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Question 144ptsA first-order reaction has a half-life of 30 minutes. How long will it take for the reactants to drop to 6.25% of their initial concentration?1.0 hou...

Question

Question 144ptsA first-order reaction has a half-life of 30 minutes. How long will it take for the reactants to drop to 6.25% of their initial concentration?1.0 hour15hours1.67hours175 hours20hours

Question 14 4pts A first-order reaction has a half-life of 30 minutes. How long will it take for the reactants to drop to 6.25% of their initial concentration? 1.0 hour 15hours 1.67hours 175 hours 20hours



Answers

What is the half-life (in minutes) of the reaction in Problem
12.50$?$ How many minutes will it take for the concentration
of cyclopropane to drop to 6.25$\%$ of its initial value?

This is actually really is a question once you know what tools union so intimidated by it. So remember that the half life is equal to the natural log of two over the rate constant for four first or decays be derived from the integrated rate laws. But it's best just know it only uses pretty often now, we were told back A is 5.11 time sensitive fifth inverse seconds, and so you'll get a value for the half life, which I got in seconds first. But to go two hours of divide by 3600. And when you do that, you'll get, uh, 3.76 hours is the half life. Now this inelegant start questions nice where they're asking us to go to 12.5% now think about what happens. We're starting at 100%. They go through 1/2 life, we're gonna 50% another one. We go to 25 and then I mean, we go directly to 12.5, which we're looking for us. We don't have to use anything formless. We can just see that it's exactly 3/2 lives. 123 That we're looking for in some multiplied 3.76 by three, and you'll get, uh, right around 11 11.3 hours as our answer.

We're trying to figure out how many half lives have to pass in order for us. Toe have 1/8 of our original. So 1/8 is just gonna be equal two point 1 to 5. So let's say if we start with ah 100 Adams after 1/2 life, we know that we have half of those Adams left. The other half have deteriorated or decayed after another half life, we have 25 Adams and then after another half life, we're down 12.5. So that just matches are 1/8. 1/8 of 100 is 12.5 Adams. And so how maney half lives had to pass. Before that, we got to 12.5 being countem up 123 So there were 3/2 lives. Now we can look at our equation to determine toe, but that in terms of K so if we know that 1/2 life so t 1/2 equals 0.693 over K, that means if we have 3/2 lives, we have to multiply both sides by three. So we get 0.693 over. Okay, times three, we can simplify that by just multiplying 0.693 by three. And so it's gonna give us 2.79 over K.

So the question here is thinking about how long it takes for, um for us to have 1/8 of our original value of some sort of reactant so we can pretend that we start with 100 and 100 Adams. And so after 1/2 life, we're down to 50. After another half life, we're down to 25. And then after another half life, we're down to 12.5. So 1/8 of 100 is 12.5. So we can just see that how maney half lives have passed one Teoh three. So three half lives have passed in order for us to get to 18 of our original value. Now, now we want to put that in terms of K. So we know t of 1/2 equals 180.693 over K. And so that's one ah, half life. So that means if I have 3/2 lines, I have to multiply both sides 53 And so that's just gonna give me if I multiply 0.693 by three. It's gonna give me 30.202 point 079 over K

Well, so today we're going to be talking about a situation where we have a first order reaction and we want to know when there will only be one days of the reactant left, while one thing that we should probably use is the half life. So if you are familiar with it, the half life is how long it takes for half of it to decay. So we'll go from one to a half, and that will be one half life. But now we're going to decay, and we're gonna have half of the half, so that would be 1/4. So the first half life we go from 1 to 1 half, so only 50% off the reactant will remain. Then we're gonna go toe one force that's another half life, Then half of 1/4 would be an ace, and we would have another half life. So as we can see the time it takes to get Teoh, 1/8 of the reactant left is three half life. But say we don't know the half life of this reaction. So how else could we find the time it takes to get to 1/8 of the reactant left. Well, hopefully you should remember that half life for first over order action is a natural log of two over the rate constant. So if time it takes to get to 1/8 of the act left is equal to three half life stun. It would just be three natural log of to over K. Or we could rewrite this as the natural log of eight because that's two to the third Power is eight natural log of eight over K. And so we see we have several different ways of finding out that time it takes for it to decay toe 1/8.


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