The data in the tablebe ow were obtained for the reaction: A + B -> P. What is the rate law for this reaction?

Mar 16, 2017

$\textsf{\text{Rate} = k {\left[A\right]}^{2} {\left[B\right]}^{0} = k {\left[A\right]}^{2}}$

Explanation:

This type of question concerns the method of initial rates to determine the rate expression.

$\textsf{A + B \rightarrow P}$

The general rate expression can be written:

$\textsf{\text{Rate} = k {\left[A\right]}^{x} {\left[B\right]}^{y}}$

At the start of the reaction when t = zero, we can regard the concentrations of the reactants to be constant.

By varying the initial concentration of one reactant while keeping the other constant we can see what effect that has on the initial rate of the reaction.

If you look at trials (1) and (2) you can see that $\textsf{\left[A\right]}$ is held constant while $\textsf{\left[B\right]}$ is increased.

The factor by which $\textsf{\left[B\right]}$ is increased = 1.526/0.763 = 2.

As you can see there is no effect on the initial rate which stays at 0.83 M/s.

This means that the order with respect to B must be zero i.e y = 0

Now look at trials (1) and (3). Here you can see that $\textsf{\left[B\right]}$ is held constant and $\textsf{\left[A\right]}$ is increased.

The factor by which $\textsf{\left[A\right]}$ is increased = 0.819/0.273 = 3.

How does this affect the initial rate?

The rate has increased by a factor of 25.47/2.83 = 9

This means that the order with respect to $\textsf{A}$ must be 2 since $\textsf{{3}^{2} = 9}$.
$\therefore$ x = 2.

The overall rate expression, therefore is:

$\textsf{\text{Rate} = k {\left[A\right]}^{2} {\left[B\right]}^{0}}$

Since $\textsf{{\left[B\right]}^{0} = 1}$ you can simply write:

$\textsf{\text{Rate} = k {\left[A\right]}^{2}}$