lan418 Posted March 19, 2004 Share Posted March 19, 2004 A(aq) + 2B(aq) ---> 3C (aq) + D (aq) For the reaction above, carried out in solution at 30deg C, the following kinetic data were obtained: Experiment Int. concentration of reactants mol/L Ao Bo 1 .24 .48 2 .24 .12 3 .36 .24 4 .12 .12 5 .24 .06 6 .140 1.35 Int. rate of reaction mol/L * Hours 8 2 9 .5 1 ? Assume that the reaction goes to completion. Under the conditions specified for experiment 2, what would be the final molar concentration of C? How do you find it? what does it mean when it goes into completion? does it mean that all is consumed? Link to comment Share on other sites More sharing options...
lan418 Posted March 20, 2004 Author Share Posted March 20, 2004 anyone? Link to comment Share on other sites More sharing options...
Skye Posted March 21, 2004 Share Posted March 21, 2004 Well all I can add is that the rate equation is: v=289.35[A]^2 i.e. A is second order, B is first order, k is around 289.35. Link to comment Share on other sites More sharing options...
lan418 Posted March 21, 2004 Author Share Posted March 21, 2004 Oh, maybe i did something wrong? for the constant i got around 69.6.... but my major concern is the consumption, i dont see how you would solve for the final Molar concentration of C. Link to comment Share on other sites More sharing options...
lan418 Posted March 21, 2004 Author Share Posted March 21, 2004 i'm sry, i did get your answer. Must of read the wrong thing Link to comment Share on other sites More sharing options...
Mediculous Posted March 21, 2004 Share Posted March 21, 2004 I know it's always fun to find the Rate of rxn., but I don't think it's necessary in this problem. -d[A]/dt = (1/3)d[C]/dt, so when you go to completion, (using exp.#2) your change in [A] is -.24M, and when you plug that in, you get your change of [C] to be .72M. You begin with [C]=0M, so your final [C] would be .72M. I'm curious, was this from a practice MCAT where they always try to waste your time by giving you too much info? Link to comment Share on other sites More sharing options...
Mediculous Posted March 21, 2004 Share Posted March 21, 2004 I forgot to add that this is assuming that there is enough B to react. The molarity of B is less than A in exp.#2 but there aren't values given for amounts of either A or B, so you must assume that B does not limit the rxn. If there actually were amounts of A and B given, then you would have to check if either limited the rxn., and the you would just use the rate of change for that reactant (if it was B, the you would use -(1/2)d = (1/3)d[C]/dt). Hope that helps. Link to comment Share on other sites More sharing options...
Skye Posted March 21, 2004 Share Posted March 21, 2004 Boy that was alot easier than it looked. Link to comment Share on other sites More sharing options...
lan418 Posted March 21, 2004 Author Share Posted March 21, 2004 lol nah. just a problem from a general chem class assignment. There were actually 4 parts to this question but i didnt bother to put them since i didnt think that they were relevent to the last part the question was asking. Mediculous, i dont understand how the alegbra works for this problem when you plug it in. what does dt and -d denote? Can you show it out for me please? Link to comment Share on other sites More sharing options...
Mediculous Posted March 21, 2004 Share Posted March 21, 2004 d=difference=change=delta(the greek letter). dt=change in time. d[anything]/dt= change in molarity over change in time= rate of change And in my last post, it should read -(1/2)d/dt = (1/3)d[C]/dt. Sorry. Ok, so the rate of change in the reactants and products are all related, and that does depend on stoichiometric coefficients. The formula for rates of rxn. for the reaction aA+bB-->cC, where all lower-case letters are coefficients, is: -(1/a)d[A]/dt = -(1/b)d/dt = (1/c)d[C]/dt There is a negative sign in front of the reactants because they are being used up(rate of consumption) as opposed to the products which are being produced(rate of formation). So for the original problem, we have: -(1/1)d[A]/dt = (1/3)d[C]/dt and since the change in time is the same for both, we have -(1/1)d[A] = (1/3)d[C] since we start with .24M A and end with 0M A, the change is -.24M, so plugging that in gives (-1)( -.24M)=(1/3)d[C] then d[C] = 3(.24M)= .72M Now, if the amounts of A or B are given anywhere, that adds an extra step of determing the limiting reagent before you choose which reactant to use in the formula(in this case, either A or B). It doesn't apply in this case, but you should be aware of the relation just in case it ever came up. When you get the actual answer from class, do you think you could post it, just so I know if I I was right, or was a complete jag-off by misleading you(I really do think I'm right, though...I think)? Link to comment Share on other sites More sharing options...
lan418 Posted March 21, 2004 Author Share Posted March 21, 2004 yea sure, thanks for the help Link to comment Share on other sites More sharing options...
Mediculous Posted March 21, 2004 Share Posted March 21, 2004 Oh crap, there is something wrong here. Give me a minute to check it over again. Link to comment Share on other sites More sharing options...
Mediculous Posted March 22, 2004 Share Posted March 22, 2004 Nevermind, I'll just have to wait with you and see. Link to comment Share on other sites More sharing options...
lan418 Posted March 22, 2004 Author Share Posted March 22, 2004 What is wrong? this assignment is a heavy burden on my grade : ( Link to comment Share on other sites More sharing options...
Mediculous Posted March 22, 2004 Share Posted March 22, 2004 Nevermind, I was just second-guessing myself, but it still looks alright to me. Anyone else care to offer what they think? Link to comment Share on other sites More sharing options...
Mediculous Posted March 22, 2004 Share Posted March 22, 2004 Ok, so I do think that there actually is a limiting reactant. AND you need to know the rate law to do this problem. The key to this is figuring out how long(change in time= dt) it takes for the molarity of A and B, individually, to reach zero. If one reaches zero before the other, then it limits the reaction. You get the time it takes for the limiting reactants to be completely consumed using the rate order in respect to each reactant. Because A is second order, d[A]/dt= k[A]^2 and because B is first order, d/dt= k. Now, using the formula I gave in an earlier post and the k you found, we have: -289.35(-.24^2)= -[.24]/dt and solving for dt, we get dt= .014hrs. Now for B, using: -289.35(-.12)= -.12/dt, and solve for dt, dt= .0035 hrs. SO, the B is used up sooner and is the limiting reagent. Because of this, we use: -(1/2)d/dt = (1/3)d[C]/dt Substituting all our known values we have: -(1/2)(-.12)/.0035 = (1/3)d[C]/.0035 multiplying both sides by .0035, .12/2 = d[C]/3 and solving for d[C]=.18 I knew something smelled fishy. But after thinking I was right at first, and now thinking I was wrong at first....I wouldn't bet my life that it is right. I don't see any flaws in my latest reasoning---I accounted for the limiting reagent, but I would get a second opinion if I were you. When is this due? Link to comment Share on other sites More sharing options...
lan418 Posted March 22, 2004 Author Share Posted March 22, 2004 heh, tommorrow, but i posted this topic a few days ago. so i didnt wait till the last minute at least Link to comment Share on other sites More sharing options...
lan418 Posted March 26, 2004 Author Share Posted March 26, 2004 Well, mediculous you got the right answer. Your work came out to the right value. However, the teacher just did it with 2 simple steps. 1) find the limiting reageant. 2) use the limiter to the ratio of C using stiochiometry and eventually the units cancels out and you end up with .18 mol. heh, the teacher went up to me and said i confused the heck out of him. but thanks for the help. Link to comment Share on other sites More sharing options...
Mediculous Posted March 26, 2004 Share Posted March 26, 2004 Wow, what a relief! I hope you learned from your teacher and not from me how to do it! Hey... I was just being "thorough"?Maybe? I feel like that question was more for me than it was for you. Next... Link to comment Share on other sites More sharing options...
lan418 Posted March 26, 2004 Author Share Posted March 26, 2004 your absolutely right, i did not have one clue what you did. (although i did kind of with your explanations but first time coming across it for me). But if you asked me to do it your way again, i would be confused. i guess this can be exposure for future reference? Link to comment Share on other sites More sharing options...
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