What is the correct way to calculate the rate constant for the reaction of NO and O2? The following data are given for the reaction of $\ce{NO}$ and $\ce{O2}$: 

$$
\ce{2NO + O2  -&gt;  2NO2}
$$

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 $\ce{NO}$:

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

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

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