Half-life from graphs

Watch
Announcements
#1
Hey, I was wondering how to work out three successive half-lives from the following graphs (I don't know why its coming up huge). I tried reading the textbook but I'm still so confused!

Is it just t1/2= 50s i.e is half life basically the same as t1/2?
Or is the half life k = (ln 2) / 50.
Also how can you tell from the graph the order of the reaction.

Thanks.
0
4 years ago
#2
Hi. The equation k=In2/t1/2 is only used to calculate the rate of a first order reaction ONLY. you can't use this equation for any other order. This equation is used, because as you probably know, the half life remains constant in a first rate reaction, so basically the half life will always be 50s each time.

In a zero order reaction, the half life decreases as the reaction goes on.
And in a second, the half life increases when the reaction goes on.

I hope I've explained your question to some extent. Try checking out the link below:
http://www.knockhardy.org.uk/sci_htm_files/15kinet.pdf
1
#3
(Original post by greentron6)
Hi. The equation k=In2/t1/2 is only used to calculate the rate of a first order reaction ONLY. you can't use this equation for any other order. This equation is used, because as you probably know, the half life remains constant in a first rate reaction, so basically the half life will always be 50s each time.

In a zero order reaction, the half life decreases as the reaction goes on.
And in a second, the half life increases when the reaction goes on.

I hope I've explained your question to some extent. Try checking out the link below:
http://www.knockhardy.org.uk/sci_htm_files/15kinet.pdf
Thanks, I thought I had to work something out. Is the half life still the same if you have to use the starting and end mass/concentration?
0
4 years ago
#4
(Original post by alde123)
Thanks, I thought I had to work something out. Is the half life still the same if you have to use the starting and end mass/concentration?
I'm not quite sure what you're trying to say. Could you elaborate?
0
#5
(Original post by greentron6)
I'm not quite sure what you're trying to say. Could you elaborate?
The question on my worksheet says to use the graphs (the ones I attached at the top of the page) to work out three successive half-lives based on the starting and final mass for the first graph and the starting and final concentration for the second. I was wondering if this changes anything or is irrelevant.

Thanks again
0
4 years ago
#6
For 1st order, yes, the half life will be the same for each mass/conc.
But for zero, the half life will decrease and for the second order, it will increase, each time you calculate the half life as the reaction goes on. So no, the half life will not be the same for starting and final mass/concentration if it is not a first order reaction.

Therefore, for the second graph, measure the first two half lives. if it is the same then the half life of the initial and final conc will be the same, because it will be a 1st order reaction. If not, analyse whether the half life increased or decreased and carry on from there.

When measuring the half life, if the line starts for example at 50, you need to draw a line from 25, until it hits the line and draw the line down to find the time, then do the same for 12.5, then 6.25. (you may know this already but the first graph was not following this pattern, so I'm just making sure you understand).

Is this any clearer?
0
#7
(Original post by greentron6)
For 1st order, yes, the half life will be the same for each mass/conc.
But for zero, the half life will decrease and for the second order, it will increase, each time you calculate the half life as the reaction goes on. So no, the half life will not be the same for starting and final mass/concentration if it is not a first order reaction.

Therefore, for the second graph, measure the first two half lives. if it is the same then the half life of the initial and final conc will be the same, because it will be a 1st order reaction. If not, analyse whether the half life increased or decreased and carry on from there.

When measuring the half life, if the line starts for example at 50, you need to draw a line from 25, until it hit the lines and draw the line down to fine the time, then do the same for 12.5, then 6.25. (you may know this already but the first graph was not following this pattern, so I'm just making sure you understand).

Is this any clearer?
Yes - a lot clearer. Thanks!
0
4 years ago
#8
(Original post by alde123)
Yes - a lot clearer. Thanks!
No problem!
0
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

Have you experienced financial difficulties as a student due to Covid-19?

Yes, I have really struggled financially (14)
12.07%
I have experienced some financial difficulties (28)
24.14%
I haven't experienced any financial difficulties and things have stayed the same (50)
43.1%
I have had better financial opportunities as a result of the pandemic (20)
17.24%
3.45%