Unformatted text preview: due Thursday, April 22, 2010 at 2:30 pm
1. (20 points) We will calculate free energy G along a reaction coordinate ξ. Consider the ideal gas reaction 2 A → B , at constant temperature and constant pressure (P = 1 atm). Assume there is initially NA = 1 mole of A and no B present. Furthermore, µA° = 5 kcal/mol and µB° = 10 kcal/mol at this temperature. € a. (10 points) Express the partial pressures PA and PB as a function of ξ.
The reaction coordinate can be defined as ξ = range between 0 and 1. ChE 359 Homework #9 2N B , which will have a valid N Ao dξ = 2 1 dN B = − dN A N Ao N Ao € This gives: € ∫ NA N Ao dN A = ∫ ξ
0 −N Ao dξ ⇒ N A =N Ao (1 − ξ) € N Ao N dξ ⇒ N B = Ao ξ 0 02 2 ȹ ξ ȹ N t = N Aoȹ1 − ȹ ȹ 2 Ⱥ ∫ NB dN B = ∫ ξ € Now solving for the partial pressures: € pA = pB = NA 2 − 2ξ P= atm Nt 2 −ξ NB ξ P= atm Nt 2 −ξ € € b. (5 points) Express G as a function of ξ and plot G vs. ξ. ȹ o ȹ o p ȹ p ȹ G = µA N A + µB N B = ȹ µA + RT ln A ȹN A + ȹ µB + RT ln B ȹ N B ȹ ȹ P Ⱥ P Ⱥ Ⱥȹ o ȹ o 2 − 2ξ ȹ ξ ȹȹ ξ ȹ Ⱥ G = N Ao Ⱥȹ µA + RT ln ȹ(1 − ξ) + ȹ µB + RT ln ȹȹ ȹ Ⱥ 2 − ξ Ⱥ 2 − ξ Ⱥȹ 2 Ⱥ Ⱥ ȹ Ⱥȹ € € Page 1 of 7 ChE 359 SP2010  Homework #9 G vs psi 4.95 dG/dpsi (kcal/mol^2) G (kcal/mol) dG/dpsi vs psi 3 2 1 0
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3 reaction coordinate, psi (mol) 4.9 4.85 4.8 4.75 4.7 0 0.2 0.4 0.6 0.8 1 reaction coordinate, psi (mol) 0 0.2 0.4 0.6 0.8 1 NOTE: You can develop this carefully from an expression for dG. An expression for ΔG will not give you the information you need about where G is a minimum. c. (5 points) Determine the value of ξ at equilibrium. € The minimum occurs when dG = 0 dξ € Ⱥ ȹ ξ − 1 ȹ Ⱥ dG µ RT ȹ ξ ȹ = N Ao Ⱥ −µAo + Bo + lnȹ ȹ Ⱥ = 0 ȹ − RT lnȹ dξ 2 2 ȹ 4 (ξ − 2) Ⱥ ȹ ξ − 2 Ⱥ Ⱥ Ⱥ € ȹ ξ(2 − ξ) ȹ 2 ⇒ lnȹ 2 ȹ = 0 ⇒ 5ξ − 10ξ + 4 = 0 ȹ ȹ 4 (ξ − 1) Ⱥ ȹ This expression gives two roots, the only one in the valid range being €
€ ξ = 0.55 This corresponds to N A = 0.45 mol and N B = 0.28 mol . NOTE: Regardless of how you defined the reaction coordinate, ξ , you should end up with the same values of N A and N B as above. € € 2. (40 points) Hydrogen (H2) is projected to be a major component of future energy infrastructures. Hydrogen is produced using the steam reforming of methane € (CH4); the carbon monoxide€(CO) produced during that process is rapidly € converted to carbon dioxide (CO2) using the watergas shift reaction. The overall gasphase reaction mechanism is given by the following rate equations (note that the equilibrium constants are given in terms of the forward and reverse rate k k constants, so K1 = 1 and K 2 = 2 ): k−1 k−2
K1 kJ CH 4 + H 2O ←Ⱥ→ CO + 3H 2 ΔH 298 K = +206 mol K2 kJ CO € H 2O ←Ⱥ→ CO2 + H 2 + ΔH 298 K = −41 mol € € Page 2 of 7 ChE 359 SP2010  Homework #9 a. (5 points) List the reactants, intermediates, and products in the overall process.
The overall reaction is given by the sum of the individual reactions CH 4 + 2 H 2O ↔ CO2 + 4 H 2 reactants: methane, water intermediate: carbon monoxide € products: carbon dioxide, hydrogen b. (5 points) For the first reaction (steam reforming of methane), write an expression for the rate of disappearance of CH4, or −rCH 4 ,1 . (Hint: you should write the equilibrium constant in terms of gasphase partial pressures).
Written in terms of concentrations: −rCH 4 ,1 = k1 [CH 4 ][ H 2O] − k−1 [CO][ H 2 ] 3 €
[CO] stands for the concentration of carbon monoxide, and likewise for other €
€
species. Recalling that [CO] = pressures: pCO lets you express the rate in terms of partial RT NOTE: All the −rCH 4 ,1 = k1 pCH 4 p H 2O − k −1 pCO p 3 2 H €1
RT terms can be absorbed by the rate constants. Just make sure € that the rate constants have the right units! c. (5 points) Use the expression for −rCH 4 ,1 derived in part (b) to obtain rH 2 ,1 , the € of appearance of H2 in the first reaction. rate
From the stoichiometry of the first reaction you know that rCH 4 ,1 −1 = rH 2 ,1 3 € € This shows mathematically how hydrogen is being produced three times as fast as methane is being consumed. This gives € rH 2 ,1 = 3k1 pCH 4 p H 2O − 3k −1 pCO p 3 2 H
d. (5 points) For the second reaction (watergas shift), write an expression for rH 2 ,2 , the rate of appearance of H2. Then combine the expressions in parts (c) and (d) to get rH 2 , which is the overall rate of appearance of H2. rH 2 ,2 = k 2 pCO p H 2O − k −2 pCO2 p H 2 The overall rate of hydrogen production is then €
€ € € rH 2 = rH 2 ,1 + rH 2 ,2 = 3k1 pCH 4 p H 2O − 3k−1 pCO p 3 2 + k 2 pCO p H 2O − k −2 pCO2 p H 2 H € Page 3 of 7 ChE 359 SP2010  Homework #9
e. (10 points) The pseudo steady state hypothesis states that the rate of appearance of an intermediate is zero. Use the pseudo steady state approximation on the reaction intermediate CO ( rCO = 0 ) and get an expression for rH 2 in terms of the rate constants and the partial pressures of the reactants and products.
Carbon monoxide is the intermediate in the overall reaction. Setting its overall € rate to zero gives € rCO = rCO ,1 + rCO,2 = k1 pCH 4 p H 2O − k −1 pCO p 3 2 − k 2 pCO p H 2O + k −2 pCO2 p H 2 = 0 H ⇒ pCO = k1 pCH 4 p H 2O + k −2 pCO2 p H 2 k −1 p 3 2 + k 2 p H 2O H
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€ This expression gives the partial pressure of the intermediate in terms of partial pressures of the reactants and products. Plugging this expression for pCO into the expression for rH 2 will give the overall hydrogen production rate in terms of partial pressures of reactants and products only. € f. (5 points) The two steps of the reaction have standard state enthalpies of reaction at T = 298 K and P = 1 atm as given above. Consider an increase € in temperature to T = 373 K while keeping P = 1 atm. State whether K1 would increase or decrease, and whether K2 would increase or decrease. Explain your reasoning and the implications for how far both reactions would proceed toward completion (i.e., producing H2).
Reaction 1 is endothermic, so the increased energy from the raised temperature can be absorbed by the reaction moving forward and making more products. Therefore, K1 will increase and the reaction proceeds further towards completion. Reaction 2 is exothermic, so the increased energy from the raised temperature can be absorbed by the reaction moving backwards towards the € reactants. Therefore, K 2 will decrease and the reaction reverses towards the reactants. These results can also be shown mathematically by the van't Hoff equation. g. (5 points) Now, consider an increase in pressure to P = 10 atm while € keeping T = 298 K. State whether K1 would increase or decrease, and whether K2 would increase or decrease. Explain your reasoning and the implications for how far both reactions would proceed toward completion (i.e., producing H2).
The system will want to restore the original pressure and it does this by shifting the reaction towards the side whose species takes up less volumetric space. Reaction 1 produces more moles of gas as it goes to completion, increasing the volume of gas. Therefore, K1 will decrease and the reaction reverses towards the reactants. € Page 4 of 7 ChE 359 SP2010  Homework #9
Reaction 2 does not produce a net change in the volume of gas and so K 2 is unaffected by the change in pressure. NOTE: If you consider the water in this problem as a liquid instead of a gas (which is totally reasonable at room temperature and 10 atm) then Reaction 2 € will also proceed in reverse towards the reactants. 3. (20 points) We will consider polymerization reactions that happen in the gas phase. n moles of identical monomers are in equilibrium with one mole of chains Ⱥ − − ⊗ of nmers, or n ( ⊗) ← K →⊗ − ⊗ ⊗ ⊗ − . a. (5 points) Express the equilibrium constant K in terms of the partition functions.
n monomers € K= qn − mer −(ε on − nε o1 ) / kT q e = nn − mer e −(( n −1)Δε ) / kT n qmonomer qmonomer Here ε on is defined as the energy of the n
mer, ε o1 is the energy of one monomer, and Δε is the change in energy in forming a bond. € b. (5 points) Explain whether you expect typical polymerization processes to € be driven, or opposed, by enthalpy. Also, explain how these processes are € driven, or opposed, by entropy. Speculate on the physical origins of these € enthalpies and entropies.
Polymerization is a spontaneous process so ΔG = ΔH − TΔS < 0 As polymerization occurs, entropy is lost due to the polymers having less possible states available compared to all of the monomers. Therefore −TΔS > 0 , and so ΔH must be negative. Thus the process is driven by enthalpy but opposed by entropy. € € c. (5 points) Explain whether you would expect polymerizations to be € exothermic (giving off heat) or endothermic (taking up heat). Use this to explain the explosions in some polymerization processes.
that occurs leads to a release of heat (assuming PV work is negligible). For many polymer processes, n is some huge number so there is a lot of heat that can be generated as the reaction proceeds to completion, possibly leading to explosions. Additionally, the increased temperatures will speed up the rate of reaction, leading to a runaway reaction. This situation can occur for any sufficiently exothermic reaction that is allowed to proceed without careful control. ΔH < 0 so the polymerization is an exothermic process. Every bond formation € d. (5 points) Determine whether polymerization equilibrium constants for long chains are more or less sensitive to temperature than for shorter chains. Page 5 of 7 ChE 359 SP2010  Homework #9
The van't Hoff equation gives the temperature sensitivity of the equilibrium constant d ln K Δh ( n − 1) Δε o = 2= dT kT kT 2 For larger values of n (larger chains) the temperature sensitivity is greater. 4. (20 points) We will see how enzymes € accelerate chemical reactions. The figure shows an Arrhenius plot for the uncatalyzed reaction of 1methylorotic acid (OMP). a. (5 points) Estimate ΔH‡ from the figure, and estimate ΔS‡ at T = 25 °C.
From transition state theory ko = kT −ΔGo≠ / kT kT −ΔH o≠ / kT ΔSo≠ / k e = e e h h ≠ ΔH o ȹ 1 1 ȹ ȹ − ȹ k ȹ T2 T1 Ⱥ Choosing two points from the plot and dividing the results gives € ko 2 T2 − =e ko1 T1 € ȹ T k ȹ k lnȹ 2 o1 ȹ ȹ T1ko 2 Ⱥ ≠ ⇒ ΔH o = = 45 kcal / mol 11 − T2 T1 ȹ 1 ȹ ȹ 1 ȹ ko ȹ ko ȹ ȹ = 1x10 −16 s−1 ȹ = 1s−1 ȹ 298K Ⱥ ȹ 555K Ⱥ For and €
≠ ΔH o = 45 kcal / mol ≠ Now you can use this value to find ΔSo €
€ € ΔH kh ≠ ΔSo = + R ln o T kT ≠ o
≠ ΔSo = 19.3cal / mol⋅ K € Choosing some rate constant and temperature gives €
€ 1 , which is T ≠ only true if you assume that Arrhenius kinetics apply, or equivalently ΔH o = E a . ≠ ≠ kT ΔH o ΔSo 1 ln k o = ln − + If you're using , this is not linear in . However the h kT k €T
NOTE: Many of you used the fact that the semilog plot was linear in € € € Page 6 of 7 ChE 359 SP2010  Homework #9
kT term is relatively small so that you can approximate linear behavior. You h ln should make it a point to state this though! € b. (5 points) Determine the change in reaction rate as the temperature increases from T1 = 25 °C to T2 = 35 °C. (Hint: The general rule of thumb is that chemical reaction rates roughly double with a tendegree increase in temperature. How valid is this approximation?)
≠ Using the value of ΔH o you calculated So the rule of thumb doesn't hold here. ko 2 T2 − =e ko1 T1 € ≠ ΔH o ȹ 1 1 ȹ ȹ − ȹ k ȹ T2 T1 Ⱥ 308K −1.986 x10 −3 kcal / mol ⋅K ȹ 308 K − 298 K ȹ ȹ Ⱥ = e ≈ 12 298K 45 kcal / mol ȹ 1 1 ȹ € c. (5 points) At T = 25 °C, the enzyme OMP decarboxylase accelerates this reaction 1.4×1017fold. Determine the rate constant of the catalyzed reaction at 25 °C, and the new ΔH‡.
At 298K ' ko = 1.4 x1017 k o = 1.4 x1017 (1x10 −16 s−1 ) = 14 s−1 ≠ ≠ Assuming ΔSo is independent of temperature, the new ΔH o is given by € ≠ ΔH o = RT ln kT ≠ + TΔSo ' koh €
€ € K (1.38 x10−23 J /€ )(298K ) + (298K )(0.0193kcal / mol⋅ K ) −3 = (1.986 x10 kcal / mol)(298K ) ln −1 (14 s )(6.626 x10−34 J ⋅ s) = 21.6 kcal / mol d. (5 points) Calculate the binding constant of the enzyme to the transition state of the reaction at T = 25 °C.
The binding constant is just the amount of enhancement provided by the enzyme, 1.4 x1017 . € € Page 7 of 7 ...
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This note was uploaded on 02/09/2011 for the course CHE 359 taught by Professor Cynthialo during the Spring '10 term at Washington University in St. Louis.
 Spring '10
 CynthiaLo
 Mole, Reaction, Kinetics

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