00:01
For this next problem, it asks you to use the data that was calculated in question 13 and show that the reaction obeys the rate law.
00:09
Rate is equal to k multiplied by the concentration of m205.
00:15
This suggests its first order.
00:18
So first order suggests that the rate is directly proportional to the concentration.
00:24
So if we were to prepare a plot of the average rate over...
00:36
Or as the function of the average concentration, we should get a straight line, as suggested by this equation, where the slope will be equal to k.
00:51
So we'll take the data then that we produced in question 13, and plot it as a function of the average concentration, and sure enough, we do get a straight line.
01:03
We're then asked to evaluate the rate constant as an average of the values obtained from the six intervals, so we could have done this in excel.
01:12
If we did this in excel, then we could have it plot the equation of the line, and the equation of the line would give us a slope that would be the average of the values obtained from the six intervals.
01:25
But because we didn't do this in excel, we will have to calculate the average value of the six intervals.
01:39
So for the first interval, the value that we calculated in problem 13, was 0 .23 molar per hour.
01:50
We then divide that by the average concentration...