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I am really confused about the rate law. From my understanding, the rate law gives an equation that tells us the rate of the disappearance of the reactants. So, for example, the rate law of the reaction $\ce{2 NO + O2 -> 2 NO2}$ is as follows

$$\text{rate of disappearance of reactants} = k \, [\ce{NO}]^2 [\ce{O2}]$$

My biggest confusion stems from the fact that the rate law gives one uniform rate for the disappearance of all the reactants, which makes no sense to me. For every $\ce{O2}$ used, two $\ce{NO}$ are used, so how can they be disappearing at the same rate?

Hopefully, it makes sense where I am coming from, and I would appreciate any help trying to clarify this for me.

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4 Answers 4

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The general case for a reaction $aA+bB=cC$ is

$$r=-\frac{1}{a}\frac{d[A]}{dt}=\cdots=+\frac{1}{c}\frac{d[C]}{dt}=k[A]^\alpha[B]^\beta[C]^\gamma$$

where $n=\alpha+\beta+\gamma$ is the total order and $\alpha,\;\beta $ etc are constants that can be positive, negative or fractional and are the orders of A, B etc. The rate constant has units (concentration)$^{1-n}$/time. This type of rate equation tells us nothing about the actual mechanism of the reaction.

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Rodrigo de Azevedo's answer can be stated with words by saying that the rate of the reaction is the rate of disappearance of $1$ mole $\ce{O2},$ and also the rate of disappearance of $2$ moles $\ce{NO}$. That is why the coefficient $2$ appears in the law written for $\ce{NO}:$

$$\frac{d}{dt}[\ce{NO}] = - 2k[\ce{NO}]^2[\ce{O2}]$$

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$$\begin{aligned} \frac{{\rm d}}{{\rm d}t} [\ce{NO}] &= - \color{magenta}{2} k \, [\ce{NO}]^2 [\ce{O2}]\\\\ \frac{{\rm d}}{{\rm d}t} [\ce{O2}] &= - k \, [\ce{NO}]^2 [\ce{O2}] \end{aligned}$$

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    $\begingroup$ The rate is determined by the rate limiting reagent, in this case, the oxygen species will dictate the rate. The consumption of the nitrous oxide is mediated by the reaction of the oxygen, therefore, both reagents will dissipate at the rate that oxygen is consumed. $\endgroup$
    – Moobius Strip
    Dec 3, 2020 at 19:10
  • $\begingroup$ The statement as by Rodrigo de Azevedo, demonstrated the proportion of this consumption as it pertains to the rate constant for some K. $\endgroup$
    – Moobius Strip
    Dec 3, 2020 at 19:11
  • $\begingroup$ This answer should have a short description of the equations. Maurice's answer did that very well, so why not this one. $\endgroup$
    – Antimon
    Dec 3, 2020 at 22:02
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    $\begingroup$ @Antimon Because I wrote before Maurice and assumed that the OP knows enough calculus to arrive at a conclusion. $\endgroup$
    – Rodrigo de Azevedo
    Dec 3, 2020 at 22:32
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Here is a better definition: The rate law of a reaction is the quantity which when scaled by the stoichiometric coefficient of reactant/ product (with sign) gives the rate at which that particular quantity varies with time.

Or, it's the rate of a chemical substance's concentration changing normalized by the number of its moles in the chemical reaction in which it is participating.

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