Question

Nitrogen and oxygen gas react as high temperatures to form nitric oxide, a component of photochemical smog. Net reaction: the net reaction is N₂ + O₂ --> 2NO

Proposed mechanism: O₂ <====> 2O   (fast, at equilibrium)

                                        N₂ + O <-- --> NO + N   (fast, not at equilibrium)

                                    N + O₂ ----> NO + O    (Slow)

a.) Derive the rate law for the proposed mechanism. rate = ____________________________

b.) What is the initial rate law, when very little product has formed? initial rate = __________________________

Answer 1

For the given reaction,

a) Derrivation of rate law,

From the slow step,

rate = k4[N][O2]

[N] can be sunstituted as,

k3[N2][O] = k-3[NO][N] + k4[N][O2]

[N] = k3[N2][O]/(k-3[NO] + k4[O2])

For [O],

-d[O2]/dt = k2[O] = k1[O2]^(0.5)

[O] = (k1/k2)[O2]^(0.5)

feed in above equation,

[N] = (k1k3/k2)[N2][O2]^(0.5)/(k-3[NO] + k4[O2])

So,

rate = k4[N][O2] = (k1k3k4/k2)[N2][O2]^(1.5)/(k-3[NO] + k4[O2])

b) initial rate when very littile product has formed,

initial rate = k[N2][O2]