The Rate Law is used to model a reaction's speed based on the concentration of its reactants. It's also sometimes known as the differential rate law.

You'll also come to learn that there are multiple types of Rate Laws (as hinted at in the above definition) and that these are used to categorize reactions based on something called *order*. In this lesson, we'll cover the types of Rate Laws, what order means, and what we can learn from this information.

- This lesson will help you understand the fundamental chemistry behind the
**Rate Law**. - We'll start with the basic
**Rate Law equation**, and we'll break down what it means. - We'll then cover what
**order**means, and briefly discuss the different types of reaction orders. - You'll learn that this basic equation is only the first of
*two*Rate Laws:**differential and integrated.** - Lastly, you'll learn a basic method of how to determine reaction order through experimental means using the
**Straight Line Test**.

## Rate Law Equation

First, let's cover the basic structure of the rate law equation. Let's imagine a basic reaction, where two reactants form some product. Let's name our two reactants A and B, and imagine our chemical formula to be:

$$A + B \rightarrow C + D$$

Thus, the rate law equation would be:

$$r=k[A]^x[B]^y$$

Where,

[A] and [B] represent the

**concentrations**of reactants A and B.x and y represent the reaction

**order**for each reactant*respectively*. (x is [A]'s reaction order, y is [B]'s reaction order, etc.) These are determined**through****experiments**.k is the rate constant.

r is the overall reaction rate.

### Analyzing the Rate Law Equation

Let's break this down step-by-step. This law clearly shows that we can find the overall rate of reaction by considering the concentrations of each reactant, these reactants' "orders", and some experimentally determined rate constant.

Notice how we don't consider the coefficients in our reaction: \(A + B \rightarrow C + D\).

It's implied here that the coefficients for A and B are 1, but what if they were 2, or 4? What if we had the chemical equation \(2A + 4B \rightarrow C + D\)?

If we look at the provided equation, we'll see that the coefficients aren't considered. So for now, we don't need to worry about the coefficients of our reaction.

**What about the x and y? These are reaction orders, that also have nothing to do with the coefficients of our chemical reaction or any sort of related stoichiometry.** There's no information that the chemical equation can give you on its own to help you determine these. These have to be found through experiments and can be calculated if given a proper amount of information.

You'll learn about the types of reaction orders, and briefly how to determine reactant order later in this lesson.

Lastly, we have to consider k. k is a rate constant that helps to keep the aforementioned variables proportional to the environment in which the experiment is taking place. This typically k changes depending on the temperature.

For example, a problem that provides k would most likely be stated like the following: "In an experiment, it was determined that the reaction constant k is x at y degrees."

### Differential Rate Law in Practice

Now that we know what the basic structure of the Rate Law looks like, let's try a simple example.

**Practice Problem 1:** Write the expression for the following reaction. (Assume that NO is second order and H_{2} is first order).

$$2NO_{(g)} + 2H_{2\,(g)} \rightarrow N_{2\,(g)} + 2H_2O_{(g)}$$

$$r=k[NO]^x[H_2]^y$$

$$x=\text{2nd order}\,\,y=\text{1st order}$$

$$r=k[NO]^2[H_2]^1$$

As you can see, when the order of reactants is provided, the problem is fairly straightforward. Make sure you're comfortable with this form, as it will appear often.

## Reaction Order

In order to help categorize different sorts of reactions more easily, chemists have devised a system of order to rank reactions.

### Reactant vs. Reaction Order

So far we have mentioned the concept of **reaction order** multiple times. But what does that mean? In short, the reaction order is the relationship between the concentration of reactants and the actual rate of reaction. A better way of looking at it would be:* "**how much does this reactant influence the rate of reaction?**" *Obviously, the higher the **reactant order**, the more influence its concentration has on the speed of the reaction.

Each reactant order represents the proportion of influence that that reactant's concentration has on the overall rate of reaction.

Let's refer back to our Practice Problem 1. Would increasing the concentration of NO or H_{2} have more of an influence on the overall rate of reaction? Of course, the answer would be NO, because of its higher reactant order.

We've talked about reactant order, but what about the overall reaction order? If we know the Rate Law for a certain reaction, how can we find what order the reaction is?

The overall rate of reaction can be found by summing the reactant orders.

Simple enough, right? We can find the overall rate of reaction by adding up the orders of each of our reactions. So, referring back to our trusty Practice Problem 1, what would the *reactant *orders and *reaction *order be? NO's reactant order is 2, H_{2}'s reactant order is 1, and the overall reaction order is 3. But why do we go through this process of categorizing reactions into overall orders? Enter the **integrated rate law! **

## Integrated Rate Law

Before, we mentioned the **integrated rate law**. What's that about? As it turns out, if we integrate the rate law, we can extract information that will help us to analyze experimental plots. This is why overall reactions are categorized into order: we can quickly deduce a good bit of information about a reaction through only a little experimental data.

The integrated rate law is dependent on the order of the differential rate law that is being analyzed. This means that there's a different integrated rate law for zero-order reactions, first-order reactions, second-order reactions, and so on. In AP Chemistry, you will only be dealing with these scenarios.

Let's keep things simple and assume that we only care about the concentration of a single reactant. Let's call this reactant A. Because we want to analyze the rate of change in this reactant, we want to consider the initial concentration and current concentration in question. Let's integrate the Rate Law for each order.

Reaction Order | Rate Law (Differential) | Integrated Rate Law |

0 | $$r=k[A]^0=k$$ | $$[A]=[A]_0-kt$$ |

1 | $$r=k[A]$$ | $$[A]=[A]_0e^{-kt}$$ |

2 | $$r=k[A]^2$$ | $$[A]=\frac{[A]_0}{1+kt[A]_0}$$ |

If you are curious about how the Integrated Rate Law is found from the rate law, a simple mathematical proof is given below for *first-order reactions*. This won't be necessary to memorize.

The differential rate law can be rewritten as:

$$r=k[A]=\frac{-d[A]}{dt}$$

Where \(\frac{-d[A]}{dt}\) represents the reaction rate (change in A's concentration)

This means that:

$$k[A]=\frac{-d[A]}{dt}$$

We can separate variables:

$$-kdt=\frac{d[A]}{[A]}$$

And now integrate to get our law (given that A(0)=A_{0}):

$$[A]_0e^{-kt}=[A]$$

First, let's graph what these integrated rate laws would look like on their own.

These are what zero-order, first-order, and-second order reactions would look like if you had gathered data, used the integrated rate law, and thrown it into a graphing calculator. *Notice that the slope of each of these graphs is the rate constant, k.* But first and second-order look similar, and maybe we want to be extra careful. How can we determine what order a reaction is through experimental data?

### Straight Line Test

The answer to this is by trying to set the equation up in a way so that [A] and t plot along a straight line.

Reaction Order | Integrated Rate Law | Straight Line Test |

0 | $$[A]=[A]_0-kt$$ | [A] vs t (yields -k) |

1 | $$[A]=[A]_0e^{-kt}$$ | ln[A] vs t (yields -k) |

2 | $${A]=\frac{[A]_0}{1+kt[A]_0}$$ | 1/[A] vs t (yields k) |

Now let's see what we get if we plot according to the **Straight Line Test**.

Now here's where the kicker comes in. The Straight Line Test will only work for the order that the reaction is. That means that if we have a first-order reaction, and we try to graph it for ln[A] vs t on our graphing calculator, we'll get a straight line. But if we try to graph that same reaction for [A] vs t or 1/[A] vs t, we won't get a straight line. This means that the Straight Line Test can be used to identify which order a reaction is. *Note that second-order reactions will give you a straight line with a positive slope k, while zero and first-order reactions will yield a negative straight line -k.*

### First order Integrated Rate Law in Practice

Let's try to work through an example of the Straight Line Test together.

**Practice Problem 2:** You are given a table of data that shows the concentration of reactant A with time. You remember that the Straight Line Test can help you determine reaction order, so you plot [A], ln[A], and 1/[A] against time on your graphing calculator. What is the reaction order?

Therefore, we know that this reaction is second-order because that order (graphing 1/[A] vs time) is the only one that passes the Straight Line Test.

I hope that you're now more comfortable using the Rate Law, and how it relates to reaction order. More lessons relating to specific order reactions will be published soon, so look at for those if you need additional guidance!

## Rate Law - Key takeaways

The Rate Law models a reaction's speed based on the concentration of its reactants. It consists of a rate constant, the concentration of reactants, and reactant orders.

Reactants have individual orders, and overall reactions also have orders. The overall reaction order is the sum of the reactant orders.

The Integrated Rate Law can be used to determine reaction order through experimental data.

AP Chemistry primarily deals with zero, first, and second-order reactions. Zero and first-order reactions will yield a negative slope straight line on the Straight Line Test, while second-order reactions will yield a positive slope straight line.

This slope is representative of the rate constant, k.

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