# Question #dd444

Jan 24, 2016

$\text{rate} = k \cdot {\left[A\right]}^{2}$

#### Explanation:

The most important thing to remember about reactions that take place in multiple steps is that the slowest step is the rate-determining step.

Simply put, the rate of the reaction will depend on the slowest step of the reaction mechanism.

In your case, the overall reaction

$A + B \to C + D$

is said to proceed by the following mechanism

$A + A \to C + E \text{ " " }$ slow step
$B + E \to D + A \text{ " " }$ fast step

As you can see, adding these two steps will result in the overall reaction. Note that $E$ is formed by the first step and consumed in the second step, so it will be a reaction intermediate.

Now, the rate-determining step will be the step with the highest activation energy. A general-form potential energy diagram for a reaction that proceeds in two steps, with the first step being the slow step, looks like this

As you an see, the slow step will have the higher activation energy barrier.

So, the first step in your reaction mechanism is the slow one. In essence, the rate of the reaction will depend on a successful collision between two molecules of $A$.

Once a successful collision takes place, the intermediate $E$ will react with $B$ in a fast step. This basically tells you that the molecules of $B$ are waiting for that successful collision between two molecules of $A$ to take place.

This is why the slow step is the rate-determining step.

So, the rate law for the slow step will look like this

$\text{rate" = k * [A] * [A] = k * [A]^2" }$, where

$k$ - the rate constant of the first step

The rate of the reaction will thus depend on the square of the concentration of reactant $A$.

Since the first step is the rate-determining step, doubling the concentration of $A$ will quadruple the rate of the reaction, not just the rate of the slow step.