Question

Suppose that the reaction of Fe3+ and SCN– produces Fe(SCN)2+. 5.00 mL of 2.0 mM Fe3+ (aq) is mixed with 5.00 mL of 2.0 mM SCN– (aq). The student finds the equilibrium concentration of Fe(SCN)2+ to be 0.3 mM.

1. Write a balanced chemical equation for this reaction in solution.

2. Write an equilibrium constant expression for the reaction.

3. What is the initial number of moles of each species present?

4. What is the equilibrium number of moles of each species present?

5. What is the equilibrium concentration of each species present?

6. What is the value of the equilibrium constant?

Answer 1

1. The balanced equation can be written as

Fe³⁺ + SCN^- <======> Fe(SCN)⁺²

2. Equilibrium constant is written as

K_eq = [Fe(SCN)⁺² ]/[Fe³⁺][SCN^-]

3. Initial number of moles

Molarity = moles/volume

moles = molarity x volume

moles = 2 x 10^-3 /L 0.005 L

moles = 1 x 10^-5 moles

The same value comes for SCN^- (Since the values are same)

4. For calculating the equilibrium number of moles of Fe(SCN)⁺² = Molarity x volume

= 3 x 10^-4 x 0.01 = 3 x 10^-6

Molarity is converted from 3.0mM to 3 x 10^-4 moles and total volume is 10 ml (5ml of Fe³⁺ and 5ml of SCN^-

equilibrium number of moles Fe⁺³ = initial moles of Fe⁺³- Equilibrium moles of FeSCN⁺²

= 1 x10^-5 - 3 x10^-6 = moles

5. concentration of Fe⁺³= moles of Fe⁺³ /volume = 0.7x10^{-5 /}0.01 = 0.7x10^-3

Same with SCN^-

6. K_eq = [Fe(SCN)⁺² ]/[Fe³⁺][SCN^-] = 3 x 10^-4 / 0.7x10^{-3 x} 0.7x10^{-3 =} 612