Drinking water may contain small amount of some contaminants that are harmful to living organisms. I — Organic Chemistry Chemistry Question
Iodine equilibrium
Drinking water may contain small amount of some contaminants that are harmful to living organisms. Iodine is used as a disinfectant for drinking water for the International Space Station Alpha. Aqueous I2 forms a number of inorganic derivatives, such as hypoiodous acid, HOI; iodate, IO3 – ; iodide, I– and triiodide, I3 –. An equilibrium reaction takes place involving I2, I – and I3 – in water according to the following equation;
I2(aq) + I–(aq) <=> I3 –(aq)
When dichloromethane, CH2Cl2 is added to aqueous solution of iodine, I2 is distributed in water and CH2Cl2 phases according to the following equilibrium process. The equilibrium constant for the distribution is 150.
I2(aq) <=> I2(CH2Cl2)
For the homogenous equilibrium reaction which species acts as a Lewis acid?
Model Answer
I2
One method for determining the concentration of I2 and I3 – in a solution is the titration with a standard solution of S2O3 2–. An oxidation-reduction reaction takes place when I2 or I3 - interacts with S2O3 2– yielding I– and S4O6 2–. Write the balanced equations for chemical reactions that take place during the titration of I2 and I3 – with S2O3 2–. Indicate the oxidant and the reductant in each reaction? Give the oxidation state of S in Na2S2O3.
Model Answer
2 S2O3 2– + I2 → S4O6 2– + 2 I–
2 S2O3 2– + I3 – → S4O6 2– + 3 I–
Oxidants: I2 and I3 –
Reductant: S2O3 2–
Formally judged the oxidation number of S in S2O3 2– is II. Taking into consideration the structure of S2O3 2– anion, one atom of sulphur (central) has the oxidation number VI while the other one has oxidation number –II.
In order to determine the equilibrium constant of the reaction involving I2, I – and I3 – in water the following experiments are performed at 298 K. When 50.0 cm3 of KI aqueous solution (c = 0.010 mol dm-3) is added to 25.0 cm3 solution of I2 in CH2Cl2, two separate phases, aqueous and organic, are formed. Assume that there is no volume change upon mixing. In order to determine concentrations of I2 distributed in CH2Cl2 and aqueous phases, a 5.00 cm3 aliquot of the CH2Cl2 phase is diluted to 100.0 cm3 by addition of the solvent, CH2Cl2. The visible spectrum of I2 in the diluted solution, recorded in a 1.00 cm-cell, had a band with a maximum absorbance of 0.516 at 510.0 nm. The molar absorption coefficient, ε of I2 in CH2Cl2 at 510 nm is 858 dm3 mol–1 cm–1. Calculate equilibrium concentrations of I2 in CH2Cl2 and aqueous phases.
Model Answer
I2(aq) <=> I2 (CH2Cl2)
Kd = [I2 (CH2Cl2)] / [I2 (aq)] = 150
A = [I2(aq)] ε l
[I2 (CH2Cl2)] = 0.516 / (858 × 1.00) = 6.01 · 10^–4 mol dm^–3
Before dilution of 5 cm3 solution to 100 cm3:
[I2 (CH2Cl2)]eq = 6.01 · 10^–4 × 100 / 5 = 1.20 · 10^–2 mol dm^–3
[I2 (aq)]eq = [I2 (CH2Cl2)]eq / Kd = 1.20 · 10^–2 / 150 = 8.02 · 10^–5 mol dm^–3
In order to determine the equilibrium concentrations of I– and I3 –, a 25.0 cm3 aliquot is taken from the aqueous phase. To this solution, an excess amount of KI, namely 10.0 cm3 of KI solution with c = 0.100 mol dm-3, is added to avoid evaporation of I2. Then, the final solution is titrated with a solution of Na2S2O3 (c = 0.0100 mol dm-3). The end point is reached upon addition of 3.10 cm3 of Na2S2O3 solution. Calculate the equilibrium concentrations of I–, and I3 – in the aqueous phase and the equilibrium constant at 298 K.
Model Answer
Aqueous equilibrium process is:
I2(aq) + I– (aq) <=> I3 -(aq)
[ I- ]eq = 0.0100 – [I3 – (aq)]eq
After the addition of excess KI all I2(aq) is converted to I3 –(aq). Thus,
[I3 –(aq)]total = [I3 –(aq)]eq + [I2(aq)]eq
n(I3 -) = 1/2 n(S2O3 2-) = 1/2 × 3.10 × 0.0100 = 1.55 · 10^–2 mmol
[I3 -(aq)]total = 1.55 · 10^–2 / 25.0 = 6.20 · 10^–4 mol dm^–3
[I3 -(aq)]eq = [I3 -(aq)]total – [I2 (aq)]eq
= 6.20 · 10^–4 – 8.02 · 10^–5 = 5.40 · 10^–4 mol dm^–3
[I-]eq = 0.0100 – 5.40 · 10^–4 = 9.46 · 10^–3 mol dm^–3
K = [I3 -(aq)]eq / ([I2 (aq)]eq × [I-]eq)
= 5.40 · 10^–4 / (8.02 · 10^–5 × 9.46 · 10^–3) = 712
Calculate ∆fG°[I2(CH2Cl2)], if ∆fG°[I2(aq)] is 16.4 kJ mol–1.
Model Answer
∆G° = – RTlnK = ∆Gf ° [I2(CH2Cl2)] – ∆Gf° [I2(aq)]
= –298 × 8.314 × ln(150) = ∆Gf° [I2(CH2Cl2)] – 16.4 · 10^3
∆Gf °[I2(CH2Cl2)] = 3985 J mol–1 = 3.985 kJ mol–1 = 4.00 kJ mol–1