# Rate and order confusion

I just learned about rate and order... I just want to confirm if these three ideas are correct:

1. If the first elementary step is the slow step, will the stoichiometric coefficients of the reactants in the rate determining step (RDS) be EQUAL to the order the concentration of the reactants in the rate equation is raised to? e.g. \begin{align} \ce{2A + B &-> C} &(\text{slow})\\ \ce{C + B &-> D} &(\text{fast}) \end{align}

Will the order with respect to $[\ce{A}]$ be 2 and $[\ce{B}]$ be 1 if this proposed mechanism is correct?

1. If the balanced equation is the ONLY elementary step in the reaction mechanism, the stoichiometric coefficients of the reactants will be the same as the order the concentration of the reactants are raised in the rate equation

2. If the second elementary step is the slow step, will the stoichiometric coefficients of the reactants in the rate determining step (RDS) be different to the order the concentration of the reactants in the rate equation is raised to?

• Could this answer be of any use to you? – Nicolau Saker Neto Mar 18 '14 at 23:47
• @Nicolau Saker Neto: This answer really helped. However, I am still unsure if the stoichiometric coefficients of the reactants of the first elementary step equal to the order in the rate equation if the first step is the slowest step – Eliza Mar 19 '14 at 4:18

Although Nicolau Saker Neto's answer provides the essential, I wish here to propose two emblematic examples for the situations cited in your question:

First case: two steps process, the first being the slowest.

Reaction between $\ce{H2O2}$ and $\ce{HBr}$:

\begin{align} \ce{H2O2 + H+ +Br- &-> HOBr + H2O} && \text{slow}\\ \ce{HOBr + H+ +Br- &-> Br2 + H2O} && \text{fast}\\ \end{align}

The rate of this reaction, determined by the slow step, equals:

$$\frac{d[\ce{Br2}]}{dt}=k_{1}\cdot[\ce{H2O2}]\cdot[\ce{H+}]\cdot[\ce{Br-}]$$

This is therefore a third order reaction. Stoichiometric coefficients of the slow elementary step yield the specific order in each of the three reacting species.

Second case: two steps process, the first being the fastest.

Reaction between $\ce{NO}$ and $\ce{O2}$:

The initial equilibrium is rapidly established:

\begin{align} \ce{2NO &<=> N2O2} && \text{fast}\\ \ce{N2O2 + O2 &-> 2NO2} && \text{slow}\\ \end{align}

In this case, the rate is:

$$\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot[\ce{N2O2}]\cdot[\ce{O2}]$$

This is therefore a second order reaction, with specific orders coinciding, as usual, with the stoichiometric coefficients in the slow step. Using the equilibrium constant $K$ of the first step, one can finally express the rate of the global equation as a function of $[\ce{NO}]$ and $[\ce{O2}]$, say:

$$\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot K \cdot[\ce{NO}]^{2}\cdot[\ce{O2}]$$

Summarizing, as you correctly say, the stoichiometric coefficient of a given reactant, in the rate determining elementary step ("slow" step), directly yields the order by which the concentration of that reactant is raised in the rate equation.

• Just to notice that the global reaction $\ce{2NO + O2 = 2NO_2}$ is known to be a third order reaction (very rare). Another specificity, its kinetic constant decrease with temperature (this is why it is modelled by a negative activation energy). See NIST Kinetic Database for detailed data and article references. – jlandercy Jun 12 '14 at 18:08