Question

Show a sample of the ICE table that will be used to calculate Kc. For your example, use the trial one with 2.00 mL of SCN and 5.00 mL of Fets and an arbitrary absorbance value for FeSCN+2 between 0.200 -0.400. (Note: Both reactant stock solutions are 0.00200 M) Hints to consider as you are preparing the ICE table: What is the reaction being investigated? • What is the final volume of the reaction mixture? What are the initial concentrations of SCN and Fe3+ (hint: they are not 0.00200 M) • How can the absorbance of the solution containing FeSCN2be used to calculate the [FeSCN2-) at equilibrium? Calculate the [FeSCN2.. • Use the calculated (FeSCN2) to complete your ICE table. Calculate the equilibrium constant, K. for the reaction, showing your work.

Answer 1

Equilibrium constant (Kc) for a reaction is the ratio between the multiplication of concentrations of the products, each raised to the power of their number of moles to the multiplication of concentrations of the reactants, each raised to the power of their number of moles, present at equilibrium.

Now the reaction between SCN- and Fe+3 is

SCN^- + Fe⁺³ ------------> FeSCN²⁺   

Thus here Kc = [FeSCN²⁺] / [SCN^-]*[Fe⁺³]

Now these concentration values are the concentrations at equilibrium. To find this, we have to form the ICE table

a-

Given initial concentration of Fe⁺³ (M1)= 0.00200 M

initial concentration of SCN^- (M1)= 0.00200 M

initial volume of SCN^- (V1)= 2 ml

initial volume of Fe⁺³ (V1)= 5 ml

Now after the mixing of the two, the final volume becomes- V2 = 5 ml +2 ml = 7 ml

Thus before the reaction starts ,

the new initial concentration of Fe⁺³ (M2)= M1V1/V2 = 0.00200 M * 5 ml / 7 ml = 0.0014 M

Similarly new initial concentration of SCN^- (M2)= M1V1/V2 = 0.00200 M * 2 ml / 7 ml = 0.00057 M

b-

Now lets take given the absorbance of the product FeSCN²⁺is = 0.300

Now the formula to calculate concentration from absorbance is

A = εlc where A = absorbance, ε = molar absortivity = 7.00 x 10³ L/cm.mol for FeSCN²⁺

l = path length = 1 cm

c = concentration

Now putting these values-

A = εlc

c = A/ εl

= 0.300/ 7.00 x 10³ L/cm.mol * 1 cm

= 0.0000428 L/mol

= 4.28 * 10^-5 M

That means at equilibrium, [FeSCN²⁺] = 4.28 * 10^-5 M = 0.0000428 M

c-

Thus the ICE table will be

Reaction	SCN-	Fe+3	FeSCN2+
Initial	0.00057	0.0014	0
Change	- 0.0000428	-0.0000428	+0.0000428
Equilibrium	0.0005272‬	0.0013572‬	0.0000428

d-

Thus equilibrium constant is-

Kc = [FeSCN²⁺] / [SCN^-]*[Fe⁺³]

Kc = [0.0000428] / [0.0005272‬]*[0.0013572‬] M^-1  

Kc = 59.816 M^-1