Last Name: First Name: High School Name: Individual Exam 3 Solutions: Kinetics, Electrochemistry, and Thermodynamics

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1 Last Name: First Name: High School Name: Washington University Chemistry Tournament April 2, 2016 Individual Exam 3 Solutions: Kinetics, Electrochemistry, and Thermodynamics Please write your full name and high school at the top of every page. 45 minutes is allowed for this examination. This exam is 6 questions long, and has 13 numbered pages in total. Please check to make sure that your exam is complete, and report to the exam proctors if any pages are missing. Necessary equations and constants, as well as a periodic table, can be found at the end of the exam. Do not write on the scoring sheet on the last page of the exam. Scratch paper will not be permitted during this exam. If you run out of room on the front of the page, you can continue to work on the back of the page, provided that all answers for parts are clearly labelled. Only write work on the backs of pages that correspond to the question on the front of the page. Make sure to circle or box your final answer if appropriate. Correct answers with appropriate work will receive full credit. If work contains reasoning or justification that is partially correct, partial credit may be awarded. Explanations should fully answer the question and provide supporting evidence and logic. These should also be given in complete sentences. Correct answers without reasonable supporting work will not receive credit. If you cannot answer a question or the entire question, it is advised that you do not spend too much time on that question and proceed onto other parts of the exam. No electronics of any kind are allowed during the exam, with the exception of a non-programmable scientific calculator. Cell phones must be turned off, and watches must be removed. A clock will be projected in the exam room. The scoring policy can be found on the WUCT website. Any appeals must be made in writing in the appeals room, Lab Sciences 400. Cheating will not be tolerated on this exam. The cheating policy that has been listed on the WUCT website will be followed. Violators of this policy will be referred to the directors of the competition. 1

2 1. The decomposition of hydrogen peroxide (H 2 O 2 ) to H 2 O and O 2 was studied with two catalysts: lead (IV) oxide and iron (III) oxide. The activation energies were found to be Ea (PbO 2 ) = 136 kj mol and Ea (Fe 2 O 3 ) = 297 kj kj. Without a catalyst, Ea = 300. mol mol a. At 298K, what is the ratio of the rate constant of the reaction on the PbO 2 surface compared to that of the reaction on the Fe 2 O 3 surface? k PbO2 = e Ea(Fe2O3) Ea(PbO2) RT = e = k Fe2 O 3 b. As the concentration of the hydrogen peroxide rises, and the catalyst concentration is held constant, the decomposition becomes zeroth-order with respect to the concentration of H2O2. In 3 complete sentences or fewer, explain why this change occurs on the lines provided below. As substrate concentration increases, active sites on the catalyst become saturated. Thus, adding more substrate will not increase reaction rate, so reaction becomes zeroth order. 2

3 2. Many complex chemical reactions go through multiple elementary steps, as it is often difficult for many molecules to accumulate enough energy to react together in one step. Consider the following overall reaction that goes through two elementary steps: NO 2(g) + CO (g) NO (g) + CO 2(g) Elementary step 1: NO 2 + NO 2 NO + NO 3 Elementary step 2: NO 3 + CO NO 2 + CO 2 a. Suppose the first step is slow. What is a rate law relative to the concentrations of NO2 and CO? If the first step is the slow step, rate equation is Rate = k[no 2 ] 2 since the equation involves only NO 2 and is bimolecular in it. Thus, the reaction is second order with respect to NO 2 and zeroth order with respect to CO. b. Suppose the second step is slow. In this scenario, step 1 is in fast equilibrium. What is the rate law relative to the concentrations of NO2 and CO? If the second step is the slow step, rate equation is Rate = k[no 3 ][CO] since the equation involves is first order with respect to the concentrations of NO3 and CO. However, since NO 3 is an intermediate, we must find a way to express it in terms of reactants. This can be done under the assumption that the first step is in fast equilibrium, which allows us to write an equilibrium statement: K eq,1 = [NO][NO 3] [NO 2 ] 2 [NO 3 ] = K eq,1 [NO 2 ] 2 [NO] Upon substituting back into our rate equation and absorbing the equilibrium constant into a new rate constant k, we obtain Rate = k [NO 2 ] 2 [CO] [NO] (This question continues on the next page.) 3

4 c. Consider a different reaction involving only NO2 and CO as reactants, as shown below. NO2 (g) + CO (g) products [NO2] (M) [CO] (M) Initial Rate (M/hr) Assuming that the rate is only dependent on the concentrations of reactants, determine the rate law and calculate the rate constant with proper units. Doubling [NO 2 ] quadruples the rate, indicating that the reaction is second order with respect to NO2. Tripling carbon monoxide concentration triples the rate, indicating the reaction is first order with respect to carbon monoxide. So the rate reaction is Rate = k[no 2 ] 2 [CO] To find k, we need only substitute using data from one trial: k = Rate 0.3 M s 1 [NO 2 ] 2 = [CO] (0.2 M) 3 = 37.5 M 2 hr 1 d. Suppose we were drawing a graph of the total reaction rate with respect to the concentration of NO2. What should we label our axes to obtain a linear graph? x axis: square of nitrogen dioxide concentration, [NO 2 ] 2 y axis: total reaction rate e. What will the concentration of NO2 be after 2 hours if its initial concentration was measured at 0.4 M? Assume the initial concentration of CO is 1 M and the averaged rate is the same as the initial rate (unless you want to solve a differential equation). First, we need to calculate initial rate: Rate = = 6 M hr 1 The problem then becomes straightforward: [NO 2 ] f = [NO 2 ] i Rate time = 0.40 M 6 M hr 1 2 hr = 0.20 M 60 4

5 3. A scientist is trying to create a galvanic cell under acidic conditions using the following half reactions at 298 K: Au e Au +1.50V Ni e Ni 0.25 V The galvanic cell consists of two 1 L chambers of solutions connected by a salt bridge. a. Write the balanced net reaction equation for this cell. 2 Au Ni 0 2 Au Ni 2+ b. The scientist starts with a solution containing 0.50 M Au 3+ ions in one chamber and another solution containing 0.75 M Ni 2+ ions in another chamber. Using the Nernst equation, calculate the expected voltage (in V) for this cell. E = E 0 RT nf ln(q) = E 0 RT nf ln ([Ni2+ ] 3 [Au 3+ ] 2) = (1.50 V V) J mol 1 K K C mol 1 ln ( ) = 1.748V As Faraday s constant was not given, equivalent expressions without the value of the constant were marked correct. c. Pretend the scientist had mixed up the metals in his lab, and put a solution containing 0.70 M Co 3+ ions instead of Au 3+ ions in one chamber, while all the other conditions remain the same. Only consider the half reaction of Co 3+ + e Co 2+, which has a standard reduction potential of V. What would you expect the voltage of the galvanic cell to be? E = E 0 RT nf ln(q) = E 0 RT nf ln ( [Ni2+ ] [Co 3+ ] 2) = (1.80 V V) J mol 1 K K C mol 1 ln ( ) = V 5

6 4. Potassium permanganate is one of the essential medicines that the WHO recommends for a basic health system. It can be used as an antiseptic, or a disinfectant for treating drinking water. You are trying to make a standard solution of potassium permanganate to further study its properties. a. The chemical formula for potassium permanganate is KMnO 4. i. Oxidation state of manganese in this compound: +7 ii. How does your answer to part (i) explain that potassium permanganate is a strong oxidizing agent? Explain your answer in two complete sentences on the lines provided below. In this oxidation state, Mn 7+ is isoelectronic with argon, making this the highest oxidation state, as it is extremely difficult to remove more electrons from Mn 7+ under normal conditions. As manganese can only decrease in oxidation number, it is easily reduced, so it will be a strong oxidizer. b. You forget what the concentration is of the potassium permanganate solution that you made. Luckily, there is a supply of sodium oxalate in your lab. You weigh out 1.97 grams of sodium oxalate into an Erlenmeyer flask and dilute with acid. You then pour some of the potassium permanganate solution into a buret, and begin to titrate the sodium oxalate with the permanganate solution. i. If permanganate (MnO 4 ) reacts to form Mn 2+, and oxalate (C 2 O 4 2 ) reacts to form carbon dioxide in acidic medium, write the balanced equation of the redox reaction that occurs. 2MnO 4(aq) H (aq) 2 + 5C 2 O 4(aq) 10CO 2(g) + 2Mn 2+ (aq) + 8H 2 O (l) ii. When the titration reaches an equivalence point, ml of permanganate solution has been added. Calculate the concentration of MnO4 (in M) in the original permanganate solution. 2 1 mol C 2 O 4 2 mol MnO 4 1 mol Na 2 C 2 O mol C 2 O 4 = mol MnO g Na 2 C 2 O 4 1 mol Na 2C 2 O g Na 2 C 2 O mol MnO 4 = M L (This question continues on the next page.) 6

7 c. Confident in the concentration of permanganate in your standard solution, you move on to test an unknown mixture containing iron (II) ions. Assume that only the iron (II) ions in the mixture react with the permanganate ions. i. You fully dissolve 4.34 g of the mixture in acid. If it takes ml of the standard permanganate solution to reach the equivalence point in titration, what was the mass percent of iron in the unknown mixture? + 5Fe 2+ + (aq) + 8H (aq) 5Fe 3+ (aq) + Mn 2+ (aq) + 4H 2 O (l) MnO 4(aq) ( L) ( mol MnO 4 ) 1L 5 mol Fe2+ 1 mol MnO g Fe2+ = 1.47 g Fe2+ 1 mol Fe g Fe 2+ (100%) = 33.87% iron(ii) by mass 4.34 g mixture ii. You can also determine the equivalence point of the titration by measuring the redox potential of the analyte solution. By combining the Nernst equations for each redox reaction, along with the table of reduction potentials found at the end of this exam, determine the potential that is expected at the equivalence point. Let reaction 1 be the reduction half-reaction of permanganate to manganese (II) ion and let reaction 2 be the reduction half-reaction of iron (III) to iron (II) E 1 = E 1 RT 5F ln ( [Mn2+ ] [MnO 4 ] ) E 2 = E 2 RT F ln ] ([Fe2+ [Fe 3+ ] ) 1pt for each correct equation (2pts total) At the equivalence point, the potentials of both half-reactions is equal, and let that potential be E. E = E 1 = E 2. Also at the equivalence point, [Mn 2+ ] = 5[Fe 3+ ]; [MnO 4 ] = 5[Fe 2+ ] from reverse reaction 5E = 5E 1 RT F ln ( [Mn2+ ] [MnO 4 ] ) E = E 2 RT F ln ([Fe2+ ] [Fe 3+ ] ) 7

8 6E = (5E 1 + E 2 ) RT F ln ( [Mn2+ ][Fe 2+ ] [MnO 4 ][Fe 3+ ] ) Natural log term equals zero after substituting values, so E equivalence = 5E 1 + E 2 6 ( V) V = = 1.37 V 6 8

9 5. When we add a drop of food coloring to water, why does the color spread spontaneously? Why do the dye molecules in colored water not spontaneously cluster into a single drop? To answer these questions, we can examine entropy as a probabilistic phenomenon. Consider a hypothetical system with two regions with equal volume I and II. Assume the system contains only 4 particles. Each particle has an equal likelihood of being in either region. a. How many possible ways can we have all four particles in region I? ( 4 4 ) = 1 b. How many possible ways can we have one particle in region I and three particles in region II? ( 4 1 ) = 4 c. How many possible ways can we have two particles in region I and two in region II? ( 4 2 ) = 6 d. What is the probability that we will see all molecules in one region or the other? ( 4 0 ) (1 2 )4 + ( 4 4 ) (1 2 )4 = 1, original solution (1/16) did not consider that if no particles are in one region of 8 a box, 4 regions are in the other. Furthermore, since the wording of the question was ambiguous and could feasibly be interpreted as the probability that every particle is in either in region I or region II, we must accept probability = 1 or 100% as well. (This question continues on the next page.) 9

10 e. Now assume that a 1 L container with two equal volume regions (0.5 L for each region) is completely filled with water. If we add a solution of dye to the container such that the concentration of dye particles is 1 mm, what is the probability that one region of the container will be completely colorless (no dye particles in that region)? Assume that there is no volume change of water after addition of dye solution. The particles of dye can move randomly between the two regions of the container = molecules 2 ( 1 2 ) f. In 3 complete sentences or less, explain how the description of entropy as a probabilistic phenomenon justifies that diffusion is a spontaneous process on the lines provided below. Over time, it is much, much more likely to observe the system in a state where the dye molecules are about equally distributed between the two sides than where one side has significantly more dye than the other. 10

11 6. Using the vapor pressure data of ice and liquid water at different temperatures, as shown in the table below, and the Clausius-Clapeyron equation, estimate each of the following terms. a. Heat of sublimation of ice (in kj/mol). ln ( P 2 P 1 ) = H sub R ( 1 T 1 1 T 2 ) H sub = ln ( ) ( ) = 50.9 kj mol 1 b. Heat of vaporization of water (in kj/mol). H vap = ln ( ) ( ) = 44.9 kj mol 1 c. Heat of fusion of ice (in kj/mol). H fus = H sub H vap = = 6.0 kj mol 1 (End of Exam) 11