# What is the correct way to calculate the rate constant for the reaction of NO and O2? [closed]

The following data are given for the reaction of $\ce{NO}$ and $\ce{O2}$:

$$\ce{2NO + O2 -> 2NO2}$$

The the reaction is second order in $\ce{[NO]}$ and first order in $\ce{[O2]}$, and the rate of disappearance of $\ce{NO}$ is $2.5 \times 10^{-5}~\mathrm{mol\over L\,s}$ at the instant when $\ce{[NO] = [O2]} = 0.01~\mathrm{mol\over L}$.

The question asks me to calculate the rate constant.

I've thought of two ways of approaching the calculation—which of these solutions is correct?

1) Take the rate of the reaction as one-half the rate of disappearance of $\ce{NO}$:

\begin{align} R &= {1\over 2} * 2.5 \times 10^{-5} = k \ce{[NO]^2[O2]}=k(0.01)^3 \\ k &= 12.5~\mathrm{L^2\over mol^2\,s} \end{align}

2) Take the rate of the reaction as equal to the rate of disappearance of $\ce{NO}$:

\begin{align} R &= 2.5 \times 10^{-5} = k \ce{[NO]^2[O2]} = k(0.01)^3 \\ k &= 25 ~\mathrm{L^2\over mol^2\,s} \end{align}

## closed as off-topic by paracetamol, M.A.R. ಠ_ಠ, bon, Klaus-Dieter Warzecha, Todd MinehardtJan 14 '17 at 16:26

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Either definition is acceptable, as long as you clearly identify relationship between the calculated rate value $R$ and the rates-of-change for each of the species in the system.

In case (1), you've effectively calculated the rate constant defined as:

$$R_1 = k_1\ce{[NO]^2[O2]} = -{\mathrm d \ce{[O2]}\over \mathrm dt} = -{1\over 2}{\mathrm d \ce{[NO]}\over \mathrm dt} = {1\over 2}{\mathrm d \ce{[NO2]}\over \mathrm dt}$$

In case (2), you've instead calculated the rate constant defined as:

$$R_2 = k_2 \ce{[NO^2][O2]} = -2{\mathrm d \ce{[O2]}\over \mathrm dt} = -{\mathrm d \ce{[NO]}\over \mathrm dt} = {\mathrm d \ce{[NO2]}\over \mathrm dt}$$

Both cases accurately describe the kinetics, but you have to be careful to identify which definition of the rate you've used in calculating $k$.

• @AdnanAL-Amleh Reported rate constants are often ambiguous. You're absolutely right that there is indeed only one actual, physical $\mathrm d \ce{[NO]}\over \mathrm dt$ observed in this system. But, there are an infinite number of rate expressions you can write, all related by multiplicative constants, that describe the kinetics equally well. Maybe someone else can help give you more insight, but I don't know if I can assist much beyond what's contained in my answer above. – hBy2Py Jan 16 '17 at 3:27
• The IUPAC Gold Book recommends using the definition as in R1 above and this is also what most, if not all textbooks, use. – porphyrin Jan 18 '17 at 13:06
• @porphyrin I also prefer the $R_1$ definition. Many journal articles do not follow this recommendation, however, and it can take quite a bit of work to back out the proper form of the rate expression. – hBy2Py Jan 18 '17 at 14:13
• @AdnanAL-Amleh "There is no simple correlation between the stoichiometry of the reaction and the rate law" I don't know what you mean by this; to my mind, the stoichiometry of the reaction defines the rate law. – hBy2Py Jan 19 '17 at 12:00
• @AdnanAL-Amleh "I think the rate of the reaction = experimentally determined value of the rate of disappearance of NO." Like I said, this is a valid choice of the rate expression. But, it is definitely not the only valid choice. – hBy2Py Jan 19 '17 at 12:01