TA practice quiz 5 key - Fall 2010 CH301 Practice Quiz...

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Unformatted text preview: Fall 2010 CH301 Practice Quiz 5—TA Style. Answer Key • QUESTION TYPE: Bomb calorimeter calculation When burned, ethanol (C2H5OH) has a H rxn of -1368 kJ mol 1 . Suppose that 43 grams of pure ethanol are burned in a bomb calorimeter containing 1 L of water. If the temperature increases by 50 C after the reaction occurs, what is the heat capacity of the calorimeter in J / C ? a. 5.7 J C 1 b. 23.2 J C 1 Correct c. 121.7 J C d. 4152.6 J C 1 H mcT Cb T where c is the heat capacity of water and Cb is the heat capacity of the bomb. Rearranging this expression to solve for Cb , H mcT Cb T Since approximately 1 mol of ethanol was used, H -1368 kJ . Thus, 1368 kJ 1000 g 4.18 J C 1 g 1 50 C Cb 50 C Cb 23.18 J C 1 Without a calculator, round 1368 to 1400, 4.18 to 4 and solve this way 1400 kJ 1000 g 4 J C 1 g 1 50 C Cb 50 C 1200 kJ Cb 50 C Cb 24 J C 1 • QUESTION TYPE: Hess’ Law and calculating enthalpy changes Given the following data N 2 H 4 (l ) CH 4 O (l ) CH 2 O ( g ) N 2 ( g ) 3H 2 g H 37 kJ N2H4 N2 l g H2 3H 2 g 2 NH 3 g 2NH 3 g g H 18 kJ H 46 kJ Calculate the enthalpy change of the following reaction. CH 4 Ol CH 2O g H 2 g Correct a. 65 kJ b. 65 kJ c. 101 kJ d. 101 kJ e. 27 kJ Label each given reaction as reaction 1, reaction 2, and reaction 3. If reaction 1, reaction 3, and the reverse of reaction 2 are all added together, then the overall reaction is found. Thus, H 37 kJ 18 kJ 46 kJ H 65 kJ • QUESTION TYPE: Bond energy calculation The combustion of benzene in the presence of oxygen (O2) results in the production of CO2 and H2O. Write the balanced equation and determine the bond energy difference for this reaction. For this question use C=O bond energy of 799 kJ/mol, O=O bond energy of 498 kJ/mol, C‐H bond energy of 410 kJ/mol, O‐H bond energy of 498 kJ/mol, and ALL C‐C bonds (whether they are single or double bonds) as 481 kJ/mol. *Note: using a single bond energy value for C‐C single bonds and C‐C double bonds does not work in all cases. Benzene is a special case where this applies. a. ‐6990 kJ/mol Correct b. 6990 kJ/mol c. 6.99 kJ/mol d. ‐6.99 kJ/mol 2 C6H6 + 15 O2 12 CO2 + 6 H2O Benzene has 6 C‐C bonds and 6 C‐H bonds and we need two moles to react. 12(481) + 12(410) = 10692 kJ/mol O2 has one O=O bond. 15(498) = 7470 kJ/mol CO2 has two C=O bonds. 24(799) = 19176 kJ/mol H2O has two O‐H bonds. 12(498) = 5976 kJ/mol Calculating the bond energy difference by subtracting the products from the reactants gives (10692 kJ/mol + 7470 kJ/mol) – (19176 kJ/mol + 5976 kJ/mol) = ‐6990 kJ/mol • QUESTION TYPE: Work calculation 7 moles of a monatomic ideal gas are heated under constant external pressure and allowed to expand. The gas is heated from 240K to 340K. How much work is done by the gas? a. 5.8 kJ b. –5.8 kJ Correct c. .058 kJ d. ‐.058 kJ w = ‐(nRT) In this case, the number of mols is not changing, only the volume. Therefore, the equation becomes w = ‐nRT Plugging in the numbers, one will find: w = ‐(7 mols)(8.3145 J mol‐1 K‐1)(340‐240) w = ‐(7)(8.3145)(100) w = ‐(700)(8.3145) w ~ ‐5800 J = ‐5.8 kJ • QUESTION TYPE: Sign convention Dr. Laude pours liquid nitrogen on a balloon filled with air. What can you say about the signs of heat, work, entropy change and free energy change as a student examines the result and exclaims, “yes, I can see liquid in the shrunken balloon!!”? a. ‐, ‐, ‐, ‐ b. ‐, +, +, ‐ c. ‐, ‐, ‐, + d. ‐, +, ‐, ‐ Correct e. ‐, +, ‐, + f. +, ‐, +, ‐ g. ‐, +, +, ‐ h. ‐, +, +, + Physically, the process that you see is gaseous air (N2 or O2) becoming liquid air. This must mean that heat is leaving the system (negative), work is being done on the system to convert a gas into a liquid (positive), the system is becoming more ordered because the gas becomes a liquid (negative) and the scream from the student indicates something happened (free energy change is negative.) • QUESTION TYPE: Predicting entropy change Dissolved MgSO4 in water dissociates into its corresponding ions Mg2+ and SO42‐. Does the entropy increase or decrease and approximately how much more “disordered” is the system? a. Increase, four times more disordered b. Decrease, four times more disordered c. Increase, two times more disordered Correct d. Decrease, two times more disordered The entropy of the system increases. This happens because we have gone from one particle of MgSO4 to two particles composed of its associated ions, Mg2+ and SO42‐. The number of particles has doubled therefore the system is roughly twice as disordered as before. • QUESTION TYPE: Temperature dependence of reaction spontaneity A reaction occurs spontaneously at 35 C but not at 200 °C . What are the signs of G , H , and S for this reaction at 35 C and 200 °C , respectively? 35 C 200 °C G H S G H S a. + + + + + + b. ‐ ‐ ‐ + + + c. ‐ ‐ ‐ + ‐ ‐ Correct d. ‐ ‐ + + ‐ ‐ The important relationship to consider here is G H T S If the reaction occurs spontaneously, G is negative. This means that G is negative at 35 C and positive at 200 °C . Since the reaction became less spontaneous as the temperature increased, this means that T S term was making a positive contribution to G . This means that S is negative. However, for the reaction to occur at low T, this must mean that H is also negative. The only choice that meets these criteria is c). • QUESTION TYPE: Theory (laws, state functions, etc.) Consider the following statements. Which ones are true? I. Energy is conserved in an isolated system. II. Work and heat are state functions. III. The entropy of the system must increase for a reaction to happen. IV. The idea of an absolute zero temperature is associated with the third law. a. I and IV only Correct b. II only c. III only d. I and II only e. II and III only f. I, II, and III The phrases “isolated system” and “universe” can be used interchangeably, and so the first statement is simply the first law of thermodynamics and is correct. Heat and work are path dependent and are the famous example of thermodynamics quantities that are NOT state functions (note they are not capitalized, q and w, like state functions are.) The second law states that the entropy of the universe must increase. It is possible for this to happen while either the system or surrounds have a net decrease in entropy. The third law defines an absolute zero for temperature and allows us to talk about what happens to matter as temperature of a system decreases. ...
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This note was uploaded on 10/10/2011 for the course CH 50995 taught by Professor Laude during the Fall '10 term at University of Texas at Austin.

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