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Ch.14 - Chemical Kinetics

Chapter 14, Problem 83b

Consider the following concentration–time data for the decomposition reaction AB → A + B.

(b) What is the molarity of AB after a reaction time of 192 min?

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hi everyone for this problem. It reads the decomposition reaction for C. D. Has the following concentration time data. What is the malaria T. Of C. D. After 100 minutes of reaction time. So we want to know the similarity of C. D. After 100 minutes. And in this table were given time and minutes and the concentration of C. D. In malaria T. Okay so let's just recall that we have three types of orders and each order has its own integrated rate law. So let's go ahead and recall those things. So we have order integrated rate law which each has its own graph and slope. So let's just discuss this. Okay so if we have a zeroth order, the integrated rate law is the concentration at time T equals negative K. T. Where K. Is the rate constant. T. Is time plus the concentration of A. O. Or initial. So that's the initial concentration. So that's for a zero order. And the graph for this is going to be the concentration of A. Versus T. And here a. Is going to be symbolic for the concentration of C. D. Okay so for the slope that means the slope is then going to be negative K. On the other hand, if we have a first order, the integrated rate law becomes Ln of a. At time T equals negative K. T. Plus Ellen of the initial concentration. And the graph is Ln of a versus time and slope is negative K. And lastly we have second order. So the integrated law for a second order is one over the concentration of a. At times T. Is equal to K. T. Plus one over the initial concentration. And the graph for this is going to be one over the concentration of A versus T. And our slope is K. So this is important information. And as we said, 80 is the concentration at time, T. K. Is rate constant, T. Is time and A. O. Is initial concentration. So what we need to do to solve this problem is we need to plot the given data using the zero first and second integrated law to determine the order of the reaction. So at the top here we have two elements. We're going to add to our graph. The two elements are we're going to calculate the Ln of C. D. Okay, and then we're also going to calculate one over the concentration of C. D. Because these values are what we're going to need so that we can plot, so let's go ahead and find these values at time. T equals zero Ln of C. D. So we're going to take the L. N. Of the concentration and the value that we're going to get is negative 3. 492. And now we're going to take one over the concentration of C. D. And this is 26.17801. So we're going to do this for all of the times. Okay, so let's go ahead and go down. So at time is 15 minutes. L N F C D becomes negative three point 39,830 and one over c. is 29.91325 at time LNFCD is negative 3.55225 and one over c. is 34.89184 Time 45. LNFCD is negative 3. 430 and one over c. d becomes 41.85852. That time 60 Ln. is negative 3. and one over C. D. Is 52.30126. And lastly at times 75 L N F C D is negative 4.24401 and one over C. D. Is 69.68641. So with this data, what we're going to do is we're going to do the graph for each order and I just highlighted what the graph is going to be. Okay. So when we do the graph for each order for our first order, we're going to get a line. Alright? So I'm going to write out what the graph line is going to be. So for zero order, when we plot this data, we get why Is equal to negative 0. x plus 0. 200. For. And this is for the zero order. Okay. For first order. What we get for our line is Y equals -0. X -3.209397. And for the second order, the line that we get is y equals zero five x plus 21.637783. Okay so the plot of the concentration of C. D. Versus time is linear. So the reaction is zeroth order. Okay so we know that the reaction is zeroth order. Alright. And what we can do then is when we write this in linear form what we get is Y. Equals M. X. Plus B. Okay and so using the integrated rate law for zero order. The integrated rate law is the concentration at time. T. Is equal to negative K. T. Plus the concentration. The initial concentration. So Kay Is equal to 3.18 times 10 to the - meters per second. And time equals 100 minutes. Okay because we're being asked to calculate the Malaria T after 100 minutes. Okay so two equals 100 minutes. So that means then the initial concentration is equal to 0.0382. Because that's what the concentration is at time is equal to zero. Okay so this is at time equals zero. Alright so now let's go ahead and calculate the concentration at time T. So we're going to get negative 3. times 10 to the negative four meters or molar per second? Times 100 min plus 0.0382 Molar. Okay so what we're going to get then as our final answer is the concentration at time T. Which time T. is 100 minutes in this case is equal to 6.4. Or let's write it as the concentration of C. D. Okay so we'll write the concentration of C. D. At time is 100 minutes is 6. zero 6.40 times 10 to the -3 Molar. So this is going to be our final answer. 6.40 times 10 to the negative three moller is the malaria T. Of C. D. After 100 minutes of reaction time. That's it for this problem. I hope this was helpful.
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