KAR - Rates of Reactions (Lesson)

Rates of Reactions

All rates are expressed as some change over time. The most common way to describe these changes is by representing them as a change in concentration over time. Consider the representation of the hypothetical chemical reaction shown below.

A $LaTeX: \longrightarrow$ $\longrightarrow$ B

Hypothetical chemical reaction There are two ways that the rate of this reaction can be expressed. The rate can be expressed in terms of how fast the reactant A is disappearing OR how fast the product B is being produced. By convention, reaction rates are always expressed in terms of the rate of disappearance of the product. This can be seen mathematically as shown below:

$LaTeX: Rate=-\frac{\Delta\lbrack A]}{\Delta t}=-\frac{\left\lbrack A\right\rbrack_f-\left\lbrack A\right\rbrack_i}{t_f-t_i}$ $Rate=-\frac{\Delta\lbrack A]}{\Delta t}=-\frac{\left\lbrack A\right\rbrack_f-\left\lbrack A\right\rbrack_i}{t_f-t_i}$

where $LaTeX: \Delta$ $\Delta$ [A] is the concentration of the reactant A and $LaTeX: \Delta$ $\Delta$ t is the change in time. It must also be noted that the negative sign out front is to ensure that reaction rates are positive.

Average Rates of Reaction

So in the hypothetical scenario described above it is possible to determine the average rate of reaction over the 25-second time period indicated. Let's assume that the vessel indicated is a 1 L vessel which would make the [A] (read concentration of A) 1.0 M at time 0 and 0.5 M at the 25 s time point. This calculation can be performed as shown below:

$LaTeX: Rate=-\frac{\left\lbrack A\right\rbrack_{25s}-\left\lbrack A\right\rbrack_{0s}}{t_{25s}-t_{0s}}=-\frac{0.5-1.0}{25-0}=0.020 M/s$ $Rate=-\frac{\left\lbrack A\right\rbrack_{25s}-\left\lbrack A\right\rbrack_{0s}}{t_{25s}-t_{0s}}=-\frac{0.5-1.0}{25-0}=0.020 M/s$

As will be shown later, the rate of reaction is constantly changing; however, over this 25-second interval, the average rate is a decrease of the [A] by 0.020 M per second.

Instantaneous Rates of Reaction

Sometimes it is preferable, or necessary, to determine rates of reaction for a single time point. These rates are referred to as instantaneous rates of reaction because there is no time interval in these cases.

Instantaneous rates of reaction can be determined using a graph of concentration versus time by finding the slope of the tangent line to the curve at a specific point on the graph. Consider the graph shown below.

Graph of concentration vs time

For this example, we will choose the specific time point of 300 s. The next step is to find the concentration at that time: Identify the concentration of a reactant or product at the chosen time point on the graph and then draw a tangent line at that point as shown by the blue line above. This tangent line represents the slope of the curve at that particular moment. The slope of the tangent line can then be determined by calculating the change in concentration divided by the change in time.

Effect of Stoichiometry on Reaction Rate

The stoichiometry of a chemical also has an effect on rate. For example, in the equation shown for a hypothetical chemical reaction, there is a coefficient of 2 in front of the reactant B. Because it takes twice as many particles of B to react with A, the rate of the reaction with respect to B will be twice as high. Therefore, if A is disappearing at a rate of 0.25 M/s then B must be disappearing twice as fast (0.50 M/s) due to the fact that it takes twice as many moles to react with it.

A + 2 B $LaTeX: \longrightarrow$ $\longrightarrow$ C + D

Rate Laws

Calculating reaction rates, as shown above, only focuses on one reactant or product at a time. In order to get a more complete picture of the rate of the overall reaction, a rate law must be used. Let's again look at the hypothetical reaction shown below.

A + 2 B $LaTeX: \longrightarrow$ $\longrightarrow$ C + D

Because rates of reaction are always relative to the rates at which reactants are disappearing a generic rate law for this reaction can be written and expressed as:

rate = k[A]^m[B]ⁿ

In this relationship, it can be seen that the rate of the reaction depends upon several factors. First, there is a constant (k) which is unique for each chemical reaction that is dependent only upon temperature. This concept will be explored further in subsequent modules. Secondly, it can be seen that there are exponents above each of the reactant substances. These exponents are referred to as rate orders and describe what effect a change in concentration for that specific substance has on the overall rate of reaction. While there are many possibilities for these rate orders there are only three specific possibilities that need to be considered under the AP Chemistry curriculum as summarized in the table below.

Table
Rate Order	Effect on Reaction Rate
0	No effect (e.g. doubling the concentration has no effect on the rxn rate
1	Linear (e.g. doubling the concentration doubles the rxn rate
2	Exponential (e.g. doubling the concentration quadruples the rxn rate

These values for the exponents (m and n above) must be determined experimentally. They are NOT found from the coefficients of the balanced equations. Once the exponents are determined experimentally, the value of the rate constant, k, can be determined. Although you will calculate the value of k for a specific reaction at a certain temperature, know that the value of k does change with temperature. The temperature dependence of reaction rates is contained in the temperature dependence of the rate constant.

Watch the presentation below to see how a rate law is determined from experimental data. Work along with the video, as you will need to be able to do this on your own using experimental data. Be sure your volume is turned on!

Use this data table to find the rate law for the reaction and find the value of the rate constant.

The order of reaction is a term that means the sum of the coefficients found in the rate law. For the example in the first video, the order of reaction was 2 + 1 = 3 and is said to the 3rd order overall, second order in relation to nitric oxide, and first order in relation to hydrogen gas.

You Try It: Rate Laws

Complete the following self-assessment questions to ensure that you understand the material from this lesson. Click the plus sign to check your answer.

Problem One: Cyclopropane, C₃H₆, is a gas used as a general anesthetic. It undergoes a slow molecular rearrangement to propylene. At a certain temperature, the following data were obtained relating concentration and rate. What is the rate law for the reaction? What is the value of the rate constant, with correct units?

Table
Initial Concentration of C₃H₆(mol L^-1)	Rate of Formation of Propylene (mol L^-1 s^-1)
0.050	2.95 x 10^-5
0.100	5.90 x 10^-5
0.150	8.85 x 10^-5

Problem Two: The formation of small amounts of nitric oxide, NO, in automobile engines is the first step in the formation of smog. As noted earlier, nitric oxide is readily oxidized to nitrogen dioxide by the reaction 2NO + O₂ → 2NO₂. What is the rate law for the reaction? What is the rate constant with its correct units?

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