Consider the data presented in Exercise 14.19. (c) What is the half-life for the reaction?

Video transcript

Hello everyone today. We have the following problem. The gas phase reaction of a solution created by dissolving 0.45 moles of C. And a total volume of 250 mL was observed. And we have the following chemical reaction. What is the half life of this reaction? So the half life for this reaction which we have one reaction forming one product. So this is first order. Any half life for a forced order reaction is going to be one over a constant times the initial concentration. And so we have one and we need to find the concentration and we also need to find our K. Or constant. So essentially we have to plot one oversee versus our time. So we will be using this column here. So we want to take the first two times and their moles and divide them by our volume. So to get For the .045 we must do .045 moles And divide that by 0.25 L which is 250 ml divided by 1000. This is going to actually give us 0.18 Mueller. We do the same thing for the 2nd 10.35 moles over 0.25 and we get 0.14 molar secondly we must then take one over that malaria that we just found. So we then do 1/0 10.18 and 1/0 0.14 and then we do that we get 5.555 repeating and then we get 7.143. We then solve for our slope, which in this case would be K. And so slope is going to just be our X or in the situation will be okay. It'll be y two minus y one over X two minus x one For the denominator or the excess. We're gonna do 15 0 for the top. We're going to do 7.143 - our 5.555 repeating. That's going to give us our zero Our 0.1037 Polarity. So we have our 0.1037 polarity here. This is gonna be a first order. So it's going to be malaria times to the negative first and the minutes to the negative first times. This is going to be our 0.18. This is going to be the first that we found there. And so when we Settelf and cross our units out, we are going to end up with 53.57 minutes as our half life for this reaction. And with that we've answered the question overall. I hope this helped. And until next time