# Question 14f66

Jan 13, 2016

Here's what I got.

#### Explanation:

Your starting point here will be the rate law for a second-order and a third-order reaction, respectively.

As you know, the rate law for a given reaction establishes a relationship between the rate of the reaction, the rate constant, and the concentrations of the reactants.

• Second-order reaction

For a second-order reaction, the rate law takes the form

color(blue)("rate" = k * ["A"] * ["B"])" ", where

$k$ - the rate constant

SIDE NOTE Since your goal here is to determine the units of the rate constant, you don't have to worry about having

"rate" = k * ["A"]^2

since the units will come out the same regardless if you have one reactant or two reactants.

Now, the rate of a reaction is a measure of the change in concentration of partial pressure of the reactants (or products) per unit of time.

In your case, the units for the rate of a reaction will be

$\textcolor{b l u e}{{\text{M"/"s" = "mol"/("L" * "s") = "mol L"^(-1)"s}}^{- 1}} \to$ for concentrations

$\textcolor{b l u e}{{\text{kPa"/"s" = "kPa s}}^{- 1}} \to$ for partial pressures

So, rearrange the rate law equation to find the units of $k$

k = "rate"/(["A"] * ["B"]) = (color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("L"^(-1)))) "s"^(-1))/(color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("L"^(-1)))) * "mol L"^(-1)) = color(green)("L mol"^(-1)"s"^(-1))

and

k = "rate"(["A"] * ["B"]) = (color(red)(cancel(color(black)("kPa"))) "s"^(-1))/(color(red)(cancel(color(black)("kPa"))) * "kPa") = color(green)("kPa"^(-1)"s"^(-1))

• Third-order reaction

For a third-order reaction, the rate law could look like this

color(blue)("rate" = k * ["A"] * ["B"] * ["C"])

This time, the units of the rate constant will be

k = "rate"/(["A"] * ["B"] * ["C"]) = (color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("L"^(-1)))) "s"^(-1))/(color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("L"^(-1)))) * "mol L"^(-1) * "mol L"^(-1))

$k = \textcolor{g r e e n}{{\text{L"^(2) "mol"^(-2)"s}}^{- 1}}$

and

k = "rate"/(["A"] * ["B"] * ["C"]) = (color(red)(cancel(color(black)("kPa"))) "s"^(-1))/(color(red)(cancel(color(black)("kPa"))) * "kPa" * "kPa") = color(green)("kPa"^(-2)"s"^(-1))#