Question

Assume that the mechanism for the gas phase reaction, 2 NO + O2 = 2 NO2, consists of the following elementary steps. FYI, NO and NO2 have an unpaired electron on the nitrogen atom. In NO3 an unpaired electron is on one of the oxygen atoms.
(1) NO + O2 ↔ NO3 (rate coefficients k1 and k-1)
(2) NO3 + NO → 2 NO2 (rate coefficient k2)
(i)Complete the following expressions with rates of the elementary steps, derive an expression for the steady-state concentration of NO3, and derive the steady-state rate law for this mechanism.
d[NO]/dt =
d[NO2]/dt =
d[NO3]/dt =
(ii)Show that there are two limiting cases of this steady-state rate law.
(iii)Experiments show that this reaction is 2nd order in NO, 1st order in O2. What conclusion may be drawn with respect to the proposed mechanism?

Answer 1

By the second reaction the rate of the reaction = K₂ [NO3] [NO]

But according to steady state approximation d[NO3]/dt = 0

d[NO3]/dt = K₁ [NO] [O2] - K_-1 [NO3] - K₂ [NO3] [NO] = 0

[NO3] = K₁ [NO] [O2} _/ K_-1 + K₂ [NO]

substituting the concentration of [NO] in rate equation of second reaction

from the rate equation of second reaction = K₂ K₁ [NO]² [O2] / K_-1 + K₂ [NO]

By the above equation this reaction is 2nd order in NO and 1st order in O2

2. d[NO]/dt = - K₁ [NO] [O2] + K_-1 [NO3] - K₂ [NO3] [NO]

3. d[NO2] / dt = 2 K₂ [NO3] [NO]

Limiting cases:

Case I:

If the concentration [NO] is very low we neglect the concentration [NO] and then the rate equation becomes

= K₁ K₂ [NO]² [O2] / K_-1

Then the rate is second order with respect to concentration of [NO] and also first order with respect to [O2]

Case II:

If the rate riversible reaction is very low we neglect the K_-1 term and then the rate equation becomes

= K₁ K₂ [NO]² [O2] / K₂ [NO] = K₁ [NO] [O2]

Then the rate is first order with respect to concentration of [NO] and also first order with respect to [O2]