SAT Chemistry - Problem Drill 22: Thermodynamics No. 1 of 10 1. A metal with a high heat capacity is placed on top of a hot plate that is turned on. What will happen to the temperature of the piece of metal? (A) The temperature will increase quickly. (B) The temperature will increase slowly. (C) The temperature will decrease quickly. (D) The temperature will decrease slowly. (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. B. Correct. A high heat capacity means it takes a lot of energy added to the matter before temperature begins to change noticeably. Putting an object on a hot plate will always cause the temperature to increase. Putting an object on a hot plate will always cause the temperature to increase. E. Incorrect. You can determine the answer to this question from the given information 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. Heat Capacity is an important concept. You will see it again in your studies! The correct answer is (B).
No. 2 of 10 2. Determine if each of the following statements is true or false. If the second statement is a correct explanation for the first statement, mark CE as well. I. Water takes a lot of energy to change temperature. II. Water has a low heat capacity. (A) False, False (B) True, False (C) False, True (D) True, True (E) True, True, CE A. Incorrect! Water does take a lot of energy to change temperature. B. Correct! Water does take a lot of energy to change temperature because it has a high heat capacity. C. Incorrect! Water does take a lot of energy to change temperature because it has a high heat capacity. D. Incorrect! Water does take a lot of energy to change temperature. E. Incorrect! Water does take a lot of energy to change temperature. Heat capacity is the amount of energy needed for 1 gram of a substance to increase temperature by 1 degree Statement I is true water takes a lot of energy to change temperature. Statement II is false water s heat capacity is high. The correct answer is (B).
No. 3 of 10 3. 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? (A) 115 J/g (B) -115 J/g (C) 460 J/g (D) -460 J/g (E) 4.18 J/g Check you calculations again. Check you calculations again. C. Correct. You correctly found the heat of fusion of this solid. Be sure to use T 2 -T 1 for change in temperature to get the correct sign for heat of fusion. E. Incorrect That is the specific heat capacity of water necessary for these calculations, but it is not the heat of fusion of the solid. m solid = 200.g = mass of the solid m water = 400. g = mass of the water 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 Please take the time to truly understand this problem and its solution. It uses several important concepts! The correct answer is (C).
No. 4 of 10 4. Which of the following has an incorrect pairing with the sign of the entropy? (A) I 2 (g) I 2 (s) S = + (B) H 2 O(l) H 2 O(s) S = - (C) CH 3 OH(g) + 3/2 O 2 (g) CO(g) + 2H 2 O(l) S = - (D) 2O 2 (g) + 2SO(g) 2SO 3 (g) S = - (E) None of the choices 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. 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. E. Incorrect! Look closely there is an equation that is correctly paired with entropy sign. Disorder or randomness. + S is an increase in disorder. - S 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. The correct answer is (A).
No. 5 of 10 5. For 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) at 25 C, the following are relevant. What is the G? H f S 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 (A) 154 kj (B) 43678 kj (C) 84.6 kj (D)-154 kj (E) -43 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. Check your calculations again. Check your sign. E. Incorrect. Be sure to multiply each change in enthalpy by the coefficient in the balanced equation.
2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) T = 25 C+273 = 298 K H f S 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 G =? ΔH = H f prod H f react ΔS = S prod Δ G = ΔH TΔS ΔH = 4 33.15 + 1 0 2 11.29 ( ) ( ) H = 110.02 kj ΔS = 4 239.9 + 1 204.8 2 655.3 ( ) ( ) S = -146.2 J/K ( 298K 0. kj ) Δ G = 110.2kJ 1462 K G = 154 kj Not spontaneous The correct answer is (A). S react
No. 6 of 10 6. All of the following reactions illustrate an increase in entropy EXCEPT: (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) None of the above This is an increase in disorder. This is an increase in disorder. This is an increase in disorder. D. Correct. This is a decrease in disorder. E. Incorrect. Look closely There is an equation that has a decrease in entropy. 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. The correct answer is (D).
No. 7 of 10 7. For the following diagram, select the answer that best which explains the difference in length between b & c and between d & e. (A) The strength of individual intermolecular forces is stronger between b & c than between d & e. (B) The 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) There are more molecules being boiled than being melted. 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 doesn 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 doesn 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. E. Incorrect! The number of molecules is the same during both changes in state. 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. The correct answer is (C).
No. 8 of 10 8. If 1000 kj of energy is released, how many grams of water are produced for the following reaction? B 2 H 6 (g) + 3 O 2 (g) B 2 O 3 (s) + 2 H 2 O (g) H = -2035 kj (A) 8.86 g (B) 73.34 g (C) -73.34 g (D) 17.71 g (E) -17.71 g First convert energy to mole H 2 O using the balanced equation information and then convert moles to grams using the molar mass. First convert energy to mole H 2 O using the balanced equation information and then convert moles to grams using the molar mass. First convert energy to mole H 2 O using the balanced equation information and then convert moles to grams using the molar mass. D. Correct. You correctly used stoichiometry to calculate the grams of water produced. E. Incorrect. Check your sign. 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 You could solve this problem by estimation as well as by calculation. First, recognize that 1000 is about ½ of 2035. Second, recognize that there are 2 moles of H 2 O produced. Third, (1/2)*2 =1 mole. Therefore, look for an answer that is slightly less than 18.02 grams that corresponds to one mole. The correct answer is (D).
No. 9 of 10 9. For 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) find heat of reaction if: N 2 O 5 NO 2 O 2 ΔH 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) None of these options is correct. Check your sign. B. Correct. You successfully used enthalpy of formation to calculate enthalpy of a reaction. 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. E. Incorrect! Check your calculations again one of these choices is correct. 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 = 110 kj mole Although, not asked in the question, recognize that there will be a very large increase in entropy, because you are starting with 2 moles of gas and ending with 5 moles of gas. The correct answer is (B).
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 interval at 200 K? (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) More 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. 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. E. Incorrect! 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 The correct answer is (B).