Consider this reaction : `2NO_(2)(g)+O_(3)(g)rarrN_(2)O_(5)(g)+O_(2)(g)` The reaction of nitrogen diozidd and ozone represented is first order in `NO_

Question

Consider this reaction : `2NO_(2)(g)+O_(3)(g)rarrN_(2)O_(5)(g)+O_(2)(g)` The reaction of nitrogen diozidd and ozone represented is first order in `NO_

Consider this reaction :
`2NO_(2)(g)+O_(3)(g)rarrN_(2)O_(5)(g)+O_(2)(g)`
The reaction of nitrogen diozidd and ozone represented is first order in `NO_(2)(g) and "i n" O_(3)(g)`. Which of these possible reaction mechanisms is consistent with the rate law?
Mechanism I: `NO_(2)(g)+O_(3)rarrNO_(3)(g)+O_(2)("slow ")`
`NO_(3)(g)+NO_(2)(g) rarrN_(2)O_(5)(g)" "("fast")`
Mechanism II `O_(3)(g)hArrO_(2)(g)+[O]" "("fast")`
`NO_(3)(g)+[O](g)rarrNO_(3)(g)" "("slow")`
`NO_(3)(g)+NO_(2)(g)rarrN_(2)O_(5)(g)" " ("fast")`
A. I only
B. II only
C. Both I and II
D. Neither I nor II

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
Mechanism I, rate =`k[NO_(2)][O_(2)]`. Slow step is rate determine step
Mechanism II, Rate = `k[NO_(2)][O]" ".....(1)`
`K=([O_(2)][O])/([O_(3)]),[O]=K([O_(3)])/([O_(2)])`
From eq. (i) , Rate = `kK[NO_(2)[O_(3)][O_(2)]^(-1)`
`:.` Both mechanisms show that reaction is first order with respect to `NO_(2) and O_(3)`