# How do you find the rate law for the following overall reaction, given its mechanism below? 2A + C -> D

## $2 A \implies B$ (slow) $\underline{B + C \implies D}$ (fast) $2 A + C \to D$

Jun 12, 2017

Notice how the first step is slow and the second step is fast.

$2 A \implies B$ $\text{ "" "" "" ""slow}$

$B + C \implies D$ $\text{ "" "" ""fast}$

$\text{----------------------}$

$2 A + C \to D$

The slow step is also known as the rate-limiting step, i.e. the step in the mechanism that bogs down the reaction the most (takes the longest), and corresponds most directly to the overall rate law.

With the rate-limiting step in mind, as each mechanistic step is an elementary reaction (in which the coefficients correlate with reaction order with respect to each reactant), the rate law is given as

$r \left(t\right) = \boldsymbol{k {\left[A\right]}^{2}}$

with the rate $r \left(t\right)$ in $\text{M/s}$ as a function of the rate constant $k$ in ${\text{M"^(-1)"s}}^{- 1}$ and concentration $\left[A\right]$ in $\text{M}$, since the first step is slow and involves two $A$ reactants colliding with each other to form $B$ product.

Thus, the rate law is primarily weighted by the bimolecular first mechanistic step, and is second order in $A$.

As a sidenote, although $C$ is a reactant in the overall reaction, the second step is assumed fast enough that the reaction order with respect to $C$ is zero, i.e. changing the concentration of $C$ does not affect the reaction rate:

$r \left(t\right) \propto \left\{{\left[A\right]}^{2} {\left[C\right]}^{\text{small}} \approx {\left[A\right]}^{2}\right\}$