The tabulated data show the concentration of AB versus time
for t...

Question

The tabulated data show the concentration of AB versus time
for this reaction:
AB( g)¡A( g) + B( g)
Time (s) [AB] (M)
0 0.950
50 0.459
100 0.302
150 0.225
200 0.180
250 0.149
300 0.128
350 0.112
400 0.0994
450 0.0894
500 0.0812
Determine the order of the reaction and the value of the rate
constant. Predict the concentration of AB at 25 s.

Accepted Answer

Hey everyone in this example we're given the below data for the given reaction here and we need to determine a our order of the reaction, be our rate constant of a reaction and see the concentration of our product. Or sorry, our reactant cd at 80 seconds. So based on our given data, we wanna first plot for a zero order reaction. We're going to recall that. We should plot the concentration of our reactant C. D. At a given time versus time. For a first order reaction we want to recall that. We should plot our natural log Ln of our concentration of our reactant cd at a given time versus time. And then we should recall last for a second order reaction. We're going to plot one over our concentration of c. d. At a given time versus time. Now based on each of these plots first beginning with the zero order reaction, we would see that we get a correlation coefficient r squared equal to a value of 0.948126. Now moving on to our second plot for a first order reaction. Based on our data set, we would get a correlation coefficient r squared equal to a value of 0.98644. And then lastly we have for our second order reaction plot, we would get a correlation coefficient r squared value equal to 0.9997. And because we see that this value based on all the three values we've generated is closest to one, we would say that therefore Our reaction is 2nd order. And so this would be our first answer for part A. Where our reaction is second order. Now we want to move on to part B, which is finding the rate constant of our reaction. So based on our plot for the second order reaction, we should recognize that we generated our equation of our line where y equals our slope, which is a value of 0.01 times X added to the y intercept, which is given us 1.218 in our calculators. So we should recall that our rate constant K. Is equal to our slope. And so we would say that therefore our rate constant K. Is equal to our slope which is this value here being 0.19836. So this would be our second answer here for part B. As the value for our rate constant. And lastly we need to move on to part C. Which is finding the concentration of our reacting C. D at 80 seconds. So because we know that our reaction is second order, we should recognize that our concentration of C. D. Is going to be one over the concentration of C. D. Our reactant where we will use the equation of our line as our rate law. So we would have is equal to Y. Which we know is equal to our Y intercept 1.0 to 18 added to our slope 0.19836. Where X. We would plug in four time. So here we would say that the time 80 seconds. So we would plug in just 80 here and so we can simplify this so that we have one over our concentration of our reactant C. D. Is equal to eight .34906. And we went there four because we have this diagonal here replace our concentration of CD with this value here. So we would take or we would say that our concentration of our reactant CD is now equal to 1/ 0.34906, which is equal to 0.110 moller. So this would be our third final answer to complete this example as our concentration of our reactant Cd at 80 seconds. So I hope that everything I viewed was clear. Everything highlighted in yellow represents our three final answers. If you have any questions, just leave them down below and I will see everyone in the next practice video

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