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The rate law for 2NO(g) + O2(g) ====>2NO2(g) is rate=k[NO]^2[O2]. In addition to the...

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xkat123 | Student, Undergraduate | (Level 1) Honors

Posted July 13, 2013 at 4:51 AM via web

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The rate law for 2NO(g) + O2(g) ====>2NO2(g) is rate=k[NO]^2[O2]. In addition to the mechanism, the following have been proposed:

I. 2NO(g) +O2(g) ===> 2NO2(g) 

II. 2NO(g) ===> N2O2(g)    [fast]

    N2O2(g) +O2(g) ===> 2NO2(g)  [slow]

III. 2NO(g) ===>N2(g) + O2(g)  [fast]

     N2(g) +2O2(g) ===> 2NO2(g)  [slow]

Which of these mechanisms is consistent with the rate law?

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llltkl | College Teacher | (Level 3) Valedictorian

Posted July 13, 2013 at 5:34 AM (Answer #1)

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The rate of a chemical reaction is determined by the slowest step.

Rate = k[Concentration of reactants individually raised to their stoichiometric co-efficients]

In mechanism I,

Overall reaction occurs in a single step. Thus,

`rate= k*[NO]^2[O2]`

This is this consistent with the observed rate law.

In mechanism II,

The overall reaction occurs in two steps, through the involvement of an intermediate, N2O2.

Rate of the slowest step should be the overall reaction rate.

Thus, overall `rate= k*[N2O2][O2]`

Again considering non-accumulation of intermediate, N2O2 in the overall reaction,

Its rate of production will be equal to its rate of decomposition.

Thus, `k_1*[NO]^2= k*[N2O2][O2]`

`rArr [N2O2]= (k_1/k)*[NO]^2/[O2]`

Overall rate= `k*(k_1/k)*([NO]^2[O2])/[O2]=k_1[NO]^2`

So, this is not consistent with the rate law.

Similarly, considering mechanism III,

the overall rate `=k’[NO]^2`

Therefore, only mechanism I is consistent with the observed rate law.

Sources:

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redrose817 | Student, Undergraduate | (Level 1) eNoter

Posted July 13, 2013 at 2:56 PM (Answer #2)

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1) Rate law= k[NO]^2 [O2]

2)Rate law= k[N2O2][O2]   [slow eq determines rate law]

3)Rate law= k[N2][O2]^2  

so the resembling equation is 1. 

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