Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: backjames24 on October 04, 2014, 07:20:34 AM
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Please could somebody explain the steps being done to get to the final step in the image
S2O82− + 2 I−→ 2 SO42− + I2
Please see image below.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi58.tinypic.com%2Fdo6po7.png&hash=4e7fb40d75df35d834426c158ce5f80ef0d37a13)
Thanks
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Please identify which steps you don't understand.
Have you looked up integrated rate laws?
Have you looked up the meaning of "pseudo first order"?
You must show you have attempted the question, this is a Forum Rule (http://www.chemicalforums.com/index.php?topic=65859.0).
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Please identify which steps you don't understand.
Have you looked up integrated rate laws?
Have you looked up the meaning of "pseudo first order"?
You must show you have attempted the question, this is a Forum Rule (http://www.chemicalforums.com/index.php?topic=65859.0).
Thank you for prompt reply Dan,
I have looked up the integrated rate law and psuedo first order previously.. I understand that the (a-x) is because the concentration of the first reactant is a at t= 0, and amount reacted after time t is repreented by x, but where does (b-2x) come from and why is it k2?
Thanks
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I think the decision to label the rate constant k2 is arbitrary.
It looks to me as though x = [I2]
What is the relationship between [I-] and [I2]?
Can you write an expression for [I-] in terms of b and x?
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I think the decision to label the rate constant k2 is arbitrary.
It looks to me as though x = [I2]
What is the relationship between [I-] and [I2]?
Can you write an expression for [I-] in terms of b and x?
I am very confused now Dan, I have only begun learning kinetics recently on 1st year. The original document is here http://www.eckerd.edu/academics/chemistry/courses/ch321/labwork/documents/KineticsF07.pdf
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OK, so the document explains everything other than where the first equation came from.
Let's start with this: http://en.wikipedia.org/wiki/Rate_equation
Are you ok with the first paragraph on that page? Are you comfortable with constructing basic rate equations?
The reaction rate in your question is being expressed as the formation of I2 over time. That is the rate of change in concentration of I2 (which is called [I2] - the square brackets mean "concentration of"). [I2] = x.
Start by writing a general rate equation (like the one at the top of the Wikipedia page) for your reaction.
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rate = k [S2082-]^x [I-]^y
I am also okay with looking at the rate as loss of reactant such as - d[X] / dt
I understand that pseudo first order means that one reactant is in excess
thanks
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rate = k [S2082-]^x [I-]^y
Ok good.
your exponents x and y are determined experimentally and are given in the document you linked to. They are both 1.
So: rate = d[I2]/dt = k[S2O82-][I-]
In this question, we let [I2] = x
dx/dt = k[S2O82-][I-]
At the beginning of the reaction (t = 0), the concentration of S2O82-, or [S2O82-]0 = a
At the beginning of the reaction (t = 0), the concentration of I-, or [I-]0 = b
The concentration of S2O82- at any given time, [S2O82-], is (what you started with) - (what has reacted). (What has reacted) is proportional to (what has formed). We know that (what you started with) = a, and (what has formed) = x
Now look at the relationship between (what has reacted) and x. One persulfate reacts to give one iodine.
1(what has reacted) = 1x
(what has reacted) = x
So we can say that [S2O82-] = (what you started with) - (what has reacted)
= a - x
Does that make sense? Try repeating this process to write an expression for [I-] in terms of b and x.
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rate = k [S2082-]^x [I-]^y
Ok good.
your exponents x and y are determined experimentally and are given in the document you linked to. They are both 1.
So: rate = d[I2]/dt = k[S2O82-][I-]
In this question, we let [I2] = x
dx/dt = k[S2O82-][I-]
At the beginning of the reaction (t = 0), the concentration of S2O82-, or [S2O82-]0 = a
At the beginning of the reaction (t = 0), the concentration of I-, or [I-]0 = b
The concentration of S2O82- at any given time, [S2O82-], is (what you started with) - (what has reacted). (What has reacted) is proportional to (what has formed). We know that (what you started with) = a, and (what has formed) = x
Now look at the relationship between (what has reacted) and x. One persulfate reacts to give one iodine.
1(what has reacted) = 1x
(what has reacted) = x
So we can say that [S2O82-] = (what you started with) - (what has reacted)
= a - x
Does that make sense? Try repeating this process to write an expression for [I-] in terms of b and x.
So for iodide ions, at the start of the reaction there is b, and the two moles of iodide ions react, giving -2x. So it can be rewritten as (b-2x).
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So for iodide ions, at the start of the reaction there is b, and the two moles of iodide ions react, giving -2x. So it can be rewritten as (b-2x). What is the reason for putting k2 instead of k, no reason at all?
Excuse the double comment, I was trying to modify the previous one.
Thank you Dan
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So for iodide ions, at the start of the reaction there is b, and the two moles of iodide ions react, giving -2x. So it can be rewritten as (b-2x).
Yes. (iodide reacted) = 2(iodine formed) = 2x
What is the reason for putting k2 instead of k, no reason at all?
I think it's been given a 2 just because the reaction is second order. It's just a label.
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You have been of great help Dan, thank you for your time :)
Since it is pseudo first order, (b-2x) can just be written as (b) because it is in great excess. Then it is simply integrated.
This leads me to one last question. I think it is best to give an example:
- { d[H2] / dt }
- { d[H2] / dt } = k[H2],
d[H2] / [H2] = -kdt
In[H2] – In[H2]0 = -kt
Why is it that doing this with first-order reactions gives a log, but for zero and 2nd orders it would be [H2] = -kt + [H2]0 and [H2]-1 = kt + [A]0-1 respectively?
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You have been of great help Dan, thank you for your time :)
You're welcome.
Since it is pseudo first order, (b-2x) can just be written as (b) because it is in great excess. Then it is simply integrated
Yes. If b is in great excess, then b >> 2x and we can say that b - 2x ≈ b
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Last message I modified. Sorry :)
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I understand that integratingx^-1 is the same as the log of x but why is the base of the log e?
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This leads me to one last question. I think it is best to give an example:
- { d[H2] / dt }
- { d[H2] / dt } = k[H2],
d[H2] / [H2] = -kdt
In[H2] – In[H2]0 = -kt
Why is it that doing this with first-order reactions gives a log, but for zero and 2nd orders it would be [H2] = -kt + [H2]0 and [H2]-1 = kt + [A]0-1 respectively?
The only answer I can give to that is: because that's what you get when you integrate it. It's a result of the calculus. Sorry I can't help more with that one.
Maybe this will help:
https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/indefinite_integrals/v/antiderivative-of-x-1
https://www.khanacademy.org/math/differential-calculus/taking-derivatives/der_common_functions/v/proof-d-dx-ln-x-1-x
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It is a shame what happened to your new building, one of my choices was Nottingham.
In a book I was reading it said that
k = Ae –(EA/RT) or [ In k = In A – (EA/RT) ] can be linked to rate law,
what is the link to the In [a/(a-x)] = k’t from the document I sent?
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Yes, the fire was a blow for the department.
In a book I was reading it said that
k = Ae –(EA/RT) or [ In k = In A – (EA/RT) ] can be linked to rate law,
This won't help you here.
Do you understand how to integrate the second equation to get to the third equation?
what is the link to the In [a/(a-x)] = k’t from the document I sent?
k' is just a new constant. Since k2 and b are constant, we can combine them to make a new constant k' = k2b
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k' is just a new constant. Since k2 and b are constant, we can combine them to make a new constant k' = k2b
I see Dan,
What is happening with the building now?
So looking at the last equation we got of ln [ a / (a-x) ] =k 't , we can do [a / (a − x)] vs time where the gradient will be k'. How can we find the half-life for this or the so called relaxation time? Looking at some problems on the internet now, I am now able to understand. From the rate values the order in respect to each reactant
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I see Dan,
What is happening with the building now?
It will be rebuilt. There is a thread here (http://www.chemicalforums.com/index.php?topic=76507.0) about it.
So looking at the last equation we got of ln [ a / (a-x) ] =k 't , we can do [a / (a − x)] vs time where the gradient will be k'.
Almost, a plot of ln[a/(a-x)] vs t will hve gradient k'
How can we find the half-life for this
Hint: After the half life time has elapsed, what is the relationship between a and x?
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I see Dan,
What is happening with the building now?
It will be rebuilt. There is a thread here (http://www.chemicalforums.com/index.php?topic=76507.0) about it.
So looking at the last equation we got of ln [ a / (a-x) ] =k 't , we can do [a / (a − x)] vs time where the gradient will be k'.
Almost, a plot of ln[a/(a-x)] vs t will hve gradient k'
How can we find the half-life for this
Hint: After the half life time has elapsed, what is the relationship between a and x?
(a-1/2x) (b-x)
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Please show working/explanation, I don't understand what you did there.
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Please show working/explanation, I don't understand what you did there.
I was thinking that since
[S2O82-] = (what you started with) - (what has reacted)
= a - x
then first half life would be a-0.5x, then a-0.25x etc
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OK. Be careful, x is a varying parameter, not a constant value. a is a constant value, let's work with that.
So if half of the persulfate reacts, and you started with a, how much iodine has been produced (in terms of a)?
In other words: what is the value of x, in terms of a, at t1/2?
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OK. Be careful, x is a varying parameter, not a constant value. a is a constant value, let's work with that.
So if half of the persulfate reacts, and you started with a, how much iodine has been produced (in terms of a)?
In other words: what is the value of x, in terms of a, at t1/2?
1/2 a? so 1/2 of the elemental iodine has formed. by following the procedure in experiments in physical chemistry the starch is blue-black --> colourless when added at the end to stop it going lumpy. Why is it this way around , I would of thought the starch would form complex with the iodine at endpoint . Sorry if I am asking basic questions, I have catching up to do
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OK. Be careful, x is a varying parameter, not a constant value. a is a constant value, let's work with that.
So if half of the persulfate reacts, and you started with a, how much iodine has been produced (in terms of a)?
In other words: what is the value of x, in terms of a, at t1/2?
1/2 a?
Yes, when t = t1/2, x = 0.5a
Now calculate t1/2 using the appropriate equation.
by following the procedure in experiments in physical chemistry the starch is blue-black --> colourless when added at the end to stop it going lumpy. Why is it this way around , I would of thought the starch would form complex with the iodine at endpoint . Sorry if I am asking basic questions, I have catching up to do
Let's finish the half life question first.
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OK. Be careful, x is a varying parameter, not a constant value. a is a constant value, let's work with that.
So if half of the persulfate reacts, and you started with a, how much iodine has been produced (in terms of a)?
In other words: what is the value of x, in terms of a, at t1/2?
1/2 a?
Yes, when t = t1/2, x = 0.5a
Now calculate t1/2 using the appropriate equation.
by following the procedure in experiments in physical chemistry the starch is blue-black --> colourless when added at the end to stop it going lumpy. Why is it this way around , I would of thought the starch would form complex with the iodine at endpoint . Sorry if I am asking basic questions, I have catching up to do
Let's finish the half life question first.
What equation do you speak of? I attempted the problem with 1/ k[A]0 which i got from the integrated 2nd order law and got lost.
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The equation is written in your first post.
I am having great difficulty steering you in the right direction, because it's not clear to me what your logic is. Why and what are you doing with 1/k[A]0?
Please:
1. State exactly what you are trying to do.
2. Show what you are doing and explain the logic behind it.
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The equation is written in your first post.
I am having great difficulty steering you in the right direction, because it's not clear to me what your logic is. Why and what are you doing with 1/k[A]0?
Please:
1. State exactly what you are trying to do.
2. Show what you are doing and explain the logic behind it.
Forget about that Dan,
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi57.tinypic.com%2Fb65a4n.png&hash=e7b1057f9bbd430964f0801d8b1b8a1487c710c9)
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I agree, well done.
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I agree, well done.
So is this method only suitable for finding the rate of reaction, how about finding the activation enthalpy? The procedure looks at the amount of titrant required for colourless endpoint at 60 degrees and at room temperature
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So is this method only suitable for finding the rate of reaction, how about finding the activation enthalpy?
Well, what information would you need to know in order to calculate Ea?
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So is this method only suitable for finding the rate of reaction, how about finding the activation enthalpy?
Well, what information would you need to know in order to calculate Ea?
You would need k from the graph .. but two temperatures are used to get a and x I cannot see how I would use the equation.
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(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi59.tinypic.com%2F30tjji1.gif&hash=82c27fbbccff83df9386330308bb51350a7c063f)
Repeat with a different temperature instead of the 60 degrees celsius and use the equation above?