Chemistry 1A, Fall 2007 KEY Midterm Exam III November 13, 2007 (90 min, closed book)

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1 Name: SID: Chemistry 1A, Fall 2007 KEY Midterm Exam III November 13, 2007 (90 min, closed book) TA: Section: Please read this first: Write your name and that of your TA on all 8 pages of the exam Test-taking Strategy In order to maximize your score on the exam: Do the questions you know how to do first. Go back and spend more time on the questions you find more challenging. Budget your time carefully -- don't spend too much time on one problem. Show all work for which you want credit and don't forget to include units. Question Page Points Score abcd efghi Total 80 Useful Equations and constants: G = H - T S H = Σ n H f (products) - Σ n H f (reactants) S = Σ ns (products) - Σ ns (reactants) G = Σ n G f (products) - Σ n G f (reactants) S = k B lnw S = q rev /T E = q + w w = - P ext V G = - RTln K G = G + RTln Q G = - nf Єº Є = Єº - (RT/nF) lnq H 1 S ln K = + R T R N 0 = x mol -1 T (K) = T (C) F = 96,485 C / mol 1 V = 1 J / C R = J K -1 mol -1 R = x 10-2 L atm K -1 mol Page 1 of 7

2 1) Evaluating H, S, and G. Consider some reactions you have seen in demonstrations in lecture. The change in free energy under standard conditions is given. a. Place checks in the boxes if the reactions match the statements at the top of the columns. Reaction G rxn (kj) at 25 C Products favored Definitely positive S Definitely exothermic I. N 2 O 4 (g) 2NO 2 (g) (smog) II. C 12 H 22 O 11 (s) 12C (s) + 11H 2 O (l) (dehydration of sucrose, black snake) III. I 2 (s) I 2 (g) b. Determine the equilibrium constant K P at 25 C for Reaction I. Show your work. G = - RTln K 4730 J = - (8.314J/K)(298K)lnK lnk = -1.91, K P = c. Consider Reaction II. In the demonstration, Lonnie had to add concentrated H 2 SO 4 to get the reaction started. The diagram of potential energy vs. reaction progress is shown for the reaction without H 2 SO 4. Draw the diagram for the reaction with H 2 SO 4. H 2 SO 4 acts as a catalyst, lowering E act d. Consider Reaction III. Suppose you have I 2 (s) in equilibrium with I 2 (g) at the sublimation point of iodine. What is G rxn? G rxn = zero at equilibrium, Q = K so G equals zero G = - RTln K G = G + RTln Q Page 2 of 7

3 2) Bond energies. When a hydrogen (H) atom collides with a chlorine (Cl 2 ) molecule, the following reaction occurs with release of energy: H (g) + Cl 2 (g) HCl (g) + Cl (g) a. Which ONE of the following statements is true? Circle the letter of the statement that applies. A. The H - Cl bond is stronger than the Cl - Cl bond. B. The H - Cl bond is weaker than the Cl - Cl bond. C. The relative bond strengths cannot be determined from the information provided. b. Briefly explain your answer. In the reaction the Cl-Cl bond is broken and the H-Cl bond is formed. Since the net energy decreases for the reaction (energy was released) the HCl bond must be stronger (more stable) than the Cl-Cl bond. c. Your friend shows you a section of a biology textbook and asks for your help in understanding the concept. The textbook says..chemical potential energy is stored in chemical bonds. When these bonds are broken, the energy is released. There is also a picture showing a bond breaking. a bond breaking What would you say to your friend based on what you have learned in chemistry? It takes energy to break bonds, so there is an initial cost to any reaction. If the molecules formed in the product are more stable than the molecules in the reactants than a net release of energy will occur for the overall reaction. Page 3 of 7

4 3) Extraction of metals vs. recycling. Metals are often obtained by extraction from natural oxide ores. One process for the extraction of Al from Al 2 O 3 ore is: Al 2 O 3 (s) 2Al (s) + 3/2 O 2 (g) a. Given that Hrxn is kj, how much heat is required to extract 1 mole of aluminum from Al 2 O 3? Since it takes 1675 kj for 2 moles of Al to be extracted, it should take half that amount for 1 mole. heat required for extraction =_837.5 kj/mole Al b. Given S rxn is +313 J/K, make a rough plot of Grxn vs. T. G = H - T S y = b + mx intercept is H, which is positive slope is - S, which is negative c. What is the minimum temperature (in Kelvin) at which the reaction becomes product favored? The reaction will switch between being reactant favored and product favored when G = 0. G = H - T S 0 = H - T S T = H / S (1,675,000 J) (313J/K) = 5351 K d. Currently, carbon is used to lower the heat needed to extract the aluminum. I) Al 2 O 3 (s) 2Al (s) + 3/2 O 2 (g) Hrxn = kj II) Al 2 O 3 (s) + 3/2 C (s, graphite) 2Al (s) + 3/2 CO 2 (g) Hrxn = kj The diagram of Hrxn vs. reaction progress is shown for Reaction I, the Al 2 O 3 decomposition. Draw another diagram for Reaction II, the decomposition of Al 2 O 3 in the presence of carbon. Since the H f of carbon is zero, the level of the reactants shouldn t change on the plot. The E act is likely to change because the reaction is on a new path, but we don t know if it is higher or lower. The H of the products is lower for the second reaction. Page 4 of 7

5 3) Extraction of metals vs. recycling. (continued) Rather than heating the sample to extract the aluminum, currently the reaction is performed using electrolysis of molten Al 2 O 3 with a solid carbon (graphite) electrode. The relevant half reactions are shown below. Al 3+ (molten) + 3e- Al(l) E = V 2O 2- (molten) + C (s, graphite) CO 2 (g) + 4e- E = V The overall reaction: 2 Al 2 O C (s, graphite) 4Al (l) + 3CO 2 (g) e. How many electrons are transferred in the overall reaction shown? 12 f. What is the maximum amount of work (in kj) required to convert the Al 2 O 3 into Al by the overall reaction shown above? E rxn = (-1.66V) + (-0.05V) = -1.71V = -1.71J/C G = - nf Єº = (12 moles e-) (96485 C/mole e-) (-1.71 J/C) = J = 1979 kj g. How much energy in kj would it take to generate 1 mole of Al by the reaction above? Since it takes 1979 kj to extract 4 moles of Al, it will take 494 kj for 1 mole of Al h. Purifying Al from scrap metals involves liquefying the metal, which is heating the metal from 25 C to a liquid at its melting point of 660 C. Calculate the heat needed to purify 1 mole of Al at 1 atm using this process. The molar heat capacity of Al is 24 J/mole K and the heat of fusion is 10.7 kj/mole Step 1) warm the Al from 25 to 660 C, a T of 635 C or 635 K Step 2) melt the solid into liquid Step 1) q = mc p T, we re given moles q = (1 mole Al) (24 J/mole K) (635K) = J = 15.2 kj/mole Al Step 2) 1 mole Al 10.7 kj/mole = 10.7 kj/mole Al Total = = 25.9 kj/mole Al i. In one to two sentences, provide an argument based on energy as to why you should recycle Al. It takes almost 20 times as much energy to extra Al from ore as it does to recycle it from scrap. Recycling saves a lot of energy. (Also, note that one of the byproducts of the reaction is CO 2 which is a concern because of global warming.) Page 5 of 7

6 4) Purification. Nickel metal can be separated from impurities by reacting it with CO. The product is a gas, so it will separate from other solids present. Consider the following reaction between Ni (s) and CO (g). Ni (s) + 4 CO (g) Ni(CO) 4 (g) H rxn = -161 kj/mol a) Given that S rxn = -410 J/mol K, calculate G rxn at 25 C. G = H - T S G = ( J /mol) - (298K)( -410 J/mol K) G = ( J /mol) - ( J/mol) G = J = kj b) Which ONE of the following statements is true about the reaction under standard conditions at 25 C? Circle the letter of the statement that applies. A. The equilibrium constant is less than one. B. The reaction is product favored as written. C. The reactants and products are at equilibrium. D. Nothing will happen c) Suppose that you place 1 atm of Ni(CO) 4 (g) in a closed container with a trace amount of CO (P CO = 0.002atm) at 25 C and allow it to reach equilibrium. Which ONE of the following statements is true? Circle the letter of the statement that applies. A. Ni(CO) 4 (g) will decompose only if the temperature is lowered. B. Since Q > K, some Ni(CO) 4 (g) will decompose to form reactants. C. Since Q < K, Ni(CO) 4 (g) will not decompose to form reactants. D. Nothing will happen. e) Given that H = +430 kj/mol for Ni (s) Ni (g), determine the bond enthalpy of the Ni-CO bond. Assume that the bonding between the C and O in the CO molecule does not change when it combines with Ni. This is a Hess s Law problem. You are given two reactions and need to find a third. 1) Ni (s) Ni (g) H = +430 kj/mol 2) Ni (g) + 4 CO (g) Ni(CO) 4 (g) H rxn =???? 3) Ni (s) + 4 CO (g) Ni(CO) 4 (g) H rxn = -161 kj/mol H 1 + H 2 = H kj/mol + H 2 = -161 kj/mol H 2 = -591 kj/ mole of Ni (or 4 moles of CO) for the bond enthalpy of one of the Ni-CO bonds, divide by 4 591/4 = 148 kj Page 6 of 7

7 5) Combustion reactions. Consider the combustion reactions of hydrocarbon molecules (on the left) with O 2 (g). All four reactions are product favored as written under standard conditions at 25 C. H rxn at 25 C CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) 891 kj/rxn CH 3 OH (l) O 2 (g) CO 2 (g) + 2 H 2 O (l) 727 kj/rxn H 2 CO (l) + O 2 (g) CO 2 (g) + H 2 O (l) 572 kj/rxn HCOOH (l) O 2 (g) CO 2 (g) + H 2 O (l) 255 kj/rxn a. Energy from burning these fuels in O 2 (g) can be used to power a generator but the downside is the release of the greenhouse gas carbon dioxide. Which fuel releases the most CO 2 per gram of fuel burned? Circle the best answer. A) CH 4 B) CH 3 OH C) H 2 CO D) HCOOH E) CO 2 b. Which of the following would result in a faster rate for the forward reaction for any of these combustions? A) Doubling the pressure B) Raising the temperature C) Using a catalyst D) all of the above c. For the combustion of one mole of methane (CH 4 ) 891 kj of heat is released. Can all of the heat be converted to work? Explain why or why not? No. Several explanations will apply here. - Some heat is likely lost when you try to convert it to work. - S surroundings is increasing due to heat lost by the system - you must always consider the T S of the system to see how much available work can be done Page 7 of 7