Problem

Fluorine-18 is a radioactive isotope that decays …

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Problem 45 Hard Difficulty

There are two molecules with the formula ${C}_{3} {H}_{6}$ . Propene, ${CH}_{3} {CH}={CH}_{2},$ is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic: When heated to $499^{\circ} {C}$ , cyclopropane rearranges (isomerizes) and forms propene with a rate constant of $5.95 \times 10^{-4} s^{-1}$ What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499$^{\circ} \mathrm{C} ?$

Answer

20.06$\%$

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Chemistry 102

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Video Transcript

All right, So this question actually has two questions in it. Um, and so I have to separate windows for these two questions. Um, the 1st 1 asks, What is the half life of this reaction in the 2nd 1 asks what fraction of Psychlo propane remains after 3/4 of an hour at 499 degrees C. So to start with our first question, we can pull some information from our question. Um, the it describes the reaction. Ah, and it is a first order reaction. And then also it, um, tells us our rate constant at 499 degrees C, which is 5.95 times 10 to the negative four. And so I've just started here by writing out, um, the half life, um, equation for a first order reaction, which is t 1/2 equals the natural log of two over your rate constant. So from here, I can just plug in the value for a rate constant that it gives us. So the natural log of two over, um C 5.95 times 10 to the negative, four seconds to negative one, and I just plug that all into my calculator. Um, and I got 1.16 times 10 to the third power, and the units are seconds. So that's the answer to our first question. And the second part asked, What is the half life? No, that was first question. It asked what fraction of the cyclo propane remain after, Um, 3/4 of an hour at 499 degrees C. So I just wrote the, um, first order, um, expression right here, which is the natural log of the concentration of a equals negative rate, constant times time, plus the natural log of the initial concentration of A And so we don't know the concentration of A That's what we're trying to find out. So I'm just gonna rewrite that, and that equals the negative of the rate constant. So I'll just plug that in which, if you remember, was five point 95 times tend to the negative four seconds to negative one. Okay. And then our time, which in this question is asking us for 3/4 of an hour. But you see that our rate constant is sec is in seconds, so we have to just convert that 3/4 of an hour. Two seconds. So there's 60 seconds in a minute and 45 minutes and 3/4 of an hour. And so if you just do that calculation, you'll come up with 2700 seconds. Um, and then we're going to add our initial concentration, which I'm just gonna make that one, and you'll see why. And a second and so natural lug of one equals zero. So I'm just gonna cross that out, and then we can, um, cross our units out. Um, let's see. Yep. And so I'll just multiply these two things together. So the natural log of the concentration of a equals, Um, and if I put that in my calculator, that comes out to 1.6065 and to solve for a which is what we want. Um, we just rewrite this expression as a equals, um, e to the power of negative 1.6065 And that comes out to, um, 0.200 And so if her initial concentration was one and now our our concentration right now is our concentration right now is 0.200 We can see that we now have 20% of are concentration left over

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