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








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**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 $\ce{NO}$:


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




**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 $k$.


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