How can you determine the rate law of the following reaction?

$2 N O (g) + O_{2} (g) ⟶ 2 N O_{2} (g)$

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Solution

We can determine the rate of this reaction as a function of initial concentrations either by keeping the concentration of one of the reactants constant and changing the concentration of the other reactant or by changing the concentration of both the reactants. e.g.. for the given reaction,

(i) Keeping $[O_{2}]$ constant, if the concentration of NO is doubled, rate is found to become four times. This shows that,

$R a t e \propto [N O]^{2}$

(ii) Keeping [NO] constant, if the concentration of $[O_{2}]$ is doubled, rate is also found to become double. This shows that,

$R a t e \propto [O_{2}]^{2}$

Hnece, overall rate law will be

$R a t e = k [N O]^{2} [O_{2}]$

Rate law expression:

$R a t e l a w = - \frac{1}{2} \frac{Δ [N O]}{Δ t} = - \frac{Δ [O_{2}]}{Δ t} = + \frac{1}{2} \frac{Δ [N O_{2}]}{Δ t}$

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N_{2} (g) + O_{2} (g) ⇌ 2 N O (g); K_{c} = 2.7 \times 10^{- 18}

2 N O (g) + O_{2} (g) ⇌ 2 N O_{2} (g)

2 S O_{2 (g)} + 2 N O_{2 (g)} ⟶ 2 S O_{3 (g)} + 2 N O_{g}

2 N O_{(g)} + O_{2 (g)} ⟶ 2 N O_{2 (g)}

2 S O_{2 (g)} + 2 N O_{2 (g)} \to 2 S O_{3 (g)} + 2 N O_{(g)}

2 N O_{(g)} + O_{2 (g)} \to 2 N O_{2 (g)}

2 N O (g) + O_{2} (g) \to 2 {N O}_{2} (g)

= k {[N O]}^{2} [O_{2}]

2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)

N O + O_{2} k ⇌ N O_{3} (f a s t)

N O_{3} + N O_{1} N O_{2} + N O_{2} (s l o w)

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