Purely theoretical question on integrated rate laws

[DatLemonDoe]

Hi

There is a big hole in my understanding of integrated rate laws, and I've been having a hard time finding what it is that i'm not understanding. I would assume that it's a very small idea that i'm missing. Anyway, here's my problem, I would assume that I will say some false statement at some point, hence me getting weird results.

Say you have the reaction A + B --) C    And let's assume that the global order of the reaction is 2.

We are also given the initial concentration of A aswell as the contentration of A at a certain point in time (t)

The integrated rate law would be 1/[A]=1/[A]0+kt.

We can use that equation to find a value for k, since we know every other variable.

Now,  here's where I get lost. Say we do a second experiment where the intial concentration of B is different, but the initial concentration of A remains the same. We once again find the concentration of A at the sime point in time (t)

If we once again find the value of k in the integrated rate law, won't it's value be different? Changing the initial concentration of B has had for effect to change the rate of the reaction, and therefore the concentration of A at the same point in time is no longer the same. However, t and [A]0 are still the same.

Now I know that  the value of k doesn't change if you change the concentration of a reactant, so I'm confused as to why I get those results.

Hopefully It wasn't too confusing. If it was, ask me questions and I will gladly clarify.

Thanks a lot, I appreciate your time and kindness.

 

[studiot]

Good morning, datlemondoe and welcome to SF.

The second order integrated rate law is only as you state if the concentration of A is equal to that of B, otherwise it is more complicated as follows.

If    [math]\left[ A \right] = \left[ B \right][/math]

Then

[math]rate =  - \frac{{d{{\left[ A \right]}_t}}}{{dt}} = k\left[ A \right]\left[ B \right][/math]

But since      [math]\left[ A \right] = \left[ B \right][/math]  we have

[math]rate =  - \frac{{d{{\left[ A \right]}_t}}}{{dt}} = k{\left[ A \right]_t}^2[/math]

On integration

[math]\frac{1}{{{{\left[ A \right]}_t}}} = {\frac{1}{{\left[ A \right]}}_0} + kt[/math]

 

Which is the expression you have.

 

However if   [math]\left[ A \right] \ne \left[ B \right][/math]

 

Then

[math]rate =  - \frac{{d{{\left[ A \right]}_t}}}{{dt}} = k\left[ A \right]\left[ B \right][/math]

We cannot replace the and the integration is more difficult.

The result is

[math]kt = \frac{1}{{{{\left[ A \right]}_0} - {{\left[ B \right]}_0}}}\ln \frac{{{{\left[ A \right]}_t}{{\left[ B \right]}_0}}}{{{{\left[ A \right]}_0}{{\left[ A \right]}_t}}}[/math]

But k remains the same constant.

Does this help?

 

 

 

 

 

 

 

Edited March 18, 2018 by studiot

[DatLemonDoe]

Yes, that makes a lot of sense. 

Thank you very much, have a great day