Chem I Homework Page, Exam 3 Material

Homework Page Without 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 12^th 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.
   • What is the first law of thermodynamics?

   Answer

   Energy can not be created or destroyed, it can change form, but the total amount of energy in the universe remains constant. Energy is conserved.

   • What is a state function?

   Answer

   The magnitudes of state functions are independent of the path. Any property of a system, like temperature, where the value of that property doesn't depend on how it got there is a state function.

   • Write the expression of the first law of thermodynamics in terms of the internal energy and define in words and equations what each term is.

   Answer

   ΔE = q + w where ΔE is the change in internal energy, q is the heat added or taken away, and w is the work done on or by the system.

   q = CΔT and w = - PΔV

   C is the heat capcity, ΔT is the change in temperature, P is the pressure, and ΔV is the change in volume.

   • Define exothermic and endothermic. What are the mathematical signs of the internal energy and enthalpy when a process is exothermic?

   Answer

   Exothermic means that energy is given off from the system and endothermic is when energy is absorbed by the system. Both ΔE and ΔH are negative for exothermic processes.

   • Under what condition is ΔE = q? Under what condition is ΔH = q?

   Answer

   Because w = -PΔV, q = CΔT, and ΔE = q + w, at constant volume (ΔV = 0)
   ΔE = C_vΔT + 0 = q_v, where the subscript v indicates constant volume.

   Under constant pressue ΔH = C_pΔT = q_p.

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 = C_vΔT = (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 = C_vΔT ⇒ C_v = ΔE/ΔT = (2200 J)/(10 °C) = 220 J/°C

   Specific heat = C_v/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 = C_pΔT = (850 g)(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 = C_pΔT = (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 = C_pΔT = (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 - T_f)

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

   Set the two expressions equal to each other and solve for T_f = 40 °C.

   -------------------

   Notice that to calculate the total heat lost by the Ni the final temperature was subtracted from the initial temperature in order to end up with a positive number.

   Another way to look at it is to write the expressions for the ΔH of the Ni and for the ΔH of the water and then make the ΔH for the Ni the negative of the ΔH for the water. That is because the Ni gives off heat and the water absorbs the same amount of heat.

   ΔH_water = (500 mL)(1g/1mL)(4.184 J/g-°C)(T_f - 21 °C)

   ΔH_Ni = (1500 g Ni)(0.44 J/g-°C)(T_f - 100 °C)

   ΔH_water = -ΔH_Ni then results in the same expressions as above.

8. What is the chemical reaction corresponding to the ΔH^o_f for each of the following substances?
   • HClO

   Answer

   ½H₂(g) + ½Cl₂(g) + ½O₂(g) → HClO

   • C₂H₅OH

   Answer

   2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH

   • KMnO₄

   Answer

   K(s) + Mn(s) + 2O₂ → KMnO₄

9. Use Appendix C to calculate ΔH° for each of the following reactions?
   • 4FeO(s) + O₂(g) → 2Fe₂O₃(s)

   Answer

   ΔH_rxn = (2 mole)(-822.16 kJ/mole) - (4 mole)(-297.9 kJ/mole) = -452.72 kJ

   • H₂SO₄(l) + 2NaCl(s) → 2HCl(g) + Na₂SO₄(s)

   Answer

   (-1387.1 kJ) + (2)(-92.30 kJ) - (-814.0 kJ) - (2)(-410.9 kJ) = 64.1 kJ

   • 4HNO₃(g) + 2H₂O(g) → 2N₂H₄(g) + 7O₂(g)

   Answer

   (2)(95.4 kJ) + 0 - 4(-134.63 kJ) - (2)(-241.82 kJ) = 1212 kJ

   • C₂H₅OH(l) + O₂(g) → H₂O(l) + CO₂(g)

   Answer

   (-285.83 kJ) + (-393.5 kJ) - (-277.7 kJ) - 0 = -401.63 kJ

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 one mole of solid naphthalene, C₁₀H₈, is completely burned 4981 kJ of heat is given off. Assume that H₂O(g) is produced and that for CO₂, ΔH^o_f = -393.5 kJ/mole. What is ΔH^o_f for naphthalene?

    Answer

    C₁₀H₈(s) + 12O₂(g) → 10CO₂(g) + 4H₂O(g)   ΔH = -4981 kJ

    ΔH_rxn = 4ΔH^o_f,H2O + 10ΔH^o_f,CO2 - ΔH^o_f,C10H8 - 12ΔH^o_f,O2

    ΔH^o_f,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

    • How much heat is produced per mole of octane?

    Answer

    ΔH = (1.02 x 10⁴ kJ)/2 = 5.1 x 10³ kJ = 5100 kJ

    • How much heat is produced when 10 g of octane is burned?

    Answer

    (

    10 g C8H18

    )

    (

    1 mole C8H18   114 g C8H18

    )

    (

    5100 kJ   1 mole C8H18

    )

    = 447.4 kJ

    • How much energy is produced when 100 mL of octane is burned? The density of octane is 0.703 g/mL.

    Answer

    (

    100 mL C8H18

    )

    (

    0.703 g C8H18   1 mL C8H18

    )

    (

    1 mole C8H18   114 g C8H18

    )

    (

    5100 kJ   1 mole C8H18

    )

    = 3145 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°)  

    ΔH₁ = C_p,iceΔT = (100 g)(2.092 kJ/g-°C)(12°) = 2510 J

    H₂O(s, 0°) → H₂O(l, 0°)  

    ΔH₂ = ΔH_fus = (100 g)(1 mole/18 g)(6008 J/mole) = 75100 J

    H₂O(l, 0°) → H₂O(l, 100°)  

    ΔH₃ = C_p,liqΔT = (100 g)(4.184 kJ/g-°C)(100°) = 41840 J

    H₂O(l, 100°) → H₂O(g, 100°)  

    ΔH₄ = ΔH_vap = (100 g)(1 mole/18 g)(40.67 J/mole) = 226 J

    H₂O(g, 100°) → H₂O(g, 120°)  

    ΔH₅ = C_p,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

15. What is the bond energy of an O-H bond in H₂O(g)?

    Answer

    2H(g) + O(g) → H₂O(g)  (Two O-H bonds formed)

    ΔH_rxn = (-241.82 kJ) - 2(217.94 kJ) - (247.5 kJ) = -925.2 kJ

    ⇒   For one bond ΔH = (-925.2 kJ)/2 = -462.6 kJ

16. What is the bond energy of an N-O bond in N₂O(g)?

    Answer

    2N(g) + O(g) → N₂O(g)  (Two N-O bonds formed)

    ΔH_rxn = (81.6 kJ) - 2(472.7 kJ) - (247.5 kJ) = -1111.3 kJ

    ⇒   For one bond ΔH = (-1111.3 kJ)/2 = -555.65 kJ