# Assume that the mechanism for

Assume that the mechanism for the gas phase reaction, 2 NO + O2= 2 NO2, consists of the following elementary steps. FYI, NO andNO2 have an unpaired electron on the nitrogen atom. In NO3 anunpaired 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 elementarysteps, derive an expression for the steady-state concentration ofNO3, 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-staterate law.(iii)Experiments show that this reaction is 2nd order in NO, 1storder in O2. What conclusion may be drawn with respect to theproposed mechanism?

Answer:

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 ofsecond reaction

from the rate equation of second reaction = K2K1 [NO]2 [O2] / K-1 +K2 [NO]

By the above equation this reaction is 2nd order in NO and 1storder 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 theconcentration [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 theK-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]

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