Consider the following mechanism for the decomposition of NO2Cl to NO2 and Cl2: (1) NO2Cl ⇌ NO2 + Cl (2) NO2Cl + Cl → NO2 + Cl2 (a) Use steady-state approximation to express the rate of Cl2 production. Select the single best 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.

ugh fiel iente of insistí is did kiwis jaj

the answer is 75.0%.