As time passes, the flow of foam slows down until it eventually stops, and it's time to clean up all that foam. So, why does it slow down and then stop? Well, that is because of the change in **concentration **of the reactants.

A **species **** concentration **is how much of it is present within a solution. It is typically measured as

**molarity**, which is in units of

**moles/L**, (M).

As the concentration of the reactants decreases, so does the rate**. **

The **rate of a reaction** is the change in the concentration of the reactants over time. It is usually measured in units of **Molarity**** **per second, **M/s**.

In this article, we will be covering the reason why concentration changes with time, and how that can affect the rate of a reaction.

- This article is about the
**change in concentration over time** - First, we will explain why the concentration changes over time and learn the formula for it
- Next, we will cover the
**differential rate law** - Then, we will dive into the concept of
**reaction order**and cover the different types of ordered reactions - Lastly, we will look at the concentration over time graphs for each of the ordered reactions

## Change in Concentration of Reactants Over Time

Here is a general forward reaction:

$$A + B \rightarrow C$$

As time passes, the concentration of A and B is going to decrease as C is formed. The concentrations will continue to decrease until one or both of the reactants are exhausted. However, there is a case where the concentration of reactants will begin to *increase.*

$$A+B\rightleftharpoons C$$

The special arrow here indicates an **equilibrium reaction **and the reaction can go forward (from reactants to products) or it can go backward (products to reactants). As time passes and the reaction proceeds forward, there will be reached a point where we have more product than reactant. At that point, the concentration of reactant will begin to increase as the reaction shifts backward, then the concentration will become stable for both reactants and products.

Moving onward, we will only be looking at forward, non-equilibrium, reactions, but it is important to remember that the concentration won't decrease to zero for every case.

## Change in Concentration Formula

The change in concentration over time is called the **rate **of the reaction. The basic formula is:

$$\text{rate}=-\frac{\Delta \text{[reactant(s)]}}{\Delta t}$$

The brackets indicate concentration, and the delta symbol (Δ) represents change.

## Change in Concentration Over Time Differential

Note, that we begin our discussion below with a derivation of a theoretical rate law from an examination of the balanced reaction equation. However, often the actual experimentally determined rate for a given reaction will not be accurately described by the theoretical rate law. If, however, it is the case that the experimentally determined rate law is described by the theoretical rate law then the following discussions concerning reaction rates will apply.

While the above mathematical expression is our basic formula, we typically express this change using a **rate law.**

The **rate law **describes the relationship between the **rate of a reaction** and the **concentration **of its **reactants**. For a general equation:

$$A + B \rightarrow C$$

The rate law is:

$$\text{rate}=k[A][B]$$

**constant "k"**we see in the rate law is called the

**rate constant.**

The **rate constant**** (k) **is a proportionality constant that relates the **change in ****concentration** to the **reaction rate**. The constant is unique to every reaction type and its conditions (temperature, pressure, etc).

The basic formula we saw before was the **differential rate law **since it is expressed as the change in concentration over time. This differential rate law is equivalent to the rate law discussed above.

## Reaction Order

There is another factor that affects the relationship between concentration and rate, which is the reaction **order. **A given reaction's order is influenced by the magnitude of the effect of a particular reactant's concentration on the rate. There are three types of order reactions: first-order, second-order, and zero-order.

### First-order Reaction

The first type of ordered reaction is a **first-order reaction**.

A **first-order reaction **is one whose rate is dependent on the concentration of one reactant. The formula for the rate law is:

$$\text{rate}=k[A]$$

First-order reactions are often **decomposition reactio**ns since they involve one reactant breaking down into two or more products.

Here are a few examples of a first-order reaction:$$2N_2O_{5\,(g)} \rightarrow 4NO_{2\,(g)} + O_{2\,(g)}\,\,\text{rate}=k[N_2O_5]$$

$$CaCO_{3\,(s)} \rightarrow CaO_{(s)} + CO_{2\,(g)} \,\,\text{rate}=k[CaCO_3]$$

$$2H_2O_{2\,(l)} \rightarrow 2H_2O_{(l)} + O_{2\,(g)} \,\,\text{rate}=k[H_2O_2]$$

Usually, first-order reactions have only one reactant, however, there are cases where a reaction has two or more reactants called a **pseudo-first-order reaction. **

A **pseudo-first order reaction **is a reaction where the concentration change of one or more reactants is negligible, so the rate is only dependent on one reactant whose concentration changes appreciably during the progress of a reaction. This is usually because the other reactant(s) is an excess.

Here is an example of a pseudo-first-order reaction:

$$CH_3I_{(aq)} + H_2O_{(l)} \rightarrow CH_3OH_{(aq)} + H^+_{(aq)} + I^-_{(aq)}\,\,\text{rate}=k[CH_3I]$$

The reaction is taking place in an aqueous solution (i.e the whole thing is in water), so the concentration of water is *a lot *larger than the concentration of CH_{3}I. The concentration of water in an aqueous solution is about 56 M, so if we have 0.15 M of CH_{3}I, the change in the concentration of water as it reacts with CH_{3}I will be negligible and can be ignored.

When we don't know the rate of a reaction but want to solve for the rate constant, k, or the concentration at a given time, we will use the **integrated rate law**.

The **integrated rate law is **used to calculate the concentration of a reactant at a given time. It differs from the differential rate law by taking the initial concentration into account.

For a first-order reaction, we have two versions of the integrated rate law:

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

$$ln[A]=-kt+ln[A]_0$$

Where:

**[A]**, represents the**concentration****of the reactant**.**[A]**_{0}, the**initial concentration of the reactant**.**t**, the**time**.**k**, the**rate constant**.**ln**, the**natural logarithm.**

The second form of the equation is a **linear form** (y = mx+b). This means that when the natural logarithm of the reactant, ln[A], over time is graphed, the slope of the line is equal to the negative rate constant, -k.

$$y=mx+b$$

Also, the units of the rate constant differ between reaction order types. The rate of the reaction will always be in units of M/s, so the units of the proportionality constant, k, are changed, so that is true. For a first-order reaction, k's units are 1/s or s^{-1}

Here is how we get these units for k:

$$\text{rate}=k[A]$$

$$\frac{M}{s}=k*M\,\text{(converted variables to their units)}$$

$$k=\frac{1}{s}$$

### Second-order Reaction

The second type of ordered reaction is the **second-order reaction.**

In a **second-order reaction, **the rate of the reaction is dependent on either the squared concentration of one reactant or the concentrations of two reactants. The general formulas are:

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

$$\text{rate}=k[A][B]$$

There are two versions of the integrated rate law: one for each type. The integrated rate law for reactions dependent on two reactants is a bit complicated, so we will only be looking at the equation for reactions dependent on one reactant.

$$\frac{1}{[A]}=kt+\frac{1}{[A]_0}$$

When 1/[A] is graphed over time, the slope will be equal to the rate constant.

The rate constant for second-order reactions is in units of 1/Ms or M^{-1}s^{-1}, and that is the case for both types.

Here is how we get those units:

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

$$\frac{M}{s}=kM^2$$

$$k=\frac{1}{M*s}$$

$$\text{rate}=k[A][B]$$

$$\frac{M}{s}=kM*M$$

$$k=\frac{1}{M*s}$$

### Zero-order Reaction

The last type of ordered reaction is the **zero-order reaction.**

In a **zero-order reaction, **the rate is independent of the concentration of the reactant(s). The rate is solely based on the rate constant, so the formula looks like:

$$\text{rate}=k$$

Zero-order reactions are a lot less common than the other types. A reason why a reaction is zero-order is that the reaction is taking place on a solid-surface **catalyst**.

A **catalyst **is a species that speeds up a reaction. The reactant(s) binds to the surface of the catalyst, however, there are a limited number of spots on the catalyst.

Think of it like standing in line for a roller coaster. The coaster has a set number of seats, so only a few people can get on at a time. The speed at which the line moves is independent of the number of people in the line. There could be 300 people in line, but it will move at the same rate as a line with 30 people.Here are some **examples **of a zero-order reaction:

$$2NH_{3\,(g)} \xrightarrow {\text{Fe catalyst}} N_{2\,(g)} + 3H_{2\,(g)}$$

$$2N_2O_{(g)} \xrightarrow {\text{Heat and Pt catalyst}} 2N_{2\,(g)} + O_{2\,(g)}$$

$$2HI_{(g)} \xrightarrow {\text{Au catalyst}} H_{2\,(g)} + I_{2\,(g)} $$

While the rate is independent of the reactant concentration, the concentration is still changing over time. Because of this, we still have an integrated rate equation, which is:

$$[A]=-kt+[A]_0$$

So if the concentration over time is graphed, the slope is equal to the negative rate constant. The units for the rate constant are M/s, since the rate is equal to k.

## Concentration vs Time Graph for First-order Reaction

You can identify the order of a reaction based on the graph of concentration over time. For a first-order reaction, the graph will only be linear if the concentration over time is graphed as the natural log (ln) of concentration over time.

On the graph above, we are given the equation of the line. Since we know that the slope is the negative rate constant, the rate constant for this reaction is, k = 0.0569 s^{-1}

## Concentration vs. Time Graph for Second-order Reaction

For a second-order reaction, the graph of the inverse concentration (1/[A]) over time will be linear.

In cases where the reaction rate is dependent on one reactant, the relationship between the inverse concentration and time is linear. The slope of the graph is the rate constant, which in this case is, k = 0.448 M^{-1}s^{-1}.

## Concentration vs. Time Graph for Zero-order Reaction

Lastly, zero-order reactions have a linear relationship between the change in concentration and time.

Like with first-order reactions, the slope of the graph is equal to the negative rate constant, so the rate constant here is equal to, **k = 0.02 M/s.**

## The Change of Concentration with Time - Key takeaways

- The
**rate**of a reaction is the change in the concentration of the reactants over time. It is usually measured in units of Molarity per second, M/s. - The
**rate law**describes the relationship between the rate of a reaction and the concentration of its reactants. **A first-order reaction**is one whose rate is dependent on the concentration of one reactant. The formula for the rate law is: \(\text{rate}=k[A]\). The units of the rate constant, k, are per second, s^{-1}- In a
**second-order reaction,**the rate of the reaction is dependent on either the squared concentration of one reactant or the concentrations of two reactants. The general formulas are: \(\text{rate}=k[A]^2\,\text{or rate}=k[A][B]\). The units of the rate constant, k, are per Molarity per second, M^{-1}s^{-1} - In a
**zero-order reaction,**the rate is independent of the concentration of the reactant(s). The rate is solely based on the rate constant, so the formula looks like: \(\text{rate}=k\). The units of the rate constant, k, are molarity per second, M/s

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##### Frequently Asked Questions about The Change of Concentration with Time

What is change in concentration over change in time?

The **rate **of a reaction is the change in the concentration of the reactants over time. It is usually measured in units of Molarity per second, M/s.

Does concentration of reactants increase with time?

Concentration can increase for equilibrium reactions.

As time passes and the reaction proceeds forward, there will be reached a point where we have more product than reactant. At that point, the concentration of reactant will begin to increase as the reaction shifts backward, then the concentration will become stable for both reactants and products.

However, for non-equilibrium reactions, the concentration of reactants will always decrease as they are consumed to form the product(s)

What is the change in concentration per unit time?

The change in concentration per unit time is the instantaneous rate

How do you find concentration after time?

To find the concentration at a given time, we need to know the rate of the reactions. If we multiply time by the rate, we get the *change* in concentration. Adding the initial concentration to this value, we get the concentration at time "t".

How do you graph time vs concentration?

When we graph the change in concentration over time. We put concentration on our y-axis and time on the x-axis.

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