3. entropy increases as the number of accessible microstates increases and vice versa. 4. Entropy can be increased by increasing temperature, volume, number of particles, or size of molecules. 5. An elementary reaction is a reaction in which one or more reactants react directly to form products in a single reaction step and with a single transition state. 6. For a reaction to occur, the correct collision, geometry, and activation energy must be present. 7. Termolecular reactions are not possible because 3 molecules all having the right collision and geometry to react together is very highly unlikely. 8. The rate determining step of a reaction is the full elementary reaction that proceeds slower than the other(s) and limits the rate of the overall reaction. 9. Each “hump” represents a transition state/activated complex, so for the given reaction there are 2. 10. The order of a reaction is determined by adding up the exponents of the reactants in the rate equation (rate = k[A]^a + [B]^b). 11. Use the formula ∆Srxn = [(#moles x ∆S) products] - [(#moles x ∆S) reactants]. ∆Srxn = [(120.9 x 1)] - [(33.2 x 2) + (32.1 x 1)] ∆Srxn = 120.9 - 98.5 ∆Srxn = 22.4 J/K 12. The rate for reactants is negative because as the reaction proceeds, their concentrations decrease as they form products. Likewise, the concentration of the products increases as the reaction proceeds, so their rate is positive. 13. The Arrhenius equation is k = Ae^(-Ea/RT). In y-intercept form it is ln(k) = (-Ea/R)(1/T) + ln(A). 14. For zero order reactions, k is in M/s. For first order reactions, k is in 1/s. For second order reactions, k is in 1/Ms. 15. Use the first order rate law ln(A) = -kt + ln(A0). ln(0.00025) = -0.000693t + ln(0.025) *A = 0.025(0.01) = 0.00025 t = 665 s *solve for t 16. ∆Gsys = 0 indicates that the system is in chemical equilibrium, which means the rate of the forward reaction is equal to the rate of the reverse reaction. 17. The Gibbs free energy equation is ∆G = ∆H - T∆S. 18. The rate of a reaction can be increased by adding a catalyst, increasing temperature or concentration of the reactants, decreasing volume, changing the rate constant, or a mechanical process (ex. mixing). 19. Increasing temperature increases reaction rate because the number of molecules with kinetic energy is greater than the increases in activation energy. 20. A catalyst increases the rate of a reaction by decreasing the required amount of activation energy without being used up. 20. If a reaction is spontaneous at high temperatures only, we can use the Gibbs free energy equation, ∆G=∆H - T∆S, to determine the signs of ∆H and ∆S. If the reaction is spontaneous at high temperatures, ∆G is negative and T is positive. This means both ∆H and ∆S must be positive in order to have a negative ∆G at a high temperature. 21. The purpose of a driving reaction is to provide enough ∆G to make a separate reaction run. 22. The extra energy is lost to the environment and cannot be recovered to run the reaction more times; only driving reactions can be repeated. 23. Both reactions have SO3 in common, except step two has 2 moles of SO3 on the product side and step one has one mole of SO3 on the reactant side. 2x[H2SO4(l)—>SO3(g) + H2O(l)] [∆H=113. kJ]2x *Cancel SO3 Pb(s) + PbO2(s) + 2SO3(g)—>2PbSO4(s) ∆H=-775. kJ ∆H = 226 kJ - 775 kJ ∆H = -549 kJ 24. Equilibrium is a constant state of movement where the forward rate of reaction equals the reverse rate. It has nothing to do with the amount of reactants or products. 25. While K and Q are both calculated the same way, K represents the constant for a certain reaction at equilibrium. Q represents the reaction quotient at any point in a reaction an tells us the relative amounts of reactants and products. When Q = K, the system is in equilibrium. 26. ∆G˚rxn = [0(1) + 0(2)] - [159.4(1)] = -159.4 kJ/mo*Calculate ∆G˚rxn K = e^[159.4 x 10^3/(8.314 x 298)] *Rearrange ∆Grxn = -RT(lnK) to K = e^(-∆Grxn/RT) and convert kJ/mol to J/mol K = 8.76 x 10^27 27. Make ICE table: CO(g) + H2O(g) ⁶ CO2(g) + H2(g) Initial 1.0 M 2.0 M 1.0 M 0 M Change -x -x +x +x Equilibrium (1.0-x) (2.0-x) (1.0+x) x K = 0.64 = [1.0+x][x] [1.0-x][2.0-x] 0.64 = x + x^2 2 - 3x + x^2 0.64(2 - 3x + x^2) = x + x^2 -0.36x^2 - 2.92x + 1.28 = 0 *standard form ax^2 + bx + c = 0 -(-2.92) ± √(-2.92)^2 - 4(-0.36)(1.28) *quadratic formula 2(-0.36) x = -8.5, x = 0.42 *plug in concentrations to find the correct one [H2] = 0.42 M 28. 2NO2(g)⁶N2O4(g) *reverse first reaction and multiply second N2(g) + 2O2(g)⁶2NO2(g) reaction x2 to cancel NO2 N2(g) + 2O2(g)⁶N2O4(g) *overall reaction ✓ top reaction: K = 1/(1.5 x 10^6) *reverse reaction—>inverse K bottom reaction: K = (1.2 x 10^-5)^2 *multiply reaction x2—>K^2 overall reaction: K = 9.6 x 10^-17 *add reactions—>multiply K values 29. Le Chatelier’s principle states that if a dynamic equilibrium is disturbed by changing the conditions or adding stress to the system, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. 30. Yes; the K value of a reaction can be changed by increasing temperature because adding heat can act like adding a reactant or product, which will affect equilibrium. 31. Solids and pure liquids are not included in equilibrium constant expressions because they are not capable of changing concentration during a reaction, whereas aqueous and gaseous components are. 32. Use the formula ∆Grxn = -RT(lnQ). 5.4 = -(0.008314)(298.15)lnQ *Convert ˚C to K e^(-2.178) = Q *remove ln Q = 0.1132 0.1132 = 1.0/x *set up equilibrium expression x = 8.8 The partial pressure of N2O4 required to make the reaction spontaneous at 25˚C is more than 8.8 atm. 33. We neglect x only for reactants, and when [∆reactant] x 100 < 5%. [reactant] The reason for this is because (typically) for reactions with a very small K value (usually 10^-5 or smaller), the change in concentration of the reactants is small relative to the initial concentration. 34. Make ICE table: HNO2(aq) + H2O(l) ⁶ NO2-(aq) + H3O+(aq) I 1.0 M 0 M 0 M C -x +x +x E 1 - x x x 4.0 x 10^-4 = x^2 (1 - x) Neglect x? 4.0 x 10^-4 = x^2 x = 0.02 test: 0.02 M x 100 = 2% < 5% ✓ neglecting x is ok 1.0 M [H3O+] = x = 0.02 M pH = -log(0.02) = 1.70 35. Make ICE table: HClO(aq) + H2O(l) ⁶ ClO-(aq) + H3O+(aq) M 0 M 1 . M0 0 I C -x +x +x E 0.1 - x x x [H3O+] = x = 10^-4.27 = 5.37 x 10^-5 Ka = (5.37 x 10^-5)^2 (0.1 - 5.37 x 10^-5) Ka = 2.89 x 10^-8 36. At 25˚C, Kw = 1.0 x 10^-14. For any conjugate acid-base pair, Ka x Kb = Kw. 37. Use the formula Kw = Ka x Kb. Solve for new Ka for NH4+ + H2O⁶NH3 + H3O+: Ka = 1.0 x 10^-14 = 5.56 x 10^-10 1.8 x 10^-5 Solve for new Kb for C2H2O4 + H2O⁶HC2H2O4+ + OH-: Kb = 1.0 x 10^-14 = 1.79 x 10^-13 5.6 x 10^-2 Bigger K value “wins” 5.56 x 10^-10 > 1.79 x 10^-13 Since Ka > Kb, NH4HC2O4 is acidic at 25˚C. 38. 5.48 x 10^-14 = [H3O+][OH-] *Kw = [H3O+][OH-] [H3O+] = 2.34 x 10^-7 *[H3O+] = [OH-] always for pure water pH = -log(2.34 x 10^-7) = 6.63 39. pH = -log[H3O+]. pOH = -log[OH-]. [H3O+] = 10^-pH. [OH-] = 10^-pOH. 40. Bronsted-Lowry definition: acids are proton donors and bases are proton acceptors. Lewis definition: acids are electron acceptors and bases are electron donors. 41. Strong acids: HCl, HBr, HI, HNO3, HClO4, H2SO4. Strong bases: LiOH, NaOH, KOH, Ca(OH)2, Ba(OH)2, Sr(OH)2. 42. Since NH4Cl is made from a combination of a weak base, NH3, and a strong acid, HCl, the resulting solution produced from this salt would be acidic. 43. Equivalence point. At this point, all of the analyte has been neutralized. 44. pH = pKa at the half-equivalence point on a titration curve, when half of the analyte has been neutralized. x 2 x 2 x 5 CN- + H2O→CNO- *balance O atoms by adding H2O MnO4-→MnO2 + 2H2O CN- + H2O→CNO- + 2H+ *balance H atoms by adding H+ MnO4- + 4H+→MnO2 + 2H2O CN- + H2O + 2OH-→CNO- + 2H+ + 2OH- *add OH- to both sides so H+ becomes H2O MnO4- + 4H+ + 4OH-→MnO2 + 2H2O + 4OH- CN- + 2OH-→CNO- + H2O + 2e- *balance charges by adding electrons MnO4- + 2H2O + 3e-→MnO2 + 4OH- *multiply oxidation reaction x3 and reduction reaction x2 to cancel electrons x3[CN- + 2OH-→CNO- + H2O + 2e-] x2[MnO4- + 2H2O + 3e-→MnO2 + 4OH-] 3CN- + 2MnO4- + H2O→3CNO- + 2MnO2 + 2OH- *✓ atoms and charges balance 56. In an electrochemical cell, electrons flow from the oxidation side of the cell to the reduction side through the connecting wire. 57. A salt bridge is used to allow ions (not electrons) to flow between the cells and balance the charge on each side. 58. For a reaction to occur, it must be spontaneous, which means that ∆G˚ must be negative. The formula relating ∆G˚ to E˚cell is ∆G˚ = -nFE˚cell. Since n refers to the number of moles of electrons transferred in a reaction, it can never be negative. Taking the negative sign in the equation into account, if n is positive, then E˚cell must also be positive for ∆G˚ to be negative. 59. Since E˚red for Cu is more negative than the E˚red for Br, Cu will be oxidized in this case. Calculate E˚cell: E˚cell = E˚reduction - E˚oxidation = E˚Br - E˚Cu = 1.065 - 0.339 = 0.726 Use ∆G˚ = -nFE˚cell: ∆G˚ = -2(9.6485 x 10^4)(0.726) = -140. kJ *convert J to kJ 1000 *Note: we know that E˚cell must be positive for ∆G˚ to be negative, so both answers here make sense. 60. Since E˚ for Zn is more negative than the E˚ for Cu, Zn will be oxidized in this case. Calculate E˚cell: Zn(s)→Zn2+(aq) + 2e- E˚ = +0.762 *write oxidation reaction Cu2+(aq) + 2e-→Cu(s) E˚ = +0.342 and change sign of E˚ Zn(s) + Cu2+(aq)→Cu(s) + Zn2+(aq) E˚cell = 1.104 *add E˚values together Calculate ∆G˚: ∆G˚ = -2(9.6485 x 10^4)(1.104) = -213. kJ *convert J to kJ 1000


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