Answer:
k2 k1
k–1
[NO]2[Cl2]
Explanation:
Steady-state Approximation
In the steady-state approach, we write the full rate law for the intermediate, N2O2, and
set this rate equal to zero. The mechanism has N2O2 appearing in one process (the forward
direction of step (i)) and disappearing in two processes (the reverse of (i) and the forward
direction of (ii)). Thus, we write
d[N2O2]
dt = k1[NO]2 – k–1[N2O2] – k2[N2O2][Cl2] = 0 .
We solve this expression for [N2O2], finding [N2O2] = k1 [NO]2
k–1 + k2[Cl2] ,
and we substitute into the rate law as before:
1
2
d[NOCl]
dt = k2 [N2O2][Cl2] = k2
k1 [NO]2
k–1 + k2[Cl2]
[Cl2] = k2 k1
k–1 + k2[Cl2]
[NO]2[Cl2]
and find an expression that is almost the same as before, and almost the expression we
were told was observed experimentally. The difference is the presence here of the term
k2[Cl2] in the denominator. What’s it doing there, and how can we get rid of it?
It’s there because the steady-state approximation always gives a more general
expression than the prior equilibrium approximation. But we can get rid of it by
recognizing that our mechanism assumes from the start that step (ii) is slow, which means
k2 is small so that k–1 >> k2 [Cl2]. Thus, we can write
1
2
d[NOCl]
dt = k2 k1
k–1 + k2[Cl2]
[NO]2[Cl2] ≈ k2 k1
k–1
[NO]2[Cl2]
which gets us back to the experimental expression. In reality, the steady-state expression
could really be correct, but not verifiable experimentally simply because the small term in
the denominator, k2 [Cl2], is just too small to have an observable effect on measured rates.
; the reaction is product favored.
; the reaction is reactant favored.
; the reaction is in equilibrium.