Chem II 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 second 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.

Kinetics and Chemical Thermodynamics (Ch. 14 and Ch. 19)

  1. The gaseous reaction of nitric oxide, NO, with chlorine, Cl2, to form NOCl(g) has the following rate expression:

    rate = k[NO]2[Cl2]

    What are the orders of NO and Cl2 and what is the overall order of the reaction?
    Answer

    NO is second order or the order is equal to 2 for nitric oxide.

    Cl2 is first order or the order is 1 for chlorine.

    The overall order for the reaction is equal to two plus one or three. Overall order = 3.

  2. The kinetics of the hydrolysis of sucrose, C12H22O11, in water is complicated by the condition that water is both a reactant and a solvent. this means that the concentration of water appears constant throughout the reaction and it seems to always cancel out in the derivation of the rate law. The pseudo-order (the order discounting the water) is first order in sucrose. The reaction is sucrose plus water yielding glucose plus fructose, but glucose and fructose have the same chemical formula:

    C12H22O11 + H2O → C6H12O6 + C6H12O6

  3. Consider the following rate law: rate = K[A]x
  4. The following reaction was investigated as indicated in the following table. Determine the rate law for this reaction.

    2NO2 + F2 → 2NO2F

    Experiment[NO2] (M)[F2] (M)Initial Rate (M/sec)
    10.150.150.855
    20.300.151.71
    30.300.303.42
    Answer

    In both cases doubling the concentration doubles the rate and so they are both first order. The general rate expression is rate = k[NO2][F2] and the rate constant can be determined from one of the experiments. Using the first experiment gives k = (0.855 M/sec)/(0.15 M)2 = 38 M-1sec-1.

    The rate law would be rate = (38 M-1sec-1)[NO2][F2]

    Here is an illustration of how this can be done mathematically. Divide the rate expression for the second experiment by the rate expression for the first experiment, cancel those factors that are constant and then solve for the exponent (the exponent is the order for that reactant).

    rate2
     
    rate1
     ) =  
    k[NO2]2x[F2]2z
     
    k[NO2]1x[F2]1z
     ) =  
    [NO2]2x
     
    [NO2]1x
     ) =  
    [NO2]2
     
    [NO2]1
     )x 

    1.71
     
    0.855
     ) =  
    0.3
     
    0.15
     )x     
    2
     ) =  
    2
     )x 

    ⇒   ln(2) = (x)ln(2) ⇒   x = 1

  5. A reaction with only one reactant, A, was investigated as indicated in the following table. Determine the rate law for this reaction.
    Experiment[A] (M)Initial Rate (M/sec)
    10.23.04 x 10-5
    20.41.20 x 10-4
    Answer
    rate2
     
    rate1
     ) =  
    k[A]2x
     
    k[A]1x
     ) =  
    [A]2x
     
    [A]1x
     ) =  
    [A]2
     
    [A]1
     )x 

    12.01 x 10-5
     
    3.04 x 10-5
     ) =  
    0.4
     
    0.2
     )x     
    4
     ) =  
    2
     )x 

    ⇒   ln(4) = (x)ln(2) ⇒   x = 2 ⇒   rate = k[A]2

    From experiment #1:   3.04 x 10-5 = k(0.2) ⇒   k = 1.52 x 10-4

    ⇒   rate = (1.52 x 10-4)[A]2

  6. Use the data in the table below to determine the rate expression for the following reaction:

    3ArSO2H → ArSO2Ar + ArSO3H + H2O

    Experiment[ArSO2H] (M)Initial Rate (M/sec)
    10.254.75 x 10-5
    20.36.84 x 10-5
    20.359.31 x 10-5
    Answer

    rate2
     
    rate1
     ) =  
    k[ArSO2H]2x
     
    k[ArSO2H]1x
     ) =  
    [ArSO2H]2x
     
    [ArSO2H]1x
     ) =  
    [ArSO2H]2
     
    [ArSO2H]1
     )x 

    6.84 x 10-5
     
    4.75 x 10-5
     ) =  
    0.3
     
    0.25
     )x     
    1.44
     ) =  
    1.2
     )x 

    ⇒   ln(1.44) = (x)ln(1.2) ⇒   0.346 = x(0.182) ⇒   x = 2

    ⇒   rate = k[ArSO2H]2

    From experiment #1:   4.75 x 10-5 = k(0.25)2 ⇒   k = 7.6 x 10-4

    ⇒   rate = (7.6 x 10-4)[ArSO2H]2

  7. The following reaction was investigated as indicated in the following table. Determine the rate law for this reaction.

    3A + 2B → A3B2

    Experiment[A] (M)[B] (M)Initial Rate (M/sec)
    11.2001.2006.912
    21.5001.20010.800
    31.2003.10017.856
    Answer

    The general rate expression would be rate = k[A]x[B]z

    To get x, divide the rate expression for the second experiment by the rate expression for the first experiment, cancel those factors that are constant and then solve for the exponent. The same is done to find z using experiments #1 and #3. The data from one of the experiments is then used to find k and the overall rate law can then be written.

    rate2
     
    rate1
     ) =  
    k[A]2x[B]2z
     
    k[A]1x[B]1z
     ) =  
    [A]2x
     
    [A]1x
     ) =  
    [A]2
     
    [A]1
     )x 

    10.8
     
    6.912
     ) =  
    1.5
     
    1.2
     )x  ;  
    1.563
     ) =  
    1.25
     )x  ;   ln(1.563) = (x)ln(1.25) ⇒   x = 2

    17.856
     
    6.912
     ) =  
    3.1
     
    1.2
     )z  ;  
    2.58
     ) =  
    2.58
     )z  ;   ln(2.58) = (z)ln(2.58) ⇒   z = 1

    ⇒   rate = k[A]2[B]; ⇒   k = (6.912)/[(1.2)2(1.2)] = 4   (using expt. #1)

    ⇒   rate = (4)[A]2[B]

  8. How long will it take a first order reaction to decrease from an initial concentration of 0.1 M to a concentration of 0.01 M if the rate constant is 1.2 sec-1?
    Answer
    t = -[ln(0.01/0.1)]/(1.2 sec-1) = 1.92 sec
  9. What is the rate constant for a first order reaction where the concentration goes from 2.2 M to 0.1 M in 7.0 sec?
    Answer
    k = -[ln(0.1/2.2)]/(7 sec) = 0.44 sec-1
  10. What is the final concentration after 10 sec for a first order reaction that has an initial concentration of 3.1 M and a rate constant equal to 0.11 sec-1?
    Answer
    [A] = [A]0e-kt = (3.1 M)e-(0.11 sec-1)(10 sec) = 1.03 M
  11. A chemical reaction undergoes a first order reaction where the concentration after 3.2 sec is 20% of the original concentration. How long will it take for the final concentration to be 1% of the original concentration?
    Answer

    Get the rate constant: k = -[ln(0.2/1)]/3.2 sec = 0.503 sec-1

    t = -[ln(0.01/1)]/(0.503 sec-1) = 9.2 sec
  12. (a) Draw and label a potential energy diagram for the following reaction. (b) Suggest an activated complex (transition state) for this reaction. (c) Indicate how a catalyst would change the curve.

    2NO(g) + Cl2(g) → 2NOCl(g)

    Answer
  13. (a) Use the tables in the back of your book to calculate ΔH, ΔS, and ΔG for the following reaction. Indicate if the result for each one makes it more or less spontaneous. (b) Calculate ΔG from ΔH and ΔS at 25 °C. (c) Assume ΔH and ΔS don't change with temperature and find the temperature where the reaction will be at equilibrium, which is the temperature where it changes from nonspontaneous (or spontaneous) to spontaneous (or nonspontaneous).

    2NO(g) + Cl2 → 2NOCl(g)

    Answer

    (a) ΔH = (2 mole)(52.6 kJ/mole) - (2 mole)(90.37 kJ/mole) - 0 = -75.5 kJ

    ΔH is negative which suggests that the reaction goes from high energy to low energy and that according to energy the reaction should be spontaneous.

    ΔS = (2)(264 J/K) - (2)(210.62 J/K) - (1)(222.96 J/K) = -116 J/K

    ΔS is negative which suggests that it is becoming more ordered and that according to entropy the reaction should not be spontaneous.

    ΔG = (2 mole)(66.3 kJ/mole) - (2 mole)(86.71 kJ/mole) - 0 = -41 kJ

    ΔG is negative which suggests that the reaction is spontaneous. In this case the energy has more influence than the entropy.

    (b) ΔG = ΔH - TΔS = (-75.5 kJ) - (298 K)(-0.1162 kJ/K) = -41 kJ

    (c) At equilibrium ΔG = 0 = ΔH - TΔS

    ⇒   T = ΔH/ΔS = (-75.5 kJ)/(-0.116 kJ/K) = 651 K

    Below 651 K the reaction is spontaneous. Above 651 K the reaction is not spontaneous.

  14. (a) Calculate ΔS° for the following reaction: C2H4(g) + H2O(g) → C2H5OH(g). (b) Would this change in entropy support spontaneous or nonspontaneous behavior? (c) Does this make sense when looking at the number of gas particles on each side of the equation?
    Answer

    (a) ΔS = (1)(282.7 J/K) - (1)(219.4 J/K) - (1)(188.83 J/K) = -125.53 J/K

    (b) ΔS is negative which suggests that it is becoming more ordered and that according to entropy the reaction should not be spontaneous.

    (c) There are two moles of gas particles on the reactant side and only one mole of gas particles on the product side. It is going from disorder to order so that a negative change in entropy is correct.

  15. (a) Should the synthesis of acetic acid shown below be spontaneous at 25 °C? (b) At what temperature could it be made to change from spontaneous to nonspontaneous or from nonspontaneous to spontaneous (depending on your answer to part (a))? Assume that ΔH and ΔS remain constant over this temperature range.

    CH3OH(l) + CO(g) → CH3COOH(l)

    Answer

    (a) ΔH = (1 mole)(-487 kJ/mole) - (1 mole)(-238.6 kJ/mole) - (1 mole)(-110.5 kJ/mole) = -138 kJ

    ΔH is negative which suggests that the reaction goes from high energy to low energy and that according to energy the reaction should be spontaneous.

    ΔS = (1)(159.8 J/K) - (1)(126.8 J/K) - (1)(197.9 J/K) = -165 J/K

    ΔS is negative which suggests that it is becoming more ordered and that according to entropy the reaction should not be spontaneous.

    ΔG = (1 mole)(392.4 kJ/mole) - (1 mole)(-166.23 kJ/mole) - (1 mole)(-137.2 kJ/mole) = -89 kJ

    ΔG is negative which suggests that the reaction should be spontaneous. In this case the energy has more influence than the entropy.

    You could also calculate ΔG as follows:

    ΔG = ΔH - TΔS = (-138 kJ) - (298 K)(-0.165 kJ/K) = -89 kJ

    (b) At equilibrium ΔG = 0 = ΔH - TΔS

    ⇒   T = ΔH/ΔS = (-138 kJ)/(-0.165 kJ/K) = 836 K

    Below 836 K the reaction is spontaneous. Above 836 K the reaction is not spontaneous.