Nitric oxide NO reacts with oxygen to produce nitrogen dioxide-2NOg-O2g- 2NO2gIf the mechanism of reaction is-NO-O2 k- NO3fastNO3-NO k1- NO2-NO2slowThe rate law is-

Slow reaction is the rate determining step. So, for NO_3+NO k_1⟶ NO_2+NO_2(slow) Rate=k_1[NO_3][NO] (1) from NO+O_2 k⇌ NO_3(fast) we can write k=[NO_3]/[O_2][ ...

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Byju's Answer

Standard XIII

Chemistry

Rate of Reaction

Nitric oxide ...

Question

Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide:
$2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)$
If the mechanism of reaction is;
$N O + O_{2} k ⇌ N O_{3} (f a s t)$
$N O_{3} + N O_{1} N O_{2} + N O_{2} (s l o w)$
The rate law is:

Rate=

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

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Rate=

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

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Rate=

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

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Rate=

k^{'} [N O]^{3} [O_{2}]

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Solution

The correct option is C Rate= $k^{'} [N O]^{2} [O_{2}]$
Slow reaction is the rate determining step. So, for
$N O_{3} + N O_{1} N O_{2} + N O_{2} (s l o w)$
$R a t e = k_{1} [N O_{3}] [N O] (1)$
from $N O + O_{2} k ⇌ N O_{3} (f a s t)$ we can write $k = \frac{[N O_{3}]}{[O_{2}] [N O]} (2)$
On substituting value of $[N O_{3}]$ from eq.(2) to eq.(1), we get $R a t e = k^{'} [N O]^{2} [O_{2}]$

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Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide: 2NO(g)+O2(g)→2NO2(g) If the mechanism of reaction is; NO+O2k⇌NO3(fast) NO3+NOk1⟶NO2+NO2(slow) The rate law is:

The correct option is C Rate=k′[NO]2[O2]Slow reaction is the rate determining step. So, for NO3+NOk1⟶NO2+NO2(slow) Rate=k1[NO3][NO] (1) from NO+O2k⇌NO3(fast) we can write k=[NO3][O2][NO] (2) On substituting value of [NO3] from eq.(2) to eq.(1), we get Rate=k′[NO]2[O2]

Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide:
$2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)$
If the mechanism of reaction is;
$N O + O_{2} k ⇌ N O_{3} (f a s t)$
$N O_{3} + N O_{1} N O_{2} + N O_{2} (s l o w)$
The rate law is: