Consider the reaction :
$2 H_{2} (g) + 2 N O (g) \to N_{2} (g) + 2 H_{2} O (g)$
The rate law for this reaction is:
Rate $= k [H_{2}] [N O]^{2}$
Under what conditions could these steps represent the mechanism?
Step 1: $2 N O ⇌ N_{2} O_{2}$
Step 2: $N_{2} O_{2} + H_{2} \to N_{2} O + H_{2} O$
Step 3: $N_{2} O + H_{2} \to N_{2} + H_{2} O$

These steps cannot be the mechanism under any circumstances

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These steps could he the mechanism if step 1 is the slowest step

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These steps could be the mechanism if step 2 is the slowest step

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These steps could be the mechanism if step 3 is the slowest step

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Solution

The correct option is C These steps could be the mechanism if step 2 is the slowest step
The rate law will be true if $N_{2} O_{2}$ is a transition state in this reaction and also step 2 is simultaneously the slow step.

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Similar questions

2 N O \leftrightarrow N_{2} O_{2}

N_{2} O_{2} + H_{2} \to H_{2} O_{2} + N_{2}

H_{2} + H_{2} O_{2} \to 2 H_{2} O

2 N O (g) + 2 H_{2} (g) ⟶ N_{2} (g) + 2 H_{2} O (g)

2 N O (g) + H_{2} (g) s l o w - ----- \to N_{2} + H_{2} O_{2}

H_{2} O_{2} + H_{2} f a s t - ----- \to 2 H_{2} O

2 N O \to N_{2} O_{2} (s l o w)

N_{2} O_{2} + O_{2} \to 2 N O_{2} (f a s t)

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