[CHM 112] Question

Sun Jul 6 04:27:52 EDT 2014

On 07/05/14 16:14,  wrote:
> Hello Dr.Pounds,
>
> For the additional chat. 14 problems number 5, what is the formula you used for the overall order of reaction?
> And this is different than the book example, where it showed us how to divide 2 by 1 and 3 by 2 to get ratios.
> Could you please explain further?
>
> Thank you

As  I said in class, my methods differ from the book when it comes to 
finding the orders of reactions when one is given data.  I tend to use 
logarithms to solve for the orders with respect to different components.

For example, in the additional problem 5 that you mention, the rate law 
is going to be something like

if A = $\mathrm{HgCl_2}$ and B is $\mathrm{C_2O_4^{2-}}$ then the rate 
law looks like

$\mathrm{rate = k [A]^m [B]^n}$

  To find the order with respect to $\mathrm{[HgCl_2]}$ I have to use 
the concentrations and rates for two experiments where the concentration 
of $\mathrm{[HgCl_2]}$ was changing but the concentration of 
$\mathrm{C_2O_4^{2-}}$ was constant (experiments 2 and 3 in this case). 
   Similarly, to find the order with respect to $\mathrm{[C_2O_4^{2-}]}$ 
I have to use experiments where the $\mathrm{C_2O_4^{2-}}$ concentration 
was changing and the $\mathrm{[HgCl_2]}$ was constant. I chose to use 
experiments 3 and 4 (but I could have also used experiments 1 and 2).

The derivation of this method is found on page 20-21 of the lecture 
slides and should also be applicable to the book problems.
By taking a ratio of the rates

$\mathrm{\frac{rate_1=k [A]_1^m [B]_1^n}{rate_2=k [A]_2^m [B]_2^n}}$

and for example, using a case where the $\mathrm{[B]_1}$ and 
$\mathrm{[B]_2}$  concentrations are the same (and also noting that the 
rate constant, k, has to be the same in both equations) the ratio 
equation simplifies to:

$\mathrm{\frac{rate_1=[A]_1^m }{rate_2= [A]_2^m}}$

Mathematically this is equvalent to

$\mathrm{\left(\frac{rate_1}{rate_2}\right)=\left(\frac{[A]_1}{[A]_2}\right)^m}$

Using the fact that $a=b^n$ is the same as $\ln a = n \ln b$ , we can 
rewrite the equation as

$\mathrm{\ln \left(\frac{rate_1}{rate_2}\right)=m \ln 
\left(\frac{[A]_1}{[A]_2}\right)}$

or, upon rearrangement

$\frac{\mathrm{\ln \left(\frac{rate_1}{rate_2}\right)}}{\ln 
\left(\frac{[A]_1}{[A]_2}\right)}=m }$

I repeat the procedure to solve for the order with respect to the other 
component and then solve for $k$.

  If you find a problem in the book that is set up this same way and 
cannot be solved with this method then I would like to know about it.

-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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