For this problem, we have to write the rate law for this reaction in terms of -d[H_{2}]/dt using steady-state approximation

Recall that the **rate law** only focuses on the reactant concentrations and has a general form of:

$\overline{){\mathbf{rate}}{\mathbf{}}{\mathbf{law}}{\mathbf{=}}{\mathbf{k}}{\left[\mathbf{A}\right]}^{{\mathbf{x}}}{\left[\mathbf{B}\right]}^{{\mathbf{y}}}}$

k = rate constant

A & B = reactants

x & y = reactant orders

- A steady-state approach makes use of the assumption that the rate of production of an intermediate is equal to the rate of its consumption.
- We will be starting this with:

$\mathbf{Rate}\mathbf{}\mathbf{law}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\frac{\mathbf{d}\mathbf{\left[}{\mathbf{H}}_{\mathbf{2}}\mathbf{\right]}}{\mathbf{dt}}\mathbf{=}\mathbf{}{\mathbf{k}}_{\mathbf{2}}\mathbf{\left[}{\mathbf{H}}_{\mathbf{2}}\mathbf{\right]}\mathbf{\left[}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\right]}\mathbf{}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{\right)}}$

The proposed mechanism for a given reaction is as follows:

Step One: 2 NO (g) N_{2}O_{2} (g)

Step Two: H_{2}(g) + N_{2}O_{2}(g) H_{2}O(g) + N_{2}O(g) Rate - determining Step

Step Three: N_{2}O(g) + H_{2}(g) N_{2}(g) + H_{2}O(g) Fast

Use the steady-state approximation on intermediates to write the rate law for this reaction in terms of . Your law should include only elementary rate constants and species that appear in the overall reaction. Show all work to support your reasoning and answer. Note: you may Not assume step one is a fast pre-equilibrium.

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