Chem I Homework Page, Exam 3 Material

Homework Page with Visible Answers

This page has all of the required homework for the material covered in the third exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.

Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.

These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.

Thermochemistry (Ch. 5)

  1. Answer the following thermochemistry questions.
  2. A 0.88 g gummy bear is burned in a bomb calorimeter. The temperature started at 21.5 °C and leveled off at 24.2 °C. The manufacturer of the bomb calorimeter determined the heat capacity of the calorimeter to be 11.4 kJ/°C. Calculate the heat of combustion per gram of gummy bear.
    Answer
    ΔE = CvΔT = (11.4 kJ/°C)(24.2-21.5 °C) = (11.4 kJ/°C)(2.7 °C) = 30.78 kJ

    (30.78 kJ)/(0.88 g) = 34.98 kJ/g
  3. When a 100 g sample of methane, CH4, is burned in a bomb calorimeter the temperature changes from 21 °C to 31 °C and 2200 J of heat is given off. What is the specific heat of methane?
    Answer
    ΔE = CvΔT ⇒ Cv = ΔE/ΔT = (2200 J)/(10 °C) = 220 J/°C

    Specific heat = Cv/g = (220 J/°C)/(100 g) = 2.20 J/g-°C
  4. The temperature of a 0.85 kg block of copper was found to be 21 °C. The copper was placed in the sun and the temperature increased to 28 °C. Assume the specific heat of copper to be 0.385 J/g-°C and determine the amount of heat absorbed by the block of copper.
    Answer
    ΔH = CpΔT = (850 kg)(0.385 J/g-°C)(7 °C) = 2291 J = 2.3 kJ
  5. A 0.258 g piece of potassium solid is placed into water inside of a coffee cup calorimeter resulting in a vigourous reaction. Assume a total volume of 100 mL for the resulting solution. The temperature of the solution changes from 22 °C to 25.1 °C due to the reaction. How much heat is generated per gram of potassium for this reaction? Assume the density of the solution after the reaction is the density of water and that the heat capacity of the solution and reaction vessel is only due to the water that has a specific heat of 4.184 J/g-°C.
    Answer
    ΔH = CpΔT = (100 g)(4.184 J/g-°C)(3.1 °C) = 1297 J

    (1297 J)/(0.258 g) = 5027 J/g.
  6. A 22.99 g sample of sodium is reacted with 1.0 L of water in a constant pressure calorimeter as follows:

    2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

    The temperature of the water goes from 20 °C to 65 °C. Assume that the volume of the solution remains 1.0 L, but that the density changes to 1.02 g/mL and the specific heat changes to 4.00 J/g-°C. How much heat is generated for every mole of H2(g) produced?
    Answer
    ΔH = CpΔT = (1000 mL)(1.02 g/mL)(4.00 J/g-°C)(45 °C) = 183600 J = 183.6 kJ

    22.99 g Na(s)  )
    1 mole Na
     
    22.99 g Na(s)
     )
    1 mole H2(g)
     
    2 mole Na
     )  = 0.5 mole H2(g)

    (183.6 kJ)/(0.5 mole H2(g)) = 367.2 kJ per mole of H2(g).

  7. A 1.5 kg block of Ni at 100 °C is placed into 500 mL of water that has a temperature of 21 °C. What is the final temperature assuming the specific heat of Ni is 0.44 J/g-°C and the specific heat of water is 4.184 J/g-°C. Hint: the total heat lost is equal to the total heat gained!
    Answer
    Total heat lost by Ni = (1500 g Ni)(0.44 J/g-°C)(100 °C - Tf)

    Total heat gained by H2O = (500 mL)(1g/1mL)(4.184 J/g-°C)(Tf - 21 °C)

    Set the two expressions equal to each other and solve for Tf = 40 °C.
  8. What is the chemical reaction corresponding to the ΔHof for each of the following substances?
  9. Use Appendix C to calculate ΔH° for each of the following reactions?
  10. Calculate ΔH for

    O3(g) + 2NO2(g) → N2O5(g) + O2(g)

    Given the following equations:

    O3(g) + NO2(g) → NO3(g) + O2(g)     ΔH = 10.1 kJ

    N2O5(g) → NO3(g) + NO2(g)     ΔH = 48.3 kJ

    Answer
    Switch the second equation and add to the first one.

    O3(g) + NO2(g) → NO3(g) + O2(g)     ΔH = 10.1 kJ

    NO3(g) + NO2(g) → N2O5(g))     ΔH = -48.3 kJ

    Combining the equations gives:

    O3(g) + 2NO2(g) → N2O5(g) + O2(g)     ΔH = -38.2 kJ

  11. Calculate ΔH for

    4K2O(s) + 3O2(g) → 3K2O2(s) + 2KO2(s)

    Given the following equations:

    K2O(s) + ½O2(g) → K2O2(s)     ΔH = -132.6 kJ

    4KO2(s) → 2K2O(s) + 3O2(g)     ΔH = 411.6 kJ

    2KO2(s) → K2O2(s) + O2(g)     ΔH = 73.2 kJ

    Answer
    Multiply the first equation by 2, switch the second equation, add them to the last equation.

    2K2O(s) + O2(g) → 2K2O2(s)     ΔH = -265.2 kJ

    2K2O(s) + 3O2(g) → 4KO2(s)     ΔH = -411.6 kJ

    2KO2(s) → K2O2(s) + O2(g)     ΔH = 73.2 kJ

    Combining the equations gives:

    4K2O(s) + 3O2(g) → 3K2O2(s) + 2KO2(s)

        ΔH = -265.2 kJ - 411.6 kJ + 73.2 kJ = -603.6 kJ

  12. When solid naphthalene, C10H8, is completely burned 4981 kJ of heat is given off. Assume that H2O(g) is produced and that for CO2, ΔHof = -393.5 kJ/mole. What is ΔHof for naphthalene?
    Answer

    C10H8(s) + O2(g) → 10CO2(g) + 4H2O(g)   ΔH = -4981 kJ

    ΔHrxn = 4ΔHof,H2O + 10ΔHof,CO2 - ΔHof,C10H8

    ΔHof,C10H8 = 4(-241.82 kJ) + 10(-393.5 kJ) + 4981 kJ = 78.7 kJ/mole

  13. A common gasoline additive is octane, C8H18. When octane is burned it produces heat according to the following equation.

    2C8H18(l) + 25O2(g) → 18H2O(g) + 16CO2(g)   ΔH = 1.02 x 104 kJ

  14. How much heat is needed to change 100 g of water from ice at -12 °C to gas at 120 °C?
    Answer

    H2O(s, -12°) → H2O(s, 0°)  

    ΔH = Cp,iceΔT = (100 g)(2.092 kJ/g-°C)(12°) = 2510 J

    H2O(s, 0°) → H2O(l, 0°)  

    ΔH = ΔHfus = (100 g)(1 mole/18 g)(6008 J/mole) = 75100 J

    H2O(l, 0°) → H2O(l, 100°)  

    ΔH = Cp,liqΔT = (100 g)(4.184 kJ/g-°C)(100°) = 41840 J

    H2O(l, 100°) → H2O(g, 100°)  

    ΔH = ΔHvap = (100 g)(1 mole/18 g)(40.67 J/mole) = 226 J

    H2O(g, 100°) → H2O(g, 120°)  

    ΔH = Cp,gasΔT = (100 g)(1.841 kJ/g-°C)(20°) = 3682 J

    ΔH = 2510 J + 75100 J + 41840 J + 226 J + 3682 J = 123358 J = 123 kJ