The Change of Concentration with Time

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 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.

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.

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 CH₃I. The concentration of water in an aqueous solution is about 56 M, so if we have 0.15 M of CH₃I, the change in the concentration of water as it reacts with CH₃I 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.

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.

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^-1s^-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.

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:

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.

Fig. 1: The change in the natural log of concentration over time is a linear relationship. The slope is equal to the negative rate constant. StudySmarter original.

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.

Fig. 2: In a second-order reaction dependent on one reactant, the change in inverse concentration over time is linear. StudySmarter Original.

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^-1s^-1.

Concentration vs. Time Graph for Zero-order Reaction

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

Fig. 3: The graph of concentration over time for zero-order reactions is linear. StudySmarter original.

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.