Consider the following reactions: R2 R3 RA RS R6 R+02 + M -> RO2 + M RO2 + NO RO
ID: 580783 • Letter: C
Question
Consider the following reactions: R2 R3 RA RS R6 R+02 + M -> RO2 + M RO2 + NO RO+NO2 NO2 + hv--> NO +O, RO+02 R, CHO + HO, HO2 + NO OH +NO2 Net: RH + 402 R, CHO + 203 + H,O Using reactions R1 to R6, determine the rate equation for the concentration of NO. Hints: 1. Use the steady state approximation to simplify the rate equation so you can solve directly for the concentration of NO. Calculate the steady state concentration of NO assuming: 2. . k3 7.8 x 10-12 cm3 molecule-1s1 k4 = 9.0 ×10-3 s-1 ks = 8.6 × 10-12 cm3 molecule-1 s-1 [NO2] = 1.23 x 1011 molecules cm-3 [RO2] = 0.5 x 105 molecules cm-3 [H02] = 2.0 × 106 molecules cm" · · · ·Explanation / Answer
From the 6th reaction rate of the reaction R6 = K6 [HO2] [NO]
But according to steady state approximation d[NO] /dt = 0
- K3 [NO] [RO2] + K4 [NO2] - K6 [HO2] [NO] = 0
[NO] = K4 [NO2] / K3 [RO2] + K6 [HO2]
Substituting [NO] value in rate of the 6th reaction
R6 = K6 K4 [NO2] [HO2] / K3 [RO2] + K6 [HO2]
= (8.6 x 10-12 x 9 x 10-3 x 1.23 x 1011 x 2.0 x 106) / (7.8 x 10-12 0.5 x 105 + 8.6 x 10-12 x 2 x 106)
= 190.404 x 102/ (0.39 x 10-6 + 17.2 x 10-6)
= 190.4 x 102/17.59 x 10-6
= 10.8 x 108 = 1.08 x 109 cm3 mol-1sec-1