OAT General Chemistry Problem Drill 15: Thermochemistry & Thermodynamics Question No. 1 of 10 1. A metal with a high heat capacity is put on a hot plate. What will happen? Question #01 (A) The temperature will increase quickly. (B) The temperature will increase slowly. (C) The temperature will decrease. (D) The temperature will remain the same. (E) You cannot determine. A high heat capacity means it takes a lot of energy added to the matter before temperature begins to change noticeably. A high heat capacity means it takes a lot of energy added to the matter before temperature begins to change noticeably. C. Correct. A high heat capacity means it takes a lot of energy added to the matter before temperature begins to change noticeably. A high heat capacity means it takes a lot of energy added to the matter before temperature begins to change noticeably. A high heat capacity means it takes a lot of energy added to the matter before temperature begins to change noticeably. Heat capacity: The amount of energy that can be absorbed before temperature increases. A lot of energy can be absorbed before the temperature will increases It will take longer to change temperature because it can absorb a lot of energy before changing temperature. Answer: (C)
Question No. 2 of 10 2. For an exothermic process where entropy increases, it is spontaneous at: Question #02 (A) All temperatures (B) Low temperatures only (C) High temperatures only (D) Never (E) It cannot be determined A. Correct. Exothermic and entropy are both favorable for free energy change to be negative (for a spontaneous reaction), at any all temperatures. Exothermic and entropy are both favorable for free energy change to be negative (for a spontaneous reaction), at any all temperatures. C. Incorrect. Exothermic and entropy are both favorable for free energy change to be negative (for a spontaneous reaction), at any all temperatures. Exothermic and entropy are both favorable for free energy change to be negative (for a spontaneous reaction), at any all temperatures. Exothermic and entropy are both favorable for free energy change to be negative (for a spontaneous reaction), at any all temperatures. Exothermic: Energy is released. ΔH = - Entropy Increase: ΔS = + Exothermic is favorable Increase in entropy is favorable If both are favorable, it will be spontaneous at all temperatures. Answer: (A)
Question No. 3 of 10 3. A 200. g of a solid is allowed to melt in 400. g of water. The water temperature decreases from 85.1 C to 30.0 C. What is the heat of fusion for the solid in J/g? Question #03 (A) 115 J/g (B) -115 J/g (C) 460 J/g (D) -460 J/g (E) 0 Use the heat capacity formula. Use the heat capacity formula.. C. Correct. You correctly found the heat of fusion of this solid. Use the heat capacity formula. This cannot be zero, otherwise it won t a solid.
m solid = 200.g m water = 400. g T 1 water = 85.1 C T 2 water = 30.0 C Cp water = 4.18 J/g C H fus solid =? J/g ΔH water = m Cp ΔT ΔH solid = m H fus ΔH water = - ΔH solid (m Cp ΔT) water = (m H fus ) solid ( 30.0 C 85.1 C) = 200 g H fus 400. g 4.18 J. g C 400. g 4.18 J g C 200. g ( 30.0 C 85.1 C) = H fus H fus = 460. J/g Use estimations: 400g / -200g = -2-2 4.18 = -8-8 -55 is going to be the +460 out of the options given Answer: (C) 460 J/g
Question No. 4 of 10 4. Which of the following has an incorrect pairing with the sign of the entropy? Question #04 (A) I 2 (g) I 2 (s) ΔS = + (B) H 2 O(l) H 2 O(s) ΔS = - (C) CH 3 OH(g) + 3/2 O 2 (g) CO(g) + 2H 2 O(l) ΔS = - (D) 2O 2 (g) + 2SO(g) 2SO 3 (g) ΔS = - (E) All of them A. Correct. Moving from gas to solid is a decrease in disorder this is a mis-match in change in entropy sign. Moving from a liquid to a solid is a decrease in disorder. C. Incorrect. Moving from 5/2 gas molecules to 1 gas and 2 liquid molecules is a decrease in disorder. Moving from 4 gas molecules to 2 gas molecules is a decrease in disorder. There is only one that mis-matches. Disorder or randomness + ΔS is an increase in disorder. - ΔS is a decrease in disorder. A: Gas solid is a decrease in disorder. The sign is incorrect. B: Liquid solid is a decrease in disorder. The sign is correct. C: 5/2 of gas molecules 1 gas and 2 liquid molecules is a decrease in disorder. The sign is correct D: 4 gas molecules 2 gas molecules is a decrease in disorder. The sign is correct. Answer: (A)
Question No. 5 of 10 5. What is the ΔG? 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) at 25 C ΔH f S N 2 O 5 11.29 kj/mole 655.3 J/K*mole Question #05 NO 2 33.15 kj/mole 239.9 J/K*mole O 2 0 kj/mole 204.8 J/K*mole (A) 154 kj (B) 43678 kj (C) 84.6 kj (D) -154 kj (E) 0 kj A. Correct. You correctly calculated change in free energy. Be sure to change the entropy quantities to kj to match the enthalpy of fusion and the answer units. C. Incorrect. Check your calculations again. Check your sign. This is not a spontaneous reaction. Without any calculation, you can rule this out, since the reaction indeed occurs, the change of free energy cannot be zero.
2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) T = 25 C+273 = 298 K ΔH f ΔS N 2 O 5 11.29 kj/mole 655.3 J/K*mole NO 2 33.15 kj/mole 239.9 J/K*mole O 2 0 kj/mole 204.8 J/K*mole ΔG =? ΔH = ΔS = S H f prod prod Δ G = ΔH TΔS ΔH = H f S react react ( 4 33.15 + 1 0) ( 2 11.29) ΔH = 110.02 kj ΔS = ( 4 239.9 + 1 204.8) ( 2 655.3) ΔS = -146.2 J/K ( 298K 0. kj ) Δ G = 110.2kJ 1462 ΔG = 154 kj Answer: (A) 154 kj Not spontaneous K
Question No. 6 of 10 6. All of the following reactions illustrate an increase in entropy except: Question #06 (A) N 2 O 4 (g) 2 NO 2 (g) (B) C 6 H 6 (l) C 6 H 6 (g) (C) 2 KClO 3 (s) 3 O 2 (g) + 2 KCl (s) (D) 3 Fe (s) + 2 O 2 (g) Fe 3 O 4 (s) (E) All of them This is an increase in disorder. This is an increase in disorder. C. Incorrect. This is an increase in disorder. D. Correct. This is a decrease in disorder. There is only one correct answer above. Entropy is disorder or randomness. A. 1 gas molecule 2 gas molecules. This is an increase in disorder B. 1 liquid molecule 1 gas molecule. This is an increase in disorder C. 2 solid molecules 3 gas molecules and 2 solid molecules. This is an increase in disorder. D. 3 solid and 2 gas molecules 1 solid molecule. This is a decrease in disorder. Answer: D
Question No. 7 of 10 7. For the following diagram, which explains the difference in length between b & c and between d & e. Question #07 (A) Strength of individual intermolecular forces is stronger between b & c than betwee d & e. (B) Strength of individual intermolecular forces is weaker between b & c than between d & e. (C) There are fewer intermolecular forces broken between b & c than between d & e. (D) There are more intermolecular forces broken between b & c than between d & e. (E) None of them The strength of the individual intermolecular forces is not a function of what state the matter is in, but the type of molecule. The strength of individual intermolecular forces don t change. The strength of the individual intermolecular forces is not a function of what state the matter is in, but the type of molecule. The strength of individual intermolecular forces don t change. C. Correct. There are fewer intermolecular forces broken between b & C than between d & e and therefore it takes less added energy. There are fewer intermolecular forces broken from solid to liquid than from liquid to gas. There is indeed one above that explains correctly. During phase changes, the energy being added to the system is used to break intermolecular forces. Since it s the same molecule during both intervals, the strength of intermolecular forces will be the same. Since the d-e range requires more energy to be added in order to complete the phase change but the strength of the intermolecular forces is the same during both changes, there must be more intermolecular forces broken between d & e. Answer: C
Question No. 8 of 10 8. If 1000 kj of energy is released, how many grams of water are produced? B 2 H 6 (g) + 3 O 2 (g) B 2 O 3 (s) + 2 H 2 O (g) ΔH = -2035 kj Question #08 (A) 8.86 g (B) 73.34 g (C) 17.71 g (D) -17.71 g (E) 0 First convert energy to mole H2O using the balanced equation information and then convert moles to grams using the molar mass. First convert energy to mole H2O using the balanced equation information and then convert moles to grams using the molar mass. C. Correct. You correctly used stoichiometry to calculate the grams of water produced. Use the basic logic, the grams cannot be a negative number. Since there is water produced, the grams cannot be zero. Use the energy change of the reaction along with the mole ratio from the balanced equation. -2035 kj = 2 mole H 2 O (molar mass of H 2 O) 1 mole H 2 O = 18.02 g H 2 O -1000 kj 2 mole H 2 O 18.02 g H 2 O = 17.71 g -2035 kj 1 mole H 2 O H 2 O Use estimations: -1000 / -2000 =.5.5 2 = 1 1 18 = 18 Answer: C. 17.71 g H 2 O
Question No. 9 of 10 9. Find heat of reaction for given equation in the figure. 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) Question #09 N 2 O 5 NO 2 O 2 Δ H f 11.29 kj/mole 33.15 kj/mole 0 kj/mole (A) -110.02 kj (B) 110.02 kj (C) 21.86 kj (D) -21.86 kj (E) 0 Check your sign. B. Correct. You successfully used enthalpy of formation to calculate enthalpy of a reaction. C. Incorrect. Make sure to multiply the enthalpy of formation values by the moles in the balanced equation before calculating enthalpy of reaction. Make sure to multiply the enthalpy of formation values by the moles in the balanced equation before calculating enthalpy of reaction. The heat of reaction is not zero, from the given. To find heat of reaction from heat of formation, find the sum of the products and subtract the sum of the reactants. ΔH rxn = H f prod H f react [( 4 33.15kJ ) + ( 1 0kJ )] [ 2 11. kj ] Δ H rxn = 29 mole mole mole Use estimations since the options in the questions are so different 4 30 = 120 120 + 1 = 121 2 + 10 = 20 121 20 = 100 This is clearly closer to 110.02 than 21.86! Answer: B. 110.02 kj
Question No. 10 of 10 10. For the following, N 2 + 3 H 2 2 NH 3 ΔH = -92 kj ΔS = -199 J/K mole ΔG is between which intervals at 200 K? Question #10 (A) Less than -100 kj (B) Between -100 kj and -40 kj (C) Between -40 kj and +40 kj (D) Between -40 kj and +100 kj (E) Greater than +100 kj Find change in free energy by multiplying temperature times entropy and subtracting that value from enthalpy. B. Correct. You correctly estimated the free energy. C. Incorrect. Find change in free energy by multiplying temperature times entropy and subtracting that value from enthalpy. Find change in free energy by multiplying temperature times entropy and subtracting that value from enthalpy. # This is an over-estimation. Find change in free energy by multiplying temperature times entropy and subtracting that value from enthalpy. Δ G = ΔH TΔS ΔG = 92 kj 200K 0. 199kJ K mole -92+40 = -52 kj Use estimations: 200-0.2 = -40-92 (-40) = -52 Answer: B. between 100 kj and 40 kj