Domain: Statistical thermodynamics

Consider the reaction I₂ → 2I at 1000 K and 1 atm. The fundamental vibration frequency ν̄ of I₂ is 214.4 cm⁻¹, and the internuclear distance is 0.2667 nm. Calculate the equilibrium constant Kp for the reaction if the ground state of I is a doublet ²P₃⁄₂, ¹⁄₂ with a separation of 7603 cm⁻¹. The dissociation energy of I₂ is 1.542 eV.

Grading rubric:
criterion: The answer explicitly uses the equilibrium constant expression Kp = ([q⁰(I)]² / ([q⁰(I₂)] N)) exp(−Δε₀ / kT).
weight: 20
criterion: The answer explicitly writes the translational contribution in the form (h² / 2πkT)³ᐟ² (mI₂³ᐟ² / mI³) (P / NRT).
weight: 10
criterion: The answer explicitly writes qrot = (8π²IkT) / (σh²) with σ = 2.
weight: 10
criterion: The answer explicitly writes qvib = (1 − e^(−hcν̄/kT))⁻¹ with ν̄ = 214.4 cm⁻¹.
weight: 10
criterion: The electronic partition function is written as qelec = 4e⁰ + 2e^(−hc(7603 cm⁻¹)/kT).
weight: 10
criterion: The dissociation energy term is explicitly evaluated, yielding exp(−1.542 eV / kT) = 1.68 × 10⁻¹⁰.
weight: 20
criterion: The answer explicitly reports Kp = 1.57 × 10⁻³.
weight: 20

Answer:
Kp = 1.57 × 10^-3