# Why is 1/t an estimate of the Rate of Reaction?

Watch
Announcements
#1
..
0
10 years ago
#2
Because if a reaction takes more time to complete, it's obviously a slower reaction. If a reaction takes less time to complete, then it's a fast reaction. 1/t just gives a quantitative value to comparing the rates of reaction.

i.e. if a reaction finishes in 1 second, then the rate = 1
if a reaction finishes in 3 seconds, then the rate = 1/3

Obviously the one that finished in less time is quicker, 3 times quicker, which is shown by 1/t.
2
#3
Sorry the question is why does 1/t ONLY give an estimate? So why is it not very accurate?
0
10 years ago
#4
(Original post by simoncb)
Sorry the question is why does 1/t ONLY give an estimate? So why is it not very accurate?
Ah right I see now, you're after the answer to one of the questions on the OCR Chemistry A Assessed Practicals for this year. The answer actually has nothing to do with 1/t being an estimate, it's worked out practically.
0
10 years ago
#5
frequency=1/t , hence frequency is the number of oscillations per second. Therefore as the frequency increases, the number of time decreases. This is because their would be less successful collisions between the reacting particles.
1
8 years ago
#6
(Original post by Noble.)
Ah right I see now, you're after the answer to one of the questions on the OCR Chemistry A Assessed Practicals for this year. The answer actually has nothing to do with 1/t being an estimate, it's worked out practically.
Hi, I'm really confused about the 1/t business. I'm doing the iodine clock reaction for my A2 chemistry coursework, and I thought that in order to work out the rate I would need to draw a concentration-time graph, draw a tangent to work out the rate and then from there draw a rate-concentration graph to check the order of the reaction... or have I just complicated things beyond comprehension
0
8 years ago
#7
I believe that 1/T gives you an average rate for a reaction. This is because T is the overal time for the reaction to finish however, the formula doesnt take into account that the rate of reaction is likely to be quicker at the start rather than the end (assuming its 1st/2nd order)

.....having said that im no expert so i might be wrong.
2
8 years ago
#8
I'd expect the last post to be right, they'd be less atoms left to react therefore the RoR would be less. so as 1/T is the time for all the reactions it won't give the right RoR for a certain point
0
8 years ago
#9
The real formula of rate of reaction is mols produced divided by time. In an iodine clock reaction the same amount of mols are produced every time so to simplify the calculation you can cancel out the mols produced on top of the reaction and leave it as 1/t and this will let you compare rates. But if you are planning to work out the rate constant ,k, or other reaction kinematics it can lead to problems, if I remember correctly.
2
8 years ago
#10
k, pka, kw, kc etc should all just disappear. No-one likes them.
0
7 years ago
#11
May I ask, what is the unit for the initial rate of reaction (1/T)? Is itmoldm-3s-1 or s-1 ? Thanks
0
7 years ago
#12
(Original post by TTNN)
May I ask, what is the unit for the initial rate of reaction (1/T)? Is itmoldm-3s-1 or s-1 ? Thanks
T is a time, so 1/T must have units of s-1.
0
1 year ago
#13
Stop jokes guys
0
1 year ago
#14
Stop jokes guys
0
11 months ago
#15
It is an approximation for the rate of reaction as it doesn't include concentration.
1
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### If you do not get the A-level grades you want this summer, what is your likely next step?

Take autumn exams (53)
43.8%
Take exams next summer (23)
19.01%
Change uni choice through clearing (29)
23.97%
Apply to uni next year instead (13)
10.74%
I'm not applying to university (3)
2.48%