AP* Chemistry: Thermochemistry and Thermodynamics PRACTICE QUESTIONS [Version Map]
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1 AP* Chemistry: Thermochemistry and Thermodynamics PRACTICE QUESTIONS [Version Map] A B C D MC MC MC MC MC MC MC MC MC MC Please note there are 4 versions of the PRACTICE MC to deter cheating should you decide to use it as a quiz. Please print only what you need. DO NOT LET THESE Question Sets OR THE SETS OF RATIONALES LEAVE YOUR CLASSROOM. APPROXIMATELY 600 AP CHEMISTRY TEACHERS ARE DEPENDING ON THESE DOCUMENTS REMAINING SECURE. The rationales are written for students, make a class set and number it so that you can pass out the rationales and let the students see what they missed and why they missed it; BUT collect and account for all of the sets once that is accomplished. Your cooperation is most enthusiastically appreciated! 1
2 ID: A AP* Chemistry: Thermochemistry and Thermodynamics PRACTICE QUESTIONS Answer Section MULTIPLE CHOICE 1. ANS: D Realize this is a combustion reaction and heat energy is lost. Vaporization is an endothermic process and requires energy. That means some of the energy lost by the combustion system is subsequently reused by the system to vaporize water if H 2 O(g) is formed. This lowers the total energy yield. As a result, more total energy is lost from the combustion reaction that produces liquid water compared to gaseous water. So, expect that this reaction would yield 2( 44 kj mol 1 ) more energy if liquid water is formed rather than water vapor. Both answers C and D yield more energy, but C didn t factor in that two moles of water are involved. DIF: Hard OBJ: 5.6 TOP: Thermochemistry MSC: 1999 #61 NOT: 31% answered correctly 2. ANS: D A catalyst lowers the activation energy [top of the hump ] thus it lowers any measurement that includes that portion of the graph, so I and II are affected by the presence of a catalyst. DIF: Medium OBJ: 5.18 TOP: Thermodynamics MSC: 1994 #30 NOT: 55% answered correctly 3. ANS: D A negative DS value implies a decrease in disorder, therefore a more ordered product or set of products is/are formed. Look for reactions that produce more condensed states and/or fewer moles of product. Reaction 1: 2 moles solid 4 moles solid...therefore, more disorder moles gas...again, more disorder Reaction 2: 3 moles of ions 1 mole of solid...far more ordered Reaction 3: 2 moles of gas 1 mole of gas...more ordered even though all are gases DIF: Medium OBJ: 5.12 TOP: Thermodynamics MSC: 1994 Q#35 NOT: 54% answered correctly 1
3 ID: A 4. ANS: B They told us the reaction is thermodynamically favorable at 298 K, so we know DG is negative. If favorableness changes with changing temperature, then go directly to DG = DH TDS. For the reaction to be favorable, DG must be negative so, DG is switching signs or becoming positive as the TDS term is increasing with increasing temperatures. Examine the reaction, 4 moles of gas are converted to 2 moles of gas, a more ordered state, so we know that DS is negative, AND we know that T must be a positive term since it is a Kelvin temperature. That means the TDS term is a negative term, so when it is subtracted, it becomes a positive term...increasing the temperature increases the magnitude of the positive term which overtakes the negative DH term. How do we know the DH term is negative? If it were positive, and we subtract the negative TDS term, we would always have a +DG and the reaction would never be thermodynamically favorable. MSC: 1994 #58 NOT: 29% answered correctly 5. ANS: A The evaporation of water is thermodynamically favorable at room temperature and requires thermal energy to do so, therefore + DH and since a liquid is converted into a vapor, +DS. MSC: 2008 #13 NOT: 41% answered correctly 6. ANS: C Clearly combustion reactions (fire!) give off heat energy so DH and since a liquid reactant is forming gaseous products +DS. MSC: 2008 #14 NOT: 34% answered correctly 7. ANS: D The classic definition for activation energy. Remember that it can be decreased by the addition of a catalyst. Why? The catalyst provides a surface for more effective collisions. DIF: Easy OBJ: 5.18 TOP: Thermodynamics MSC: 1999 #4 NOT: 71% answered correctly 8. ANS: B The heat of formation of water is negative, no matter the phase of the water. The most organized or condensed state of water releases the most energy. So, solid water releases most, liquid next and vapor last. Therefore, the magnitude (absolute value) of the H o f is smaller for liquid water than solid water (ice). DIF: Easy OBJ: 5.10 TOP: Thermochemistry MSC: D&S 5th ed. Exam I #51 2
4 ID: A 9. ANS: D 230 K? That s cold! How cold? Below freezing 43 C. Water will freeze, so the change is entropy is negative since a more ordered state if formed. Freezing is exothermic, since heat is lost from the system, so enthalpy is negative. A lot of energy is lost from this system, so G is also negative. DIF: Medium OBJ: 5.3, 5.6, 5.12 TOP: Thermodynamics MSC: D&S 5th ed. Exam III # ANS: D Endothermic processes have + enthalpy. Dissolving generally brings more disorder to the system, especially in the case of ionic compounds since more than one ion is formed per formula unit. DIF: Easy OBJ: 5.12 TOP: Thermodynamics MSC: D&S 5th ed. Exam I #66 3
5 ID: B AP* Chemistry: Thermochemistry and Thermodynamics PRACTICE QUESTIONS Answer Section MULTIPLE CHOICE 1. ANS: D Realize this is a combustion reaction and heat energy is lost. Vaporization is an endothermic process and requires energy. That means some of the energy lost by the combustion system is subsequently reused by the system to vaporize water if H 2 O(g) is formed. This lowers the total energy yield. As a result, more total energy is lost from the combustion reaction that produces liquid water compared to gaseous water. So, expect that this reaction would yield 2( 44 kj mol 1 ) more energy if liquid water is formed rather than water vapor. Both answers C and D yield more energy, but C didn t factor in that two moles of water are involved. DIF: Hard OBJ: 5.6 TOP: Thermochemistry MSC: 1999 #61 NOT: 31% answered correctly 2. ANS: A The evaporation of water is thermodynamically favorable at room temperature and requires thermal energy to do so, therefore + DH and since a liquid is converted into a vapor, +DS. MSC: 2008 #13 NOT: 41% answered correctly 3. ANS: D Endothermic processes have + enthalpy. Dissolving generally brings more disorder to the system, especially in the case of ionic compounds since more than one ion is formed per formula unit. DIF: Easy OBJ: 5.12 TOP: Thermodynamics MSC: D&S 5th ed. Exam I #66 4. ANS: D 230 K? That s cold! How cold? Below freezing 43 C. Water will freeze, so the change is entropy is negative since a more ordered state if formed. Freezing is exothermic, since heat is lost from the system, so enthalpy is negative. A lot of energy is lost from this system, so G is also negative. DIF: Medium OBJ: 5.3, 5.6, 5.12 TOP: Thermodynamics MSC: D&S 5th ed. Exam III #75 1
6 ID: B 5. ANS: B They told us the reaction is thermodynamically favorable at 298 K, so we know DG is negative. If favorableness changes with changing temperature, then go directly to DG = DH TDS. For the reaction to be favorable, DG must be negative so, DG is switching signs or becoming positive as the TDS term is increasing with increasing temperatures. Examine the reaction, 4 moles of gas are converted to 2 moles of gas, a more ordered state, so we know that DS is negative, AND we know that T must be a positive term since it is a Kelvin temperature. That means the TDS term is a negative term, so when it is subtracted, it becomes a positive term...increasing the temperature increases the magnitude of the positive term which overtakes the negative DH term. How do we know the DH term is negative? If it were positive, and we subtract the negative TDS term, we would always have a +DG and the reaction would never be thermodynamically favorable. MSC: 1994 #58 NOT: 29% answered correctly 6. ANS: D The classic definition for activation energy. Remember that it can be decreased by the addition of a catalyst. Why? The catalyst provides a surface for more effective collisions. DIF: Easy OBJ: 5.18 TOP: Thermodynamics MSC: 1999 #4 NOT: 71% answered correctly 7. ANS: D A catalyst lowers the activation energy [top of the hump ] thus it lowers any measurement that includes that portion of the graph, so I and II are affected by the presence of a catalyst. DIF: Medium OBJ: 5.18 TOP: Thermodynamics MSC: 1994 #30 NOT: 55% answered correctly 8. ANS: C Clearly combustion reactions (fire!) give off heat energy so DH and since a liquid reactant is forming gaseous products +DS. MSC: 2008 #14 NOT: 34% answered correctly 2
7 ID: B 9. ANS: D A negative DS value implies a decrease in disorder, therefore a more ordered product or set of products is/are formed. Look for reactions that produce more condensed states and/or fewer moles of product. Reaction 1: 2 moles solid 4 moles solid...therefore, more disorder moles gas...again, more disorder Reaction 2: 3 moles of ions 1 mole of solid...far more ordered Reaction 3: 2 moles of gas 1 mole of gas...more ordered even though all are gases DIF: Medium OBJ: 5.12 TOP: Thermodynamics MSC: 1994 Q#35 NOT: 54% answered correctly 10. ANS: B The heat of formation of water is negative, no matter the phase of the water. The most organized or condensed state of water releases the most energy. So, solid water releases most, liquid next and vapor last. Therefore, the magnitude (absolute value) of the H o f is smaller for liquid water than solid water (ice). DIF: Easy OBJ: 5.10 TOP: Thermochemistry MSC: D&S 5th ed. Exam I #51 3
8 ID: C AP* Chemistry: Thermochemistry and Thermodynamics PRACTICE QUESTIONS Answer Section MULTIPLE CHOICE 1. ANS: B The heat of formation of water is negative, no matter the phase of the water. The most organized or condensed state of water releases the most energy. So, solid water releases most, liquid next and vapor last. Therefore, the magnitude (absolute value) of the H o f is smaller for liquid water than solid water (ice). DIF: Easy OBJ: 5.10 TOP: Thermochemistry MSC: D&S 5th ed. Exam I #51 2. ANS: A The evaporation of water is thermodynamically favorable at room temperature and requires thermal energy to do so, therefore + DH and since a liquid is converted into a vapor, +DS. MSC: 2008 #13 NOT: 41% answered correctly 3. ANS: D The classic definition for activation energy. Remember that it can be decreased by the addition of a catalyst. Why? The catalyst provides a surface for more effective collisions. DIF: Easy OBJ: 5.18 TOP: Thermodynamics MSC: 1999 #4 NOT: 71% answered correctly 4. ANS: C Clearly combustion reactions (fire!) give off heat energy so DH and since a liquid reactant is forming gaseous products +DS. MSC: 2008 #14 NOT: 34% answered correctly 5. ANS: D 230 K? That s cold! How cold? Below freezing 43 C. Water will freeze, so the change is entropy is negative since a more ordered state if formed. Freezing is exothermic, since heat is lost from the system, so enthalpy is negative. A lot of energy is lost from this system, so G is also negative. DIF: Medium OBJ: 5.3, 5.6, 5.12 TOP: Thermodynamics MSC: D&S 5th ed. Exam III #75 6. ANS: D A catalyst lowers the activation energy [top of the hump ] thus it lowers any measurement that includes that portion of the graph, so I and II are affected by the presence of a catalyst. DIF: Medium OBJ: 5.18 TOP: Thermodynamics MSC: 1994 #30 NOT: 55% answered correctly 1
9 ID: C 7. ANS: D A negative DS value implies a decrease in disorder, therefore a more ordered product or set of products is/are formed. Look for reactions that produce more condensed states and/or fewer moles of product. Reaction 1: 2 moles solid 4 moles solid...therefore, more disorder moles gas...again, more disorder Reaction 2: 3 moles of ions 1 mole of solid...far more ordered Reaction 3: 2 moles of gas 1 mole of gas...more ordered even though all are gases DIF: Medium OBJ: 5.12 TOP: Thermodynamics MSC: 1994 Q#35 NOT: 54% answered correctly 8. ANS: D Endothermic processes have + enthalpy. Dissolving generally brings more disorder to the system, especially in the case of ionic compounds since more than one ion is formed per formula unit. DIF: Easy OBJ: 5.12 TOP: Thermodynamics MSC: D&S 5th ed. Exam I #66 9. ANS: D Realize this is a combustion reaction and heat energy is lost. Vaporization is an endothermic process and requires energy. That means some of the energy lost by the combustion system is subsequently reused by the system to vaporize water if H 2 O(g) is formed. This lowers the total energy yield. As a result, more total energy is lost from the combustion reaction that produces liquid water compared to gaseous water. So, expect that this reaction would yield 2( 44 kj mol 1 ) more energy if liquid water is formed rather than water vapor. Both answers C and D yield more energy, but C didn t factor in that two moles of water are involved. DIF: Hard OBJ: 5.6 TOP: Thermochemistry MSC: 1999 #61 NOT: 31% answered correctly 2
10 ID: C 10. ANS: B They told us the reaction is thermodynamically favorable at 298 K, so we know DG is negative. If favorableness changes with changing temperature, then go directly to DG = DH TDS. For the reaction to be favorable, DG must be negative so, DG is switching signs or becoming positive as the TDS term is increasing with increasing temperatures. Examine the reaction, 4 moles of gas are converted to 2 moles of gas, a more ordered state, so we know that DS is negative, AND we know that T must be a positive term since it is a Kelvin temperature. That means the TDS term is a negative term, so when it is subtracted, it becomes a positive term...increasing the temperature increases the magnitude of the positive term which overtakes the negative DH term. How do we know the DH term is negative? If it were positive, and we subtract the negative TDS term, we would always have a +DG and the reaction would never be thermodynamically favorable. MSC: 1994 #58 NOT: 29% answered correctly 3
11 ID: D AP* Chemistry: Thermochemistry and Thermodynamics PRACTICE QUESTIONS Answer Section MULTIPLE CHOICE 1. ANS: A The evaporation of water is thermodynamically favorable at room temperature and requires thermal energy to do so, therefore + DH and since a liquid is converted into a vapor, +DS. MSC: 2008 #13 NOT: 41% answered correctly 2. ANS: C Clearly combustion reactions (fire!) give off heat energy so DH and since a liquid reactant is forming gaseous products +DS. MSC: 2008 #14 NOT: 34% answered correctly 3. ANS: D Realize this is a combustion reaction and heat energy is lost. Vaporization is an endothermic process and requires energy. That means some of the energy lost by the combustion system is subsequently reused by the system to vaporize water if H 2 O(g) is formed. This lowers the total energy yield. As a result, more total energy is lost from the combustion reaction that produces liquid water compared to gaseous water. So, expect that this reaction would yield 2( 44 kj mol 1 ) more energy if liquid water is formed rather than water vapor. Both answers C and D yield more energy, but C didn t factor in that two moles of water are involved. DIF: Hard OBJ: 5.6 TOP: Thermochemistry MSC: 1999 #61 NOT: 31% answered correctly 4. ANS: D The classic definition for activation energy. Remember that it can be decreased by the addition of a catalyst. Why? The catalyst provides a surface for more effective collisions. DIF: Easy OBJ: 5.18 TOP: Thermodynamics MSC: 1999 #4 NOT: 71% answered correctly 1
12 ID: D 5. ANS: B They told us the reaction is thermodynamically favorable at 298 K, so we know DG is negative. If favorableness changes with changing temperature, then go directly to DG = DH TDS. For the reaction to be favorable, DG must be negative so, DG is switching signs or becoming positive as the TDS term is increasing with increasing temperatures. Examine the reaction, 4 moles of gas are converted to 2 moles of gas, a more ordered state, so we know that DS is negative, AND we know that T must be a positive term since it is a Kelvin temperature. That means the TDS term is a negative term, so when it is subtracted, it becomes a positive term...increasing the temperature increases the magnitude of the positive term which overtakes the negative DH term. How do we know the DH term is negative? If it were positive, and we subtract the negative TDS term, we would always have a +DG and the reaction would never be thermodynamically favorable. MSC: 1994 #58 NOT: 29% answered correctly 6. ANS: D A negative DS value implies a decrease in disorder, therefore a more ordered product or set of products is/are formed. Look for reactions that produce more condensed states and/or fewer moles of product. Reaction 1: 2 moles solid 4 moles solid...therefore, more disorder moles gas...again, more disorder Reaction 2: 3 moles of ions 1 mole of solid...far more ordered Reaction 3: 2 moles of gas 1 mole of gas...more ordered even though all are gases DIF: Medium OBJ: 5.12 TOP: Thermodynamics MSC: 1994 Q#35 NOT: 54% answered correctly 7. ANS: D Endothermic processes have + enthalpy. Dissolving generally brings more disorder to the system, especially in the case of ionic compounds since more than one ion is formed per formula unit. DIF: Easy OBJ: 5.12 TOP: Thermodynamics MSC: D&S 5th ed. Exam I #66 8. ANS: D 230 K? That s cold! How cold? Below freezing 43 C. Water will freeze, so the change is entropy is negative since a more ordered state if formed. Freezing is exothermic, since heat is lost from the system, so enthalpy is negative. A lot of energy is lost from this system, so G is also negative. DIF: Medium OBJ: 5.3, 5.6, 5.12 TOP: Thermodynamics MSC: D&S 5th ed. Exam III #75 2
13 ID: D 9. ANS: B The heat of formation of water is negative, no matter the phase of the water. The most organized or condensed state of water releases the most energy. So, solid water releases most, liquid next and vapor last. Therefore, the magnitude (absolute value) of the H o f is smaller for liquid water than solid water (ice). DIF: Easy OBJ: 5.10 TOP: Thermochemistry MSC: D&S 5th ed. Exam I # ANS: D A catalyst lowers the activation energy [top of the hump ] thus it lowers any measurement that includes that portion of the graph, so I and II are affected by the presence of a catalyst. DIF: Medium OBJ: 5.18 TOP: Thermodynamics MSC: 1994 #30 NOT: 55% answered correctly 3
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