# Homework Help: Beer's Law

1. Jul 4, 2010

### Whalstib

1. The problem statement, all variables and given/known data
Hi,

I've run into a problem and fear I'm just approaching from the wrong angle.

From Ebbing Experiments in General Chemistry 9e 14B if you have it to reference..

It's the classic Fe + SCN > Fe(SCN)

In our experiment we are using SCN as the limiting reactant and overwhelming amts of Fe so that the initial [SCN] should equal the final [Fe(SCN)] at equalibrium. We are using a spectophotometer to measure absorbance at 450nm

2. Relevant equations
Our Keq is obtained by: [Fe(SCN)]/[Fe] [SCN] resulting in m=∑xy/∑x^2 ; K = 2.704 10^4

Our fist 5 series we varied [SCN] keeping [Fe] constant.

6-10 we varied [SCN] and keeping [Fe] to a different constant.

11-15 we varied [SCN] and keeping [Fe] to yet a different constant.

3. The attempt at a solution

So we have 3 constant [Fe].

Since the Keq = [Fe(SCN)]/[Fe] [SCN] and [Fe(SCN)] =[ SCN] the results should be 1/[Fe]...right?

[Fe] in 5-10 = 3.57E-4 ..... inv= 2.8E-3 so K = 2.8E-3
[Fe] in 5-10 = 7.14E-4 ..... inv= 1.4E-3 so K = 1.4E-3

Using Beer's Law A=kc I have from for example #5 A=.216 (@450nm) K = 2.8E-3
Solving for c [Fe(SCN)]? gives me 605 moles!

Subsequent equations yield equally baffling results (to me baffling!)

Via plotting data from the first 5 experiments in Excel and crunching number in a linear regression analysis formula:

Since m=∑xy/∑x^2; K = 2.704 10^4 so my k values are reasonable but the 605 moles is not! ...(is it?)

What am I not seeing here?

Perhaps I'm not grasping what the "c" in A=kc is for as I already know the Keq for [FE(SCN)] and in fact ALL my concentrations.....and A's and K's.........

Thanks
Whalstib

Last edited: Jul 4, 2010
2. Jul 6, 2010

### chemisttree

No. Given the initial concentration of thiocyanate is equal to [SCNi], let x equal to the amount of [SCN] converted to [Fe(SCN)]. The concentration of thiocyanate in the equilibrium expression is therefore [SCNi-x] and the concentration of [Fe(SCN)] you measure is simply 'x'. Substitute these values into your equilibrium equation and solve for x.