Integrated Rate Law Example 3

by Jules Bruno
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here we're told that the reacting concentration for a second order reaction was 0.7670 moller after 300 seconds and 7.3 times 10 to the negative to moller after 750 seconds. What is the rate constant K. For this reaction? Alright so they tell us it's second order. So that means 1/18 equals K. T. Plus one over A. L. Here we're told that we initially have .670 moller after 300 seconds. That's gonna be our initial. Okay. No one says that initial has to start at zero seconds. It's just in this case we're starting at 300 seconds And then one over the final concentration is 7.3 times 10 to the -2. Here we're looking for. Kay but we need to figure out what our time is, how much time has elapsed. Well we're starting at 300 seconds and this is the initial concentration And we go to 750 seconds. If we subtract those two numbers that tells us how much time has elapsed And that's 450 seconds have elapsed. Alright so we're going to subtract one over .670 from both sides. Alright so then when we do that we're gonna get 12. equals K. 4 50. So then we're gonna divide both sides now by 4 50. And when we divide both sides by 4 50 that's gonna give us our K value K. Here will equal 2.71 times 10 to the -2. Now here's the thing where the units for K. Remember for K it's M to the negative end plus one times time inverse since its second order and is too so negative two plus one times time inverse. Here. The units for a time where we see are in seconds. So this would be seconds inverse. So negative two plus one is negative one, so M to the negative one times seconds to the negative one. So this would represent the value for our rate constant K as well as its units.