Solutions to: Thermodynamics HomeworkProblem Set S.E. Van Bramer 1/11/97

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Miles Hardy
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1 Solutions to: Thermodynamics HomeworkProblem Set S.E. Van Bramer 1/11/97 1. Nitrogen Oxides are very important species in atmospheric chemistry. They are critical for a number of reactions that contribute to photochemical smog. The major anthropogenic (man made) source of NO is from the following reaction: N 2 + O 2 <--> 2 NO In this problem you will study the temperature dependence of this reaction to learn how the emission of NO may be reduced. Answer the following questions (assume that H is constant at each temperature): a. What is DH rxn, DS rxn, DG rxn, and K (We will assume that H rxn and S rxn do not change with temperature. This is not strictly true and in more advanced courses you will learn how to account for these changes.) at 25, 1000, 2000, and 3000 o C Constants: 10 3 R K H rxn S rxn ( 2 ) 90.4 ( 2 ) ( 1 ) 0 ( 1 ) ( 1 ) 0 + ( 1 ) i. 25 C T ( )K H rxn = S rxn = 24.59K 1 G rxn H rxn T S rxn G rxn = G rxn = RT lnk ( eq ) G rxn K 25 exp K 25 = RT hwkc17_a.xmcd S.E. Van Bramer 1/3/2017, 1

2 ii C T ( )K H rxn = S rxn = 24.59K 1 G rxn H rxn T S rxn G rxn = G rxn K 1000 exp K 1000 = RT iii C T ( )K H rxn = S rxn = 24.59K 1 G rxn H rxn T S rxn G rxn = G rxn K 2000 exp K 2000 = RT iv C T ( )K H rxn = S rxn = 24.59K 1 G rxn H rxn T S rxn G rxn = G rxn K 3000 exp K 3000 = RT hwkc17_a.xmcd S.E. Van Bramer 1/3/2017, 2

3 b. The atmospheric pressure of N2 is 0.80 atm and O2 is 0.20 atm. Determine the equilibrium pressure of NO at each temperature. K= 2 P N2 P O2 = KP N2 P O2 P N2 0.80atm P O2 0.20atm i. 25 C T ( )K K 25 P N2 P O2 = atm ii C T ( )K K 1000 P N2 P O2 = atm iii C T ( )K K 2000 P N2 P O2 = 0.015atm iv C T ( )K K 3000 P N2 P O2 = 0.063atm hwkc17_a.xmcd S.E. Van Bramer 1/3/2017, 3

4 c. At high temperatures this reaction fast and reaches equilibrium quickly, but when exhaust gases cool the reaction is very slow and the system stays at the high temperature equlibrium. Would lowering the combustion temperature from 3000 C to 2000 C have any effect on the NO emission? Yes at lower temperatures, the equilibrium is much less favorable. At higher temperatures more NO is produced. Since the system does not equilibrate quickly at ambient temperatures, this high NO concentration remains for an extended time period after the exhaust leaves the combustion chamber. Although the equilibrium favors lower NO concentrations at ambient temperatures, the return to equilibrium is slow. As we will discuss in the next chapter, the Kinetics (or rate) of the NO returning to equilibrium may be increased by using a catalyst. This is one of the functions of the catalytic converter in automobiles. d. The concentration of NO is normally determined with an instrument that first converts it to NO2. Based upon the Gibbs free energy, could the following reaction be used to convert NO into NO2 at ambient temperatures? i. NO + O 3 <--> 2 NO 2 + O 2 G rxn ( 1 ) G rxn = ( 1 ) 0 ( 1 ) ( 1 ) G values are from Kask and Rawn. Since G rxn is negative, the reaction is spontaneous as written so this mechanism is one possible mechanism. hwkc17_a.xmcd S.E. Van Bramer 1/3/2017, 4

5 2. Calculate G at 20 C for the reaction H 2 + I 2 <--> 2 HI. Starting with 1.0 atm H 2 and 1.0 atm I 2. What is G when 0.1%, 1%, 10%, 50%, 90%, 99% and 99.9% has reacted: G o ( 2 ) 1.30 G o = 16.8 T ( )K ( 1 ) ( 1 ) 0 a. 1.0 % has reacted P H2_initial 1atm P I2_initial 1atm reacted ( 1% )P I2_initial Qa ( ) ( 2a ) 2 ( P H2_initial a) P I2_initial a ( ) G rxn G o + RT ln( Q( reacted) ) G rxn = c. 10% has reacted reacted ( 10% )P I2_initial G rxn G o + RT ln( Q( reacted) ) G rxn = e. 90% has reacted reacted ( 90% )P I2_initial G rxn G o + RT ln( Q( reacted) ) G rxn = 2.71 f. 99% has reacted reacted ( 99% )P I2_initial G rxn G o + RT ln( Q( reacted) ) G rxn = hwkc17_a.xmcd S.E. Van Bramer 1/3/2017, 5

6 This exercise was to show how the Gibbs free energy changes with the concentrations of reactants and products. From this you should see that there is a point when this reaction is between 90% and 99% complete that the free energy is 0. At this point the reaction is at equilibrium. Below is a graph that shows this relationship. react 0.1%, 0.2% % Ga ( ) G o + RT ln ( 2a P I2_initial ) 2 ( P H2_initial ap I2_initial ) P I2_initial ap I2_initial ( ) G( react) react % hwkc17_a.xmcd S.E. Van Bramer 1/3/2017, 6

7 For those of you who have taken calculus, the curve above is actually the derivative of the free energy as a function of the "reaction coordinate" Or how much has reacted. It is possible to derive the free energy (not G) vs. reaction coordinate by integrating the above expression. Since G is the change in G, it is the equivalent to the slope of a graph of the free energy curve. If you have taken enough calculus you will recognize that if I integrate the above curve it will give me the free energy curve. This will give: Ga ( ) a G o + RT ln 0.1 ( ap I2_initial ) 2 ( P H2_initial ap I2_initial ) P I2_initial ap I2_initial ( ) da react 0%, 1%.. 100% 5 0 Greact ( ) Zoom in on minimum: react 90%, 90.1% % 13.4 react % Greact ( ) react % The minimum of the free energy curve is the equlibrium condition. Recall that since this is a minimum, it's slope is zero. Which is why this corresponds to the zero crossing of the earlier graph. hwkc17_a.xmcd S.E. Van Bramer 1/3/2017, 7

8 3.Using values from your textbook calculate S o and G o for the following reactions: a.co 2 (g) + H 2 (g) --> CO + H 2 O (g) S rxn ( 1 ) ( 1 ) ( 1 ) ( 1 ) G rxn ( 1 ) ( 1 ) ( 1 ) ( 1 ) 0... S rxn = K 1 G rxn = b.4 CO 2 (g) + 2 H 2 O (g) --> 2 C 2 H 2 (g) + 5 O 2 (g) S rxn ( 2 ) ( 4 ) ( 5 ) ( 2 ) G rxn ( 2 ) ( 5 ) 0... ( 4 ) ( 2 ) S rxn = K 1 G rxn = hwkc17_a.xmcd S.E. Van Bramer 1/3/2017, 8

Chapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase

Chapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase changes Apply the second law of thermodynamics to chemical