Question

5. A student does an experiment to determine the equili but at a higher temperature. ermine the equilibrium constant for the same reaction that you will study, Fe** (aq) + SCN (aq) FeSCN2+ (aq) The student prepares solution 1, by mixing 5. s Solution, by mixing 5.00 ml of 0.002 M Fe(NO2)2 solution with 5.00 mL of .00011M KSCN solution, heats the mixture, and finds that the equilibrium concentration of F and finds that the equilibrium concentration of FeSCN2+ in the mixture is .00005M. Calculate the equilibrium constant for the reaction under the conditions in this expo r the reaction under the conditions in this experiment and answer the following questions. in solution 1. the initial concentration of Te+ is about 1800 times the initial SCN concentration, we can assume that virtually all of the thinevanate is converted to Fe(SCN). Therefore, the equilibrium concentration of Fe(SCN) is equal to the initial concentration of SCN. The initial concentration of the KSCN, in the reaction solution needs to be calculated since SmL of two solutions are mixed, by using, MiVi = M V The final volume is 10mL. (a) Calculate the initial concentration of KSCN in solution 1. b. So what is the equilibrium concentration of Fe(SCN)?" based on the above calculation? The absorbance at 447 nm is due completely to Fe(SCN)". (Use Lambert Beer's law to calculate. A-cle) What is the absorbance in solution 1 at 447 nm (ife - 29000 and 1cm cell is used)? What is the concentration of Fe(SCN) in solution 1 if absorbance is 0.455?

Answer 1

The given reaction is Fe⁺³ + SCN^- -------------> Fe(SCN⁾²⁺   

Now we get this Fe⁺³ from Fe(NO₃)₃ as Fe(NO₃)₃ ----------------> Fe⁺³ + 3NO₃^-

Again we get SCN^- from KSCN as KSCN ----------------> K⁺ + SCN^-

Thus we can write euqaation 1 as-

Fe(NO₃)₃ + KSCN -------------> Fe(SCN⁾²⁺   

That means to get 1 mole of Fe(SCN⁾²⁺ we have to react 1 mole of Fe(NO₃)₃ and 1 mole of KSCN

a-

Now given KSCN taken = concentration (M1) = 0.00011 M, volume (V1) = 5.00 ml

Similarly Fe(NO₃)₃ taken = concentration (M1) = 0.002 M, volume (V1) = 5.00 ml

Thus after mixing the two, the final volume = V2 = 5.00 ml + 5.00 ml = 10 ml

Then after mixing the two, the new initial concentration of KSCN (M2) = M1V1/V2 = 0.00011 M* 5.00 ml / 10 ml = 0.000055 M

And the new initial concentration of Fe(NO₃)₃ (M2) = M1V1/V2 = 0.002 M* 5.00 ml / 10 ml = 0.001 M

b-

Again the given Fe(SCN⁾²⁺ at equilibrium = 0.00005 M

Thus to find the equilibrium constant (Kc), we have to form the ICE table-

Reaction	Fe(NO3)3	KSCN	Fe(SCN)2+
Initial	0.001 M	0.000055 M	0M
Change	-0.00005 M	-0.00005 M	0.00005 M
Equilibrium	0.00095‬ M	0.000005‬	0.00005 M

Now the equilibrium constant (Kc) is the ration between product of concentration of products to the product of concentration of reactants at equilibrium. i.e for the reaction

Fe(NO₃)₃ + KSCN -------------> Fe(SCN⁾²⁺

Kc = [Fe(SCN⁾²⁺] / [Fe(NO₃)₃] * [KSCN]

Now putting the values from ICE table-

Kc = [Fe(SCN⁾²⁺] / [Fe(NO₃)₃] * [KSCN]

Kc = [0.00005] / [0.00095‬ ] * [0.000005]

=10526

c-

According to Beer's law- A = ECl  

where A = absorbance, E = specific absorptivity, C = concentration of solution, l = length of cell

Now from the above calculation, we have found at equilibrium, C = [Fe(SCN⁾²⁺] = 0.00005 M

Given E = 29000, l = 1 cm

Thus putting the values-

Absorbance (A) = ECl  

= 29000 * 0.00005 M * 1 cm

= 1.45 L mol^-1 cm^-1

d-

Here given Absorbance (A) = 0.455 L mol^-1 cm^-1 , E = 29000, l = 1 cm

Then  Absorbance (A) = ECl  

C = A / El

= 0.455 L mol^-1 cm^-1 / 29000 * 1 cm

= 0.000015 L mol^-1