Sulfuryl dichloride (SO2Cl2) is a compound of industrial, environmental and scientific interest and — Physical Chemistry — Thermodynamics Chemistry Question
Decomposition kinetics of sulfuryl dichloride
Sulfuryl dichloride (SO2Cl2) is a compound of industrial, environmental and scientific interest and widely used as chlorinating/sulfonating agent or as component of the catholyte system in batteries. At room temperature, SO2Cl2 is a colorless liquid with a pungent odor; its boiling point is 70 °C. It decomposes to SO2 and Cl2 when heated to or above 100 °C.
SO2Cl2(g) → SO2(g) + Cl2(g)
An empty container is filled with SO2Cl2. Its decomposition to SO2 and Cl2 is followed by monitoring the change in total pressure at 375 K. The following data are obtained.
[VISUAL]
Time (s) 0 2500 5000 7500 10000
ptotal (atm) 1.00 1.05 1.105 1.152 1.197
By graphical approach, show that the decomposition is a first order reaction and calculate the rate constant at 375 K.
Model Answer
SO2Cl2(g) → SO2(g) + Cl2(g)
ptotal = 1.0 − x + x + x = 1.0 + x
pSO2Cl2 = 1 – x
After 2500 s: x = 0.053, pSO2Cl2 = 0.947 atm
After 5000 s: x = 0.105, pSO2Cl2 = 0.895 atm
After 7500 s: x = 0.152, pSO2Cl2 = 0.848 atm
After 10000 s: x = 0.197, pSO2Cl2 = 0.803 atm
Time(s) p(SO2Cl2) ln(p)
0 1.000 0.000
2500 0.947 –0.05446
5000 0.895 –0.11093
7500 0.848 –0.16487
10000 0.803 –0.2194
Since lnp vs time plot is linear, decomposition reaction is first order.
[VISUAL]
Rate constant from the slope is 2.2 ⋅ 10–5 s–1
When the same decomposition reaction is carried out at 385 K, the total pressure is found to be 1.55 atm after 1 hour. Calculate the activation energy for the decomposition reaction.
Model Answer
At 385 K: ptotal = 1.55 atm
1.55 = 1.0 + x ⇒ x = 0.55
pSO2Cl2 = 0.45 atm
ln(1.00 / 0.45) = k * 3600
k = 2.2 ⋅ 10–4 s–1
Using Arrhenius equation:
ln(k2 / k1) = ln(2.2 ⋅ 10^-4 / 2.2 ⋅ 10^-5) = (Ea / 8.314) * (1/375 - 1/385)
Ea = 276 kJ
There will be a negligible amount of SO2Cl2(g) in the reaction vessel after a long period of time. Therefore, the content of the vessel might be considered to be a mixture of SO2 and Cl2 gases. SO2(g) is separated from Cl2(g) as H2SO4 and Cl2(g) is used to construct a Cl2/Cl– electrode. This electrode is combined with a Cu2+/Cu electrode to make a Galvanic cell. Which electrode is the cathode?
E°(Cu2+/Cu) = +0.36 V and E°(Pt/Cl2, Cl–) = +1.36 V
Model Answer
Cu2+/Cu electrode ⇒ anode
Pt/Cl2, Cl– electrode ⇒ cathode
E°cell = 1.36 - 0.36 = 1.00 V
Calculate the ∆G° for the cell reaction given in 13.3.
Model Answer
∆G0 = –2 × 96485 × 1.00 = –1.93 ⋅ 10^2 kJ
A possible way for separating SO2 and Cl2 from each other is to pass the mixture over solid CaO which will convert all SO2 to CaSO3, a strong electrolyte. Calculate the pH of a CaSO3 solution when c = 0.020 mol dm-3. For H2SO3 Ka1 = 1.7 ⋅ 10-2 and Ka2 = 6.4 ⋅ 10-8.
Model Answer
SO3^2– + H2O ⇌ HSO3^– + OH^–
Kb = Kw / Ka2 = 1.0 ⋅ 10^-14 / 6.4 ⋅ 10^-8 = 1.56 ⋅ 10^-7
Since Kb is small, Kb = x^2 / c = x^2 / 0.020
x = [HSO3^-] = [OH^-] = 5.6 ⋅ 10^-5 mol dm^-3
pOH = 4.25 and pH = 9.75