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Whalstib
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Homework Statement
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
Homework 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.
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
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