Texas Tech University
The Integrated Rate Laws - Example 3


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Now let's solve and every question We have a reaction, A goes to products and this reaction is a first order. The composition reaction. So they're giving us the K read Constant is 0.1386 years to the port of negative one. How much time needs to pass to consume 85 87.5% off a. Now there are two ways to solve this question. First of all, we'll see you look at the, um que The unit is year to the poor of native one. You don't need to convert this two seconds. This is just fine. What is our integrated rate law? Ln now, before actually going for the integrated rate law, we need to find if 8 to 7.5% is going to get consumed. How much percent off A and we're going toe have left. So it was 100%. Now we're consuming 87.5% and we get 12.5. Her scent left. So integrated right law Ellen, 12.5 over 100. This is our 12.5% A zero equals two minus k zero point 13 86 plus Ellen 100 over 100 easier. So we can just simply put the same terms on the same side. And there's a TV air. Soling 40. What we're going to have is going to be Ellen off. 12 point five or 100. Do I buy 100? 100. A zeros canceled equals two minus zero point 13 86 team. And now, if you sold for this question T is going to be 15 years now. This is the first way off solving this question, really just right to integrated. Read law and we go for him. So what else can we do? If you see this 12.5% then you need to think off if this number is some, uh, multiplier off 100. So we had 100% at first after some hall. Five times. We have 50%. After some hall. Five times we have 25%. And after some hall, five times. Well, 12.5% left. So they're 123 hall fly IVs. So what is the whole Fife again? 0.693 divided by 0.13 86 which is r. K. equals are half life and our hall Fife is going to be equal to five years. So if tree five years has passed, we get the same result 15 years.

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