What is the half-life (in minutes) of the reaction in Problem
14...

Question

What is the half-life (in minutes) of the reaction in Problem 
14.74 when the initial C4H6 concentration is 0.0200 M? 
How many minutes does it take for the concentration of 
C4H6 to drop from 0.0100 M to 0.0050 M?

Accepted Answer

Hello. Everyone in this video. We're considering a hypothetical reaction. So when A reacts with the South products, B. Is formed. If we're given the initial concentration of A. We're being told to determine the half life of the reaction. Ross being asked. How long does it take for the concentration of A. To drop from 0.0 to 50 to 0.125 molars. Were assuming that the re constant. Is this right here? So let's first determine the order of your action. Which is N. Based on the rate constant K. So we have M. To the power of negative N plus one times as to power of negative one. We're going to M. To the negative one, multiplied by S. Two negative one. So here we can see that if we divide both sides by S. Two negative one. We basically cancel those out. Then we're just left with this here and this here. We're just gonna go ahead and focus on the exponents here. So that's negative one plus one. We're going to negative one. We can see that if we just subtract both sides by one we'll get negative N. Equal to -2. And then divide both sides by -1. To kind of isolate this end by itself. We get a positive two. So if you get the end is equal to two. We can see that this reaction then is second order. Alright so then our rate equation is equal to R. K. Constant, multiplied by the concentration of a raise to the power of two. So the integrated rate law for a second order reaction is equal to one Over the concentration of eight. Or the final concentration of a. You going two K. of T. Plus one over the initial concentration of a. Alright let's go ahead and scroll down for more space. So we're dividing deriving are half reaction. So we're gonna go ahead and let so let's try this on the side here, we're gonna let the concentration of A. And we'll have this half the bottom here to equal to 1/2 times a knot our sheets. The other way around. We have are a not. Alright so basically what we have right here so we're gonna go ahead and simplify to T. Half Equal to one over K. Times A. Not. Alright so scrolling down again, we're gonna go ahead and solve for our half life reaction. So T. Two half as you go to again, one over K. Times the concentration of A. Not. We're putting in numerical values for this, we know that K. Is equal to eight times 00 times 10 to the negative two Units being em to a super of -1 times S 2 -1. And then our concentration for this is 0.0500 molar. And again it's all over one. So this year and this year will cancel, leaving us with the final answer To be 250 units being seconds. So t. half here is equal to 250 seconds. So that's our first answer for this problem again scrolling down for more space. Now we're gonna go ahead and solve for tea When the initial concentration of a. So a not is equal 0.0-50 moller. And let's break this. And a little bit better when they find a concentration of A. Is equal to 0.0125 Molar. Alright again we'll scroll down. Okay so we have this second or this integrated rate law for a second order reaction of one over the final concentration of a. Equalling two K times T. Plus one over the initial concentration of A. So A. Not. So we're gonna go ahead and just isolate T. Here. So get that T. Is equal to one over K. Times one over the final concentration of a minus the initial concentration of A. So then again we're going to go ahead and plug in actual values So we know the cape is eight times 00 times 10 to the negative two. And it's being M. So moller, two negative one times as two negative one for seconds and then go ahead and multiply this by One over the final concentration which is 0.0125 molar. And this is subtracted by one over the initial concentration which is 0.0-50 moller. So once you put everything into a calculator. We see that T. Here is equal to 500 seconds, so this is going to bring me a last and final answer for this problem, which is going to be just these two highlighted here.

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