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Standardized Test Preparation. Chapter 16. Preview. Multiple Choice Short Answer Extended Response. Standardized Test Preparation. Chapter 16. Multiple Choice. 1. Which of the following two conditions will favor a spontaneous reaction?

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  1. Standardized Test Preparation Chapter 16 Preview • Multiple Choice • Short Answer • Extended Response

  2. Standardized Test Preparation Chapter 16 Multiple Choice • 1. Which of the following two conditions will favor a spontaneous reaction? • A.an increase in entropy and a decrease in enthalpy • B.an increase in entropy and an increase in enthalpy • C.a decrease in entropy and a decrease in enthalpy • D.a decrease in entropy and an increase in enthalpy

  3. Standardized Test Preparation Chapter 16 Multiple Choice • 1. Which of the following two conditions will favor a spontaneous reaction? • A. an increase in entropy and a decrease in enthalpy • B.an increase in entropy and an increase in enthalpy • C.a decrease in entropy and a decrease in enthalpy • D.a decrease in entropy and an increase in enthalpy

  4. Standardized Test Preparation Chapter 16 Multiple Choice • 2. The gasification of coal is a method of producing methane by the following reaction. • C(s) + 2H2(g) → CH4(g) ∆H = ? • Find ∆H by using the enthalpy changes in the following combustion reactions: • C(s) + O2(g) → CO2(g) ∆H = –394 kJ • H2(g) + ½O2(g) → H2O(l) ∆H = –286 kJ • CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆H = –891 kJ • A.75 kJ • B.–75 kJ • C.1856 kJ • D.–1856 kJ

  5. Standardized Test Preparation Chapter 16 Multiple Choice • 2. The gasification of coal is a method of producing methane by the following reaction. • C(s) + 2H2(g) → CH4(g) ∆H = ? • Find ∆H by using the enthalpy changes in the following combustion reactions: • C(s) + O2(g) → CO2(g) ∆H = –394 kJ • H2(g) + ½O2(g) → H2O(l) ∆H = –286 kJ • CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆H = –891 kJ • A.75 kJ • B.–75 kJ • C.1856 kJ • D.–1856 kJ

  6. Standardized Test Preparation Chapter 16 Multiple Choice • 3. Two metals of equal mass but different specific heats absorb the same amount of heat. Which metal undergoes the smaller change in temperature? • A.The metal with the higher specific heat. • B.The metal with the lower specific heat. • C.Both undergo the same change in temperature. • D.Cannot determine from the information given.

  7. Standardized Test Preparation Chapter 16 Multiple Choice • 3. Two metals of equal mass but different specific heats absorb the same amount of heat. Which metal undergoes the smaller change in temperature? • A. The metal with the higher specific heat. • B.The metal with the lower specific heat. • C.Both undergo the same change in temperature. • D.Cannot determine from the information given.

  8. Standardized Test Preparation Chapter 16 Multiple Choice • 4. Which of the following processes has a negative ∆S? • A.evaporating 1 mol of a liquid • B.raising the temperature of 1 L of water from 295 K to 350 K • C.freezing of 1 mol of a liquid • D.None of the above.

  9. Standardized Test Preparation Chapter 16 Multiple Choice • 4. Which of the following processes has a negative ∆S? • A.evaporating 1 mol of a liquid • B.raising the temperature of 1 L of water from 295 K to 350 K • C. freezing of 1 mol of a liquid • D.None of the above.

  10. Standardized Test Preparation Chapter 16 Multiple Choice • 5. At a constant pressure, the following reaction is exothermic: 2NO2(g) → N2O4(g). Which of the following statements is true about the reaction (as written)? • A.The reaction is always spontaneous. • B.The reaction is spontaneous at low temperatures, but not at high temperatures. • C.The reaction is spontaneous at high temperatures, but not at low temperatures. • D.The reaction is never spontaneous.

  11. Standardized Test Preparation Chapter 16 Multiple Choice • 5. At a constant pressure, the following reaction is exothermic: 2NO2(g) → N2O4(g). Which of the following statements is true about the reaction (as written)? • A.The reaction is always spontaneous. • B. The reaction is spontaneous at low temperatures, but not at high temperatures. • C.The reaction is spontaneous at high temperatures, but not at low temperatures. • D.The reaction is never spontaneous.

  12. Standardized Test Preparation Chapter 16 Short Answer • 6. For a reaction in which a diatomic molecule dissociates into atoms, what are the signs of ∆S and ∆H? Explain your answer. You may use the reaction N2(g) → 2N(g) as an example of such a reaction.

  13. Standardized Test Preparation Chapter 16 Short Answer • 6. For a reaction in which a diatomic molecule dissociates into atoms, what are the signs of ∆S and ∆H? Explain your answer. You may use the reaction N2(g) → 2N(g) as an example of such a reaction. • Answer:∆S is positive, because the reaction is producing more moles of gas than were originally present. Therefore, there is more randomness in the sample. ∆H is positive, because bonds are being broken and it takes energy to break a bond.

  14. Standardized Test Preparation Chapter 16 Extended Response • 7. For certain molecules, enthalpies of formation can be determined from combustion data. Using the diagram on the next slide, calculate the enthalpy of formation of methane gas, CH4(g), and the enthalpies for two of the combustion reactions shown on that slide.

  15. Standardized Test Preparation Chapter 16 Extended Response • 7. continued • C(s) + O2(g) → CO2(g) ∆H = –393.5 kJ • H2(g) + O2(g) → H2O(l) ∆H = ? • CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆H = ?

  16. Standardized Test Preparation Chapter 16 Extended Response • 7. For certain molecules, enthalpies of formation can be determined from combustion data. Using the diagram on the next slide, calculate the enthalpy of formation of methane gas, CH4(g), and the enthalpies for two of the combustion reactions shown on that slide. • Answer:The formation reaction for CH4(g) is C(s) + 2H2(g) → CH4(g). From the graph, ∆H for H2(g) + O2(g) → H2O(l) is [−965.1 kJ − (−393.5 kJ)]/2 = −285.8 kJ. • The ∆H for CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) can be read directly from the graph and is −890.2 kJ. The enthalpy of formation for CH4(g) is −965.1 kJ + 890.2 kJ = −74.9 kJ.

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