## Coefficients in pre-equilibrium

$K = \frac{k_{forward}}{k_{reverse}}$

Lucia H 2L
Posts: 43
Joined: Mon Nov 14, 2016 3:00 am
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### Coefficients in pre-equilibrium

When do you multiply by a specific rate law when finding pre-equilibrium?

For example, in the example we did in class:

2NO + O2 --> 2NO and observed rate = k[NO]^2[O2]

step 1 (fast): NO + NO --> N2O2 , rate = k1[NO]^2
step 2 (slow) N2O2 + O2 --> 2NO2 , rate = k2[N2O2][O2]

we then went on to remove the intermediate (N2O2) from the slow rate law with the equilibrium constant, using the usual steps, giving:
[N2O2] = K[NO]^2
and plugging this in,
rate = k2K[NO]^2[O2].

However, the final rate = 2x this value (2k2K[NO]^2[O2]).

Why did we multiply the rate by 2? Is it because there are 2 moles of NO in the products in the original reaction? If so, should final rates always be a specific rate with respect to the products?

Humza_Khan_2J
Posts: 56
Joined: Thu Jul 13, 2017 3:00 am

### Re: Coefficients in pre-equilibrium

I believe the problem you're talking about asks for the rate of NO production, which would be 2 times the unique rate, which is what is observed.