The oxidation of nitrogen monoxide (NO) with oxygen (O₂) to produce nitrogen dioxide (NO₂) proceeds through the following mechanism: Step 1. NO (g) + O₂ (g) ⇌ NO₃ (g) [fast, equilibrium; forward rate constant K₁, backward K₋₁]; Step 2. NO₃ (g) + NO (g) → 2 NO₂ (g) [slow; rate con

Question

The oxidation of nitrogen monoxide (NO) with oxygen (O₂) to produce nitrogen dioxide (NO₂) proceeds through the following mechanism: Step 1. NO (g) + O₂ (g) ⇌ NO₃ (g) [fast, equilibrium; forward rate constant K₁, backward K₋₁]; Step 2. NO₃ (g) + NO (g) → 2 NO₂ (g) [slow; rate constant K₂]. Based on the mechanism find out the rate law and identify the intermediate.

Accepted Answer

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> [!TLDR]
> Rate is set by the slow step (rate = K₂[NO₃][NO]); express [NO₃] from the fast equilibrium of step 1 to get rate = (K₁K₂/K₋₁)[NO]²[O₂]; the intermediate is NO₃.
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> **Step 2 is the rate-determining (slow) step**, so the rate of the overall reaction equals the rate of step 2:
> $$	ext{rate} = K_2\,[NO_3][NO]. 	ag{1}$$
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> But $NO_3$ is a reactive **intermediate** — its concentration cannot appear in the final rate law, so we eliminate it using the fast pre-equilibrium of **step 1**. At equilibrium the forward and backward rates are equal:
> $$K_1[NO][O_2] = K_{-1}[NO_3] \;\Rightarrow\; [NO_3] = \frac{K_1}{K_{-1}}[NO][O_2]. 	ag{2}$$
>
> Substituting (2) into (1): …

Q.The oxidation of nitrogen monoxide (NO) with oxygen (O₂) to produce nitrogen dioxide (NO₂) proceeds through the following mechanism: Step 1. NO

Concept understanding — Rate Determining Step

The Intuition: A Factory Assembly Line

The Precise Statement

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Q.The oxidation of nitrogen monoxide (NO) with oxygen (O₂) to produce nitrogen dioxide (NO₂) proceeds through the following mechanism: Step 1. NO

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