HW2QuestionsAnswersME525SP2011

HW2QuestionsAnswersME525SP2011 - 1 ME 525 SP2011 Homework...

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1 ME 525 SP2011 Homework #2 Questions and Answers Due Tuesday, February 8, 2011 Prof. Lucht (E-mail address: Lucht@purdue.edu) Question: When I derive the differential equation for [O], I come accorss an equation that looks just like the one developed on the example problem worked in class. This does not include any O2 concentrations. Do I use the O2 <->2O for the equilibrium equation to obtain K_p to sub in for the concentration of O in order to solve the equation for the case of constant O3 and O2 concentrations? Then also perform the same substitution into the [O]_ss in part c? If that's the case can I leave Kp as an unknown value, because I didn't want to assume to solve gibbs free energy at 800K in part b. Thanks for any help. Answer: K_p does not come into play in this problem at all. I have revised the problem statement to state “Neglect reverse reactions”. If you only consider forward reactions, and consider short times such that [O2], [O3], and [M] are constant, you should obtain a differential equation of the form dx/dt=A+Bx, where x = [O] and t is the time. Question: I guess I'm not seeing the need for constant concentrations of O2 and why the mole fraction of O2 is being supplied in the problem requirements. As you can see from the attached calculations O2 concentration is not being considered since I am only considering the forward reactions. Also, I only use the mole fraction for O3 in part d. I must be missing something or that's just extra info. being provided in the problem statement. Answer: You are correct, O2 concentration is not used directly. You are also correct to neglect the reverse reactions. Note: I have revised HW #2 on the website to advise you to neglect reverse reactions. I have also listed the answers for problems 2 and 3. Question: When I plug in t=0, and [O]=0, I obtain t=(-1/B)ln(A-B[O])+(1/B)ln(A). If I solve for t, when [O]=[O]_ss=1.732X10^-8 gmol/cm^3(very small amount of O), I will always get (A- B[O]_ss) =0, just due to the characteristics of d[O]/dt=0=A-B[O]_ss. Taking ln(0) provides a negative infinity solution, which provides an unreasonable solution for t. Big picture wise, since [O]_ss is such a small amount of [O], the time it would take to decompose ozone to monotomic oxygen of this order, in my mind, would be very quick. Instead of plugging in my solution for [O]_ss should I approach it from a specific amount from the left and/or right, say 1.731(or3)X10- 8 gmol/cm^3. This should offset the value of A-B[O] and provide a positive solution for time while giving me an estimate of time when [O]=[O]_ss. Thanks again. Answer: Do not solve the equation t=(-1/B)ln(A-B[O])+(1/B)ln(A) for t. Solve the equation for [O]. The value of [O]_ss will emerge naturally as t approaches infinity (actually it will approach the steady state value in a few microseconds).
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2 Question: On the second question in the homework, I am getting an equation for d[O]/dt=- k1^2/(2*k2)*[M]^2/[O]+k1*[M]. In order to find the concentration of M we need to have an
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HW2QuestionsAnswersME525SP2011 - 1 ME 525 SP2011 Homework...

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