Question #44804

The decomposition of nitrous oxide ($\ce{N2O}$) on a platinum surface follows the overall stoichiometry:

$\ce{2 N2O(g) -> 2 N2(g) + O2(g)}$

The proposed Langmuir-Hinshelwood mechanism is: Step 1: $\ce{N2O(g) + S <=> N2O \cdot S}$ (Fast equilibrium, adsorption constant $K$) Step 2: $\ce{N2O \cdot S -> N2(g) + O \cdot S}$ (Slow, rate-determining step, rate constant $k_2$) Step 3: $\ce{2 O \cdot S -> O2(g) + 2 S}$ (Fast) (where S represents an active site on the catalyst surface)

At $800$ K, when the initial concentration of $\ce{N2O}$ is extremely high ($[\ce{N2O}]_0 = 4.0$ M), the catalyst surface is fully saturated. Under these zero-order conditions, the rate of appearance of $\ce{O2(g)}$ is measured to be $1.5 \times 10^{-3}$ M/s.

When the concentration of $\ce{N2O}$ is very low ($[\ce{N2O}] \ll 1/K$), the reaction follows apparent first-order kinetics with respect to $\ce{N2O}$, and the half-life for the disappearance of $\ce{N2O}$ is found to be $\frac{\ln 2}{6.0 \times 10^{-4}}$ seconds.

Calculate the Langmuir adsorption constant $K$ (in M$^{-1}$) for $\ce{N2O}$ on the platinum surface.