Top Posters
Since Sunday
6
s
3
3
d
3
s
2
c
2
G
2
y
2
t
2
2
k
2
j
2
This topic contains 1 attachment.
Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
New Topic  
Doctor-2-B Doctor-2-B
wrote...
Go to Answer
Donated
Valued Member
Posts: 1634
Rep: 476 0
7 years ago
The slope of the graph of ln(k) vs 1/T represents -Ea/R so when you multiply by R (8.314), you get Ea. I just can't figure out why the Ea is actually (incorrectly) higher for the catalyzed reaction? Can someone help guide me?

I mean at (approximately) the same temperatures between uncatalyzed and catalyzed reactions, the catalyzed is MUCH faster, so the Ea should definitely follow suit and be correct.

Any help would be appreciated.

Spreadsheet located at http://expirebox.com/download/a24438f1b8786e84c7dbd64c828e4322.html
Read 320 times
3 Replies
Pretty fly for a SciGuy
Replies
wrote...
#1
Subject Expert
7 years ago
Hi,

I actually re-do your work a few times, but still come up with the same answers, so I don't think it's calculation errors. You're right that the Ea should be much smaller for the catalyzed reaction. One of the explanations off the top of my mind that can account for this mismatch of observation and calculation is human error. Regarding the calculated Ea for catalyzed reaction is higher than the uncatalyzed ones, the slope of the graph is based on the differences in y and x-axis: (y-y)/(x-x), so this means that what matters is the differences between the y axis and the x axis of each data set and not how big the number from one data set compared to other numbers from different data set. At this point, I would recommend you to ask your TA for what to do next, since each TA grades differently. In the attachment is one of the example about lab that I found online. I apologized that I couldn't be more helpful this time.
 Attached file 
(261.14 KB)
You must login or register to gain access to this attachment.
Doctor-2-B Author
wrote...
#2
Donated
Valued Member
7 years ago
I'm guessing the times are just wrong maybe? Thank you for taking a look, btw.

Quote from: Laser_3 (7 years ago)
Hi,

I actually re-do your work a few times, but still come up with the same answers, so I don't think it's calculation errors. You're right that the Ea should be much smaller for the catalyzed reaction. One of the explanations off the top of my mind that can account for this mismatch of observation and calculation is human error. Regarding the calculated Ea for catalyzed reaction is higher than the uncatalyzed ones, the slope of the graph is based on the differences in y and x-axis: (y-y)/(x-x), so this means that what matters is the differences between the y axis and the x axis of each data set and not how big the number from one data set compared to other numbers from different data set. At this point, I would recommend you to ask your TA for what to do next, since each TA grades differently. In the attachment is one of the example about lab that I found online. I apologized that I couldn't be more helpful this time.
Pretty fly for a SciGuy
Answer accepted by topic starter
Laser_3Laser_3
wrote...
#3
Subject Expert
Posts: 303
Rep: 97 0
7 years ago
Sign in or Sign up in seconds to unlock everything for free
1

Related Topics
New Topic      
Quick Reply
Explore
Post your homework questions and get free online help from our incredible volunteers
  1153 People Browsing
Start New Topic Take the Tour CoursesNew Study Tools Topics Trending Browse by Textbook
Live Chat
Related Images
  
 288
  
 324
  
 183
Your Opinion
Which 'study break' activity do you find most distracting?
Votes: 820
Previous poll results: What's your favorite coffee beverage?