1123581321 Posted May 30, 2011 Posted May 30, 2011 (edited) I was wondering if anyone would be able to explain to me how exactly the 'm' & 'n' exponents in the rate law equation work(s), because they just seem to slip my understanding.. or if not, if you would be able to provide a link of some description to a video etc which can explain them in a simple and elegant fashion, that would be much appreciated, thanks. Edited May 30, 2011 by 1123581321
CaptainPanic Posted May 30, 2011 Posted May 30, 2011 (edited) In this wikipedia article, the m and n are just the stoichiometric coefficients... but I guess you're not asking about those. But if you are, I guess you should just read this wikipedia article. It helps if you post a link, or just the formula itself next time, because now I have to guess what your question is, and we might all be wasting our time. I'm guessing that you want to know what the x and y are in the following rate equation: [math]r\; =\; k[\mathrm{A}]^x[\mathrm{B}]^y [/math] The x and y are experimentally derived values. The fact that in exercises they are often integers (i.e. values like 0, 1 or 2) are not based on a physical/chemical reason. The x or y might just as well have a value of 0.6472 or something. It just means that if you increase the concentration of A, what happens to the rate. Typically, these values are determined by many measurements. You end up with a list of concentrations and reaction rates. And then you have to apply some statistical analysis to fit a function to that data set. So, the fact that some measurements in a lab can be turned into a nice and simple equation is not because it is derived. It is only because some mathematical tricks are applied to make an equation which describes reality quite well (but not perfectly). Edited May 30, 2011 by CaptainPanic
mississippichem Posted May 30, 2011 Posted May 30, 2011 (edited) In the classroom. You often get nice integer exponents like 0,1,2,3. In those cases you get the luxury of applying one of the common integrated rate laws: [math] -\frac{d[X]}{dt}=-k_{1}[X] [/math] [math] \frac{d[X]}{[X]}=-k_{1} dt [/math] [math] \int\frac{d[X]}{[X]}=\int- k_{1} dt [/math] [math] \ln[X]=-k_{1}t+r_{0} [/math] This help you isolate the effects of one reactant's concentration on the rate. But often in reality, you get fractional order exponents that can't be integrated so nicely like Captain Panic said. More qualitatively, think about the exponents as being an expression of how strongly the rate of a reaction depends on the concentration of that species. Edited May 30, 2011 by mississippichem 1
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