J.R. S. answered • 02/02/24
Ph.D. University Professor with 10+ years Tutoring Experience
For a 1st order reaction,
ln[A] = -kt + ln[A]o
If you plot the data that is provided as ln [N2O] on the y-axis, and time (sec) on the x-axis, you will get a straight line of the form y = mx + b. It will be a straight line with a negative slope, and thus, slope = -k.
So, you can plot the data and get a linear regression equation and solve for k using slope = -k.
Since I'm not going to plot the data, we will do it here as a "guesstimate".
We'll take the change in y as the difference between ln 0.0630 M and ln 0.250 M = -1.37 M
The corresponding ∆x = 600 min - 0 min = 600 min x 60 sec/min = 36000 s
Slope =-k = ∆y / ∆x = -1.37 / 36000
-k = -3.8x10-5 s-1
k = 3.8x10-5s-1
J.R. S.
02/02/24
J.R. S.
02/02/24
Constantine S.
I see, that's what I thought, there was some difference in the standardized calculations that I did vs those they did when doing it by graphing. However, your answer helped me realize that at least my method was correct. Thanks again! Hopefully the exam questions are more straightforward.02/02/24
Constantine S.
J.R. S., This is exactly the answer I got! I also took the slope from the last point to the first, knowing the reaction was first order. Yet, the online homework that I was doing marked me wrong multiple times and said the correct answer was actually 3.92⋅10^−5 s−1. So if the correct answer has to be in seconds, then you just divide by 60 since our provided time is in minutes, getting 3.8*10^-5. So how in the world did they get 3.92? Does the time conversion have to happen earlier. I've racked my brain about this problem for a few days, and I can't tell how to get to that answer. If you can that would be awesome, but no worries if not, because at least I know I did the method correctly.02/02/24