3. [14%] Gaseous molecular and atomic bromine have the following thermodynamic properties: Species Δ — Thermodynamics Chemistry Question
Problem Context
- [14%] Gaseous molecular and atomic bromine have the following thermodynamic properties:
Species ΔH°f (kJ mol-1) S° (J mol-1 K-1)
Br2(g) 30.9 245.4
Br(g) 111.9 175.0
What is the bond dissociation enthalpy (BDE) of the bromine-bromine bond in Br2(g)?
Model Answer
Br2(g) → 2 Br(g) ∆H°rxn = BDE of Br–Br bond
∆H°rxn = 2(111.9 kJ mol-1) – 30.9 kJ mol-1 = 192.9 kJ mol-1
How many atoms of Br(g) will be present at equilibrium in a 1.00-L container with 0.100 bar of Br2(g) at 298 K?
Model Answer
For Br2(g) → 2 Br(g), ∆S° = 2(175.0 J mol-1 K-1) – 245.4 J mol-1 K-1 = 104.6 J mol-1 K-1
∆G° = ∆H° – T∆S° = (192.9 kJ mol-1) – (298 K)(0.1046 kJ mol-1 K-1) = 161.7 kJ mol-1
∆G° = –RTln(Keq)
161700 J mol-1 = –(8.314 J mol-1 K-1)(298 K)ln(Keq)
Keq = 4.5 × 10-29
At equilibrium: pBr2 / pBr2 = pBr2 / 0.100 = 4.5 × 10-29
pBr = 2.1 × 10-15 bar
In a 1.00 L container, PV = nRT:
(2.1 × 10-15 bar)(1.00 L) = n(0.08314 L bar mol-1 K-1)(298 K)
n = 8.6 × 10-17 mol
(8.6 × 10-17 mol)×(6.022 × 1023 mol-1) = 5.2 × 107 Br atoms
The ionization energy of Br(g) is 1145.9 kJ mol-1, while the ionization energy of Br2(g) is 1025.1 kJ mol-1. What is the bond dissociation enthalpy of Br2 +(g) (defined as the enthalpy of reaction of Br2 +(g) to form Br+(g) and Br(g))?
Model Answer
Using Hess’s Law (all species are gas-phase):
Br2 → 2 Br ∆H° = 192.9 kJ mol-1
Br → Br+ + e– ∆H° = 1145.9 kJ mol-1
Br2+ + e– → Br2 ∆H° = –1025.1 kJ mol-1
Br2+ → Br + Br+ ∆H° = BDE = 313.7 kJ mol-1
Propose an explanation for the difference in BDE between Br2(g) and Br2 +(g) in terms of the bonding in these two species.
Model Answer
The electron that is removed from Br2 is π* in character, so the bond order of Br2+ is 1.5 compared to 1.0 for Br2. The higher bond order results in a greater bond strength.
The vapor pressure of liquid bromine at 298 K is 0.283 bar. What is the absolute entropy S° of Br2(l)?
Model Answer
∆G°vap, 298K = –RTln(Keq) = –(8.314 J mol-1 K-1)(298 K)ln(0.283) = 3.13 kJ mol-1
∆G°vap, 298K = ∆H°vap – T∆S°vap
3.13 kJ mol-1 = 30.9 kJ mol-1 – (298 K)(∆S°vap)
∆S°vap = 93.2 J mol-1 K-1
∆S°vap = (S° of Br2(g)) – (S° of Br2(l))
93.2 J mol-1 K-1 = 245.4 J mol-1 K-1 – (S° of Br2(l))
S° of Br2(l) = 152.2 J mol-1 K-1
The absolute entropies of the gas-phase atoms of the fourth period given in the table exhibit a non-monotonic pattern. Explain why the absolute entropies of Se(g) and Br(g) are larger than the absolute entropies of either As(g) or Kr(g).
As(g) Se(g) Br(g) Kr(g)
S°, J mol-1 K-1 163.2 176.7 175.0 164.1
Model Answer
As and Kr have only one way of arranging their valence electrons (for As, each valence p orbital is half-filled, while Kr has each valence p orbital completely filled). In contrast, Se and Br have orbitally degenerate ground states, which gives them more ways of arranging their electrons. This increase in the number of possible arrangements corresponds to a higher entropy (S = kBlnΩ).