Question

In: Chemistry

Using the initial concentrations of [Fe(H2O)6]^3+, SCN^-, and the equilibrium concentration of the [Fe(H2O)5SCN]^2+ complex, calculate...

Using the initial concentrations of [Fe(H2O)6]^3+, SCN^-, and the equilibrium concentration of the [Fe(H2O)5SCN]^2+ complex, calculate the equilibrium concentrations of both [Fe(H2O)6]^3+ and SCN^-. All mol/L. Ice table is needed.

Initial concentrations of [Fe(H2O)6^3+: Initial concentrations of SCN: [Fe(H2O)5SCN]^2+ equilibrium:

Fe(H2O)6 - SCN - EQ Fe(H2O)5SCN

0.390 1.54 23.42

0.390 3.12 35.99

0.390 4.84 58.09

0.390 6.25 74.81

0.390 7.96 98.97

Solutions

Expert Solution

In order to do this, let's write the overall reaction:

[Fe(H2O)6]3+ + SCN- --------------> [Fe(H2O)5SCN]2+ + H2O

Now, let's do an ICE chart in general:

r: [Fe(H2O)6]3+ + SCN- --------------> [Fe(H2O)5SCN]2+ + H2O

i: 0.390 1.54 0

c: -x -x +x

e: 0.390-x 1.54-x x

As we already have the concentration in equilibrium of the product, this value is the value of "x" here in the ICE chart. So, all you have to do is substract the value of concentration in equilibrium of the complex with the innitial concentration. Do this for each of the measurement and you'll get the final results and answers.

NOTE: the equilibrium concentration of the product is way much bigger than the innitial concentrations, according to the data you provide, so, when you do the substract, you'll get a negative number, and this cannot be. So, check again your data, because equilibrium concentration cannot be bigger than the innitial concentrations of reactants. Follow the procedure once you correct the values.

Hope this helps.


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