Question

For an equilibrium constant lab.

2. If the reaction in question 1 is correct, the concentrations of Fe3+ at equilibrium in each sample would be the same as you calculated in part 2 of the calculations section. But the values for the equilibrium concentrations of SCN- would change. Show how they would be calculated.

3. Look at the % deviation section of the calculations table. Choose the two samples with the least % deviation. Write the values from the table for the [SCN-]eq for these two: 5.3410 x 10^-4 M, 7.0900 x 10^-4M. Use the equation you found in question 2 to show what these values would be if the reaction equation in question 1 is the actual equation for this equilibrium. (Show your work.)

Answer 1

The reaction in question 1, according to your comment is:

Fe³⁺ + 2SCN^- --------> Fe(SCN)₂⁺

It can't be as you put in your second coment because the equation would not be balanced in charges. The equation for Kc would be:

Kc = [Fe(SCN)₂⁺] / [Fe³⁺] [SCN^-]²

For part 2, Assuming this reaction is correct, then the values of SCN will change, basically because we are changing the number of moles reacting, and the contribution that is lost is 2x instead of x. we also know that number of moles relation would be:

[SCN] / [Fe] = 2/1 ----> [SCN] = 2[Fe]

So, to get the concentration in equilibrium, all you have to do is the substract of both concentrations:

[SCN]eq = [SCN]i - 2[Fe]i

Now, to do part 3, I need the values of Fe innitial to get the value of SCN at equilibrium, but use the expression I put, and calculate them.

Hope this helps