See the BOLD part below. AND IMPORTANT NOTE - Typically the reaction creates thi
ID: 1022583 • Letter: S
Question
See the BOLD part below. AND IMPORTANT NOTE - Typically the reaction creates thiocyanate iron (III) ions, NOT dithiocyanate (iron (III) ions.
Under certain conditions, the product of the reaction between iron (III) ions and thiocyanate ions is the dithiocyanate iron (III) ion, Fe(SCN)2+.
Write an equation for the reaction in which dithiocyanate iron (III) ions are produced.
Fe3+ + 2SCN- <=> Fe(SCN)2+
Write the equilibrium-constant expression for the equation written in part A.
Kc = [Fe(SCN)2+]/[Fe3+] • [SCN-]2
Explain how your experimentally determined values of Kc (which were determined based on assuming no dithiocyante iron (III) was formed) would be affected if this reaction (above chemical equation) occurred to any significant extent in your experiment instead of assuming only thiocyanate iron (III) was produced.
Explanation / Answer
First, let us state the two competing equations
Equation 1
Fe3+ + 2SCN- <=> Fe(SCN)2+
Equation 2
Fe3+ + SCN- <=> Fe(SCN)+2
Note that if we assume Equation 1 is not happening, then Fe3+ is not being consumed
Note that the Kc expressions will have both [Fe3+] ions and [SCN-]
the producto coefficient, that is Fe(SCN)2+ and Fe(SCN)+ are going to change
then
Kc1 and Kc2 are dependent of each other, since Fe+3 and SCN- are in both cases.
The only thing that will let us differ from each constant is the actual value of the products, Fe(SCN)2+ and Fe(SCN)+
If we assumed that no side reaction took place, then we get:
- lowe reactants
- higher products
- the overall effect is an increase in Kc
If we actually got a side reaction, then
- lower reactants
- lower products
- Lower Kc due to the fact that lowering products will lower the "calculated" Kc (which is wrong)