[backjames24]

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.


Thanks

[Dan]

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.

[backjames24]

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.

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

[Dan]

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?

[backjames24]

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

[Dan]

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.

[backjames24]

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

[Dan]

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[S₂O₈^2-][I^-]

In this question, we let [I₂] = x

dx/dt = k[S₂O₈^2-][I^-]

At the beginning of the reaction (t = 0), the concentration of S₂O₈^2-, or [S₂O₈^2-]₀ = a

At the beginning of the reaction (t = 0), the concentration of I^-, or [I^-]₀ = b

The concentration of S₂O₈^2- at any given time, [S₂O₈^2-], 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 [S₂O₈^2-] = (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.

[backjames24]

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[S₂O₈^2-][I^-]

In this question, we let [I₂] = x

dx/dt = k[S₂O₈^2-][I^-]

At the beginning of the reaction (t = 0), the concentration of S₂O₈^2-, or [S₂O₈^2-]₀ = a

At the beginning of the reaction (t = 0), the concentration of I^-, or [I^-]₀ = b

The concentration of S₂O₈^2- at any given time, [S₂O₈^2-], 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 [S₂O₈^2-] = (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).

[backjames24]

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

[Dan]

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.

[backjames24]

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?

[Dan]

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

[backjames24]

Last message I modified. Sorry

[backjames24]

I understand that integratingx^-1 is the same as the log of x but why is the base of the log e?