3. The initial rate of a reaction A + B C was measured for several different starting concentrations of A and B, and the results are as follows:

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1 Kinetic Chemistry 1. a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O 3 (g) 3 O 2 (g)? b) If the rate at which O 2 appears, Δ[O 2 ]/Δt, is 6.0 x 10-5 M/s at a particular instant, at what rate is O 3 disappearing at this same time, -Δ[O 3 ]/Δt? 2. Consider a reaction A + B C for which rate = k[a][b] 2. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction. 3. The initial rate of a reaction A + B C was measured for several different starting concentrations of A and B, and the results are as follows:

2 Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [A] = M and [B] = M. Solution: a) Rate= k[a] 2 [B] 0 = k[a] 2 4. The following data were measured for the reaction of nitric oxide with hydrogen: 2 NO(g) + 2 H 2 (g) N 2 (g) + 2 H 2 O(g) (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = M and [H2] = M.

3 The following data were obtained for the gasphase decomposition of nitrogen dioxide at 300 C, NO (g) NO(g) + O (g): Is the reaction first or second order in NO 2? 5. Solution: 6. How much time will be required for a sample of H-3 to lose 75% of its radioactivity? The half-life of tritium is years. Solution: If you lose 75%, then 25% (0.25 as a decimal fraction) remains. (1/2) n = 0.25 n = 2 (remember (1/2) 2 = 1/4 and 1/4 = 0.25) x 2 = years 7. The half-life of Zn-71 is 2.4 minutes. If one had g at the beginning, how many grams would be left after 7.2 minutes has elapsed?

4 Solution: 7.2 / 2.4 = 3 half-lives (1/2) 3 = (the amount remaining after 3 half-lives) g x = 12.5 g remaining 8. The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures

5 Thermochemistry 1. What is the difference between kinetic energy and potential energy? kinetic energy, the energy of motion. The magnitude of the kinetic energy, E k, of an object depends on its mass, m, and speed, v: E K = ½ mv 2 An object has potential energy by virtue of its position relative to other objects. Potential energy is, in essence, the stored energy that arises from the attractions and repulsions an object experiences in relation to other objects. 2. The SI unit for energy is joule. 1 J= 1 kg-m 2 /s 2 3. Hess s law states that if a reaction is carried out in a series of steps, ΔH for the over- all reaction equals the sum of the enthalpy changes for the individual steps. Calculate the enthalpy for this reaction: 2C(s) + H 2 (g) ---> C 2 H 2 (g) ΔH =??? kj 4. Given the following thermochemical equations: C 2 H 2 (g) + (5/2)O 2 (g) ---> 2CO 2 (g) + H 2 O(l) ΔH = kj C(s) + O 2 (g) ---> CO 2 (g) H 2 (g) + (1/2)O 2 (g) ---> H 2 O(l) ΔH = kj ΔH = kj a) first eq: flip it so as to put C 2 H 2 on the product side b) second eq: multiply it by two to get 2C c) third eq: do nothing. We need one H 2 on the reactant side and that's what we have. d) LOOK at what cancels out and sum everything up Answer: kj + (-787 kj) + ( kj) = kj 5. * The standard enthalpy change of a reaction is defined as the enthalpy change when all reactants and products are in their standard states. We de- note a standard enthalpy change as ΔH, where the superscript indicates standard-state conditions. Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl4). Answer: C(graphite) + 2 Cl2(g) CCl4(l) 6. How many kj are required to heat 45.0 g of H2O at 25.0 C and then boil it all away?

6 Solution: Comment: We must do two calculations and then sum the answers. 1) The first calculation uses this equation: q = (mass) (Δt) (Cp) This summarizes the information needed: Δt = 75.0 C The mass = 45.0 g Cp = J g 1 C 1 2) Substituting, we have: q = (45.0 g) (75.0 C) (4.184 J g 1 C 1) q = J = kj 3) The second calculation uses this equation: q = (moles of water) (ΔHvap) This summarizes the information needed: ΔHvap = 40.7 kj/mol The mass = 45.0 g The molar mass of H2O = 18.0 gram/mol 4) Substituting, we obtain: q = (45.0 g / 18.0 g mol 1) (40.7 kj/mol) q = kj 5) Adding: kj kj = 116 kj (to three sig figs) 7. When 26.7 g of H 2 S was burned in excess oxygen, 406 kj was released. What is H for the following equation: 2 H 2 S(g) + 3 O 2 (g) 2 SO 2 (g) + 2 H 2 O(g) H =??? Steps: Answer: kj 8. You are given 12.0 g of ice at C. How much energy is needed to melt the ice completely to water? Solution: 1) The first calculation: q = (mass) (Δt) (Cp) q = (12.0 g) (5.0 C) (2.06 J g 1 C 1) q = J = kj 2) The second calculation: q = (moles of water) (ΔHvap) q = (12.0 g / 18.0 g mol 1) (6.02 kj/mol) q = kj 3) Summing up the values from the two steps gives 4.14 kj, to three significant figures.

7 9. How many kj need to be removed from a g sample of water, initially at 25.0 C, in order to freeze it at 0 C? (Area three, then area two on the time-temperature graph.) Solution: 1) The first calculation: q = (mass) (Δt) (C p ) q = (120.0 g) (25.0 C) (4.184 J g 1 C 1) q = 12,552 J = kj 2) The second calculation: q = (moles of water) (ΔH vap ) q = (120.0 g / 18.0 g mol 1) (6.02 kj/mol) q = kj 3) Summing up the values from the two steps gives 52.8 kj. Review Electrochemistry BASICS Be able to determine what is oxidized or reduced in a chemical reaction with the use of oxidation numbers. Know your Oxidation Rules. Oxidation is Loss and Reduction is gain (Use whatever you like to remember this) The substance that makes it possible for another substance to be oxidized is called either the oxidizing agent or the oxidant. The oxidizing agent acquires electrons from the other substance and so is itself reduced. A reducing agent, or reductant, is a substance that gives up electrons, thereby causing another substance to be reduced. Practice Problems 1. The nickel-cadmium (nicad) battery uses the following redox reaction to generate electricity: Cd(s) + NiO 2 (s) + 2 H 2 O(l) Cd(OH) 2 (s) + Ni(OH) 2 (s) Identify the substances that are oxidized and reduced, and indicate which is the oxidizing agent and which is the reducing agent. Answer: The oxidation state of Cd increases from 0 to +2, and that of Ni decreases from +4 to +2. Thus, the Cd atom is oxidized (loses electrons) and is the reducing agent. The oxidation state of Ni decreases as NiO2 is converted into Ni(OH)2. Thus, NiO2 is reduced (gains electrons) and is the oxidizing agent. 2. Identify the oxidizing and reducing agents in the reaction 2 H 2 O(l) + Al(s) + MnO 4 - (aq) Al(OH) 4 - (aq) + MnO 2 (s) Answer: Al(s) is the reducing agent; MnO4-(aq) is the oxidizing agent.

8 HALF- REACTIONS Reaction: Sn 2+ (aq) + 2 Fe 3+ (aq) Sn 4+ (aq) + 2 Fe 2+ (aq) Split the reaction into an oxidation and reduction reaction (these are Half- Reactions) BALANCING REDOX REACTION IN ACIDIC SOLUTIONS 1. Divide the equation into one oxidation half-reaction and one reduction half- reaction Balance each half-reaction. a. First,balance elements other than H and O. b. Next, balance O atoms by adding H 2 O as needed. c. Then balance H atoms by adding H+ as needed. d. Finally, balance charge by adding e - as needed. At this point, you can check whether the number of electrons in each half-reaction corresponds to the changes in oxidation state. 3. Multiply half-reactions by integers as needed to make the number of electrons lost in the oxidation half-reaction equal the number of electrons gained in the reduction halfreaction. 4. Add half-reactions and, if possible, simplify by canceling species appearing on both sides of the combined equation. 5. Check to make sure that atoms and charges are balanced. BALANCING REDOX REACTION IN ACIDIC SOLUTIONS Follow the above steps, however the equation must be balanced by using OH - and H 2 O rather than H + and H 2 O. A way to approach these problems: 1. balance the half- reactions as if they occurred in acidic solution and the number of H + in each half-reaction and add the same number of OH- to each side of the half-reaction reaction is mass-balanced because you are adding the same thing to both sides. what you are doing is neutralizing the protons to form water (H+ + OH - H2O) on the side containing H+, and the other side ends up with the OH -. The resulting water molecules can be canceled as needed. Practice Problems 3. Balance the following redox reaction: MnO 4- (aq) + C 2 O 4 2- (aq) Mn 2+ (aq) + CO 2 (aq) Answer: 16 H + (aq) + 2 MnO 4- (aq) + 5 C 2 O 4 2- (aq) 2 Mn 2+ (aq) + 8 H 2 O(l) + 10 CO 2 (g)

9 4. Complete and balance this equation by the method of half-reactions: Cr 2 O 7 2- (aq) + Cl - (aq) Cr 3+ (aq) + Cl 2 (g) (acidic solution) Answer: 14 H + (aq) + Cr 2 O 7 2- (aq) + 6 Cl - (aq) 2 Cr 3+ (aq) + 7 H 2 O(l) + 3 Cl 2 (g) 5. Complete and balance the following equations using the method of half-reactions. Both reactions occur in acidic solution. a. Cu(s) + NO 3- (aq) Cu 2+ (aq) + NO 2 (g) b. Mn 2+( aq) + NaBiO 3 (s) Bi 3+ (aq) + MnO 4- (aq) 6. Complete and balance this equation for a redox reaction that takes place in basic solution: CN - (aq) + MnO 4- (aq) CNO - (aq) + MnO 2 (s) (basic solution)

10 7. You have the answers to check your own answers. Follow the step from above if you need help.