Chemical Reaction Rates

D. Latimer; D. Vanderwel; J. Hollett

75 Chemical Reaction Rates

Learning Objectives

By the end of this section, you will be able to:

Define chemical reaction rate
Express the rate of a reaction in terms of any of the reactants or products (i.e., derive the rate expressions from the balanced chemical equation)
Calculate reaction rates from experimental data

A rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time.

The rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution’s conductivity.

For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. For example, the concentration of hydrogen peroxide, H₂O₂, in an aqueous solution changes slowly over time as it decomposes according to the equation:

2H₂O₂(aq) → 2 H₂O(l) + O₂(g)

The rate at which the H₂O₂ reacts can be expressed in terms of the rate of change of the concentration of H₂O₂, as shown here:

$\begin{array}{cc}\hfill \text{rate of reaction of}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{O}}_{2}& \bigskip=\phantom{\rule{0.2em}{0ex}}\frac{\text{change in concentration of reactant}}{\text{time interval}}\hfill \\ & \bigskip=\phantom{\rule{0.2em}{0ex}}\frac{{\left[{\text{H}}_{2}{\text{O}}_{2}\right]}_{{t}_{2}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}{\left[{\text{H}}_{2}{\text{O}}_{2}\right]}_{{t}_{1}}}{{t}_{2}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}{t}_{1}}\hfill \\ & =\phantom{\rule{0.2em}{0ex}}\frac{\Delta \left[{\text{H}}_{2}{\text{O}}_{2}\right]}{\Delta t}\hfill \end{array}$

The brackets indicate molar concentrations, and the symbol delta (Δ) indicates “change in.” Thus, [H₂O₂]_t1 represents the molar concentration of hydrogen peroxide at some time t₁; likewise, [H₂O₂]_t2 represents the molar concentration of hydrogen peroxide at a later time t₂; and Δ[H₂O₂] represents the change in molar concentration of hydrogen peroxide during the time interval Δt (that is, t₂− t₁).

Recall that overall reaction rates can be determined either by the rate of disappearance of reactant, or the rate of appearance of product. The rate of change of the reactant would be a negative quantity; the rate of change of the product would be positive quantity. However the overall reaction rate should be the same regardless of the method by which it is determined. By convention, reaction rates are always positive quantities. Thus, the rate of change of the reactants (e.g. the rate at which the H₂O₂ “disappears”) must be multiplied by −1 in order to provide the overall reaction rate, as shown below:

$\begin{array}{cc}\hfill \text{rate of reaction } \bigskip=-\phantom{\rule{0.2em}{0ex}}\frac{\Delta \left[{\text{H}}_{2}{\text{O}}_{2}\right]}{\Delta t}\hfill \end{array}$

This mathematical representation of the change in species concentration over time is the rate expression for the reaction.

The table below provides an example of data collected during the decomposition of H₂O₂.

To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of 40 °C. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period:

$\text{rate of reaction } =\(\frac{-\Delta \left[{\text{H}}_{2}{\text{O}}_{2}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{-\left(\text{0.500 mol/L}-\text{1.000 mol/L}\right)}{\left(\text{6.00 h}-\text{0.00 h}\right)}\phantom{\rule{0.1em}{0ex}}=0.0833 mol\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{h}}^{-1}$

Notice that the reaction rates vary over time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of:

$\text{rate of reaction } =\(\frac{-\Delta \left[{\text{H}}_{2}{\text{O}}_{2}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{-\left(0.0625\phantom{\rule{0.2em}{0ex}}\text{mol/L}-0.125\phantom{\rule{0.2em}{0ex}}\text{mol/L}\right)}{\left(24.00\phantom{\rule{0.2em}{0ex}}\text{h}-18.00\phantom{\rule{0.2em}{0ex}}\text{h}\right)}\phantom{\rule{0.1em}{0ex}}=0.010\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{h}}^{-1}$

This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate. The instantaneous rate of a reaction at “time zero,” when the reaction commences, is its initial rate. Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle’s initial rate—analogous to the beginning of a chemical reaction—would be the speedometer reading at the moment the driver begins pressing the brakes (t₀). A few moments later, the instantaneous rate at a specific moment—call it t₁—would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car’s average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop (Δt). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates.

The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. In a plot of the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of H₂O₂ at any time t is given by the slope of a straight line that is tangent to the curve at that time (Figure 75-1)These tangent line slopes may be evaluated using calculus, but the procedure for doing so is beyond the scope of this chapter.

FIGURE 75-1: This graph shows a plot of concentration versus time for a 1.000 M solution of H₂O₂. The rate at any time is equal to the negative of the slope of a line tangent to the curve at that time. Tangents are shown at t = 0 h (“initial rate”) and at t = 12 h (“instantaneous rate” at 12 h).

Relative Rates of Reaction

The rate of a reaction may be expressed as the change in concentration of any reactant or product. For any given reaction, these rate expressions are all related simply to one another according to the reaction stoichiometry. The rate of the general reaction

aA → bB

can be expressed in terms of the decrease in the concentration of A or the increase in the concentration of B. These two rate expressions are related by the stoichiometry of the reaction:

$\text{rate}=-\left(\frac{1}{\text{a}}\right)\left(\frac{\Delta A}{\Delta t}\right)=\left(\frac{1}{\text{b}}\right)\left(\frac{\Delta B}{\Delta t}\right)$

Consider the reaction represented by the following equation:

2 NH₃(g) → N₂(g) + 3 H₂(g)

The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is:

$\text{rate of reaction } =\(-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{{\Delta mol}\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{{\Delta mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}{\Delta t}$

Note that a negative sign has been included as a factor to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). For homogeneous reactions, both the reactants and products are present in the same solution and thus occupy the same volume, so the molar amounts may be replaced with molar concentrations:

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{NH}}_{3}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{\Delta\left[{\text{N}}_{\text{2}}\right]}{\Delta t}$

Similarly, the rate of formation of H₂ is three times the rate of formation of N₂ because three moles of H₂ are produced for each mole of N₂ produced.

$\frac{1}{3}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{H}}_{2}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{\Delta\left[{\text{N}}_{\text{2}}\right]}{\Delta t}$

Figure 75-2 illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen and hydrogen at 1100 °C. Slopes of the tangent lines at t = 500 s show that the instantaneous rates derived from all three species involved in the reaction are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production:

$\frac{2.91\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}M\text{/s}}{9.70\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}M\text{/s}}\phantom{\rule{0.2em}{0ex}}\approx 3$

FIGURE 75-2: Changes in concentrations of the reactant and products for the reaction 2NH₃ → N₂ + 3 H₂. The rates of change of the three concentrations are related by the reaction stoichiometry, as shown by the different slopes of the tangents at t = 500 s.

Expressions for Relative Reaction Rates

The first step in the production of nitric acid is the combustion of ammonia:

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.

Solution

Considering the stoichiometry of this homogeneous reaction, the rates for the consumption of reactants and formation of products are:

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{NH}}_{3}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=-\phantom{\rule{0.2em}{0ex}}\frac{1}{5}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{O}}_{2}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[\text{NO}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{6}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{H}}_{2}\text{O}\right]}{\Delta t}$

Check Your Learning

The rate of formation of Br₂ is 6.0 x 10⁻⁶ mol/L/s in a reaction described by the following net ionic equation:

5 Br^– + BrO₃^– + 6 H⁺ → 3 Br₂ + 3 H₂O

Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.

Answer:

$\text{reaction rate}= \(-\phantom{\rule{0.2em}{0ex}}\frac{1}{5}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{Br}}^{\text{−}}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=-\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{BrO}}_{3}{}^{\text{−}}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=-\phantom{\rule{0.2em}{0ex}}\frac{1}{6}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{H}}^{\text{+}}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{3}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{Br}}_{2}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{3}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{H}}_{\text{2}}\text{O}\right]}{\Delta t}$

Reaction Rate Expressions for Decomposition of H₂O₂

The graph in (Figure) shows the rate of the decomposition of H₂O₂ over time:

2H₂O₂(aq) → 2 H₂O(l) + O₂(g)

Based on these data, the instantaneous rate of decomposition of H₂O₂ at t = 11.1 h is determined to be 3.20 x 10⁻² mol/L/h, that is:

$-\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{H}}_{2}{\text{O}}_{2}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=3.20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}{\text{mol L}}^{-1}{\text{h}}^{-1}$

What is the instantaneous rate of production of H₂O and O₂?

Solution

The reaction stoichiometry shows that

$-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{H}}_{2}{\text{O}}_{2}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{H}}_{2}\text{O}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{\Delta\left[{\text{O}}_{2}\right]}{\Delta t}$

Therefore:

$\frac{1}{2}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.2em}{0ex}}(3.20\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}{\text{h}}^{-1})=\phantom{\rule{0.1em}{0ex}}\frac{\Delta\left[{\text{O}}_{2}\right]}{\Delta t}$

and

$\frac{\Delta\left[{\text{O}}_{2}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=1.60\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{L}}^{-1}{\text{h}}^{-1}$

Check Your Learning

If the rate of decomposition of ammonia, NH₃, at 1150 K is 2.10 x 10⁻⁶ mol/L/s, what is the rate of production of nitrogen and hydrogen?

Answer: 1.05 x 10⁻⁶ mol/L/s of N₂; and 3.15 x 10⁻⁶ mol/L/s of H₂.

Key Concepts and Summary

The rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction.

Key Equations

For the reaction aA → bB, the reaction rate can be expressed in terms of either the reactants or the products, as follows: $\text{reaction rate}=-\phantom{\rule{0.2em}{0ex}}\frac{1}{a}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[\text{A}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{1}{b}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[\text{B}\right]}{\Delta t}$

End of Chapter Exercises

1. What is the difference between average rate, initial rate, and instantaneous rate?

Answer(s): The instantaneous rate is the rate of a reaction at any particular point in time, a period of time that is so short that the concentrations of reactants and products change by a negligible amount. The initial rate is the instantaneous rate of reaction as it starts (as product just begins to form). Average rate is the average of the instantaneous rates over a time period.

2. Ozone decomposes to oxygen according to the equation 2 O₃(g) → 3 O₂(g). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of O₃ and the formation of oxygen.

3. In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the following reaction: Cl₂(g) + 3 F₂(g) → 2 ClF₃(g). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl₂ and F₂ and the formation of ClF₃.

Answer(s): $\text{rate}=+\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{CIF}}_{3}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=-\frac{\Delta\left[{\text{Cl}}_{2}\right]}{\Delta t}\phantom{\rule{0.1em}{0ex}}=-\frac{1}{3}\phantom{\rule{0.2em}{0ex}}\frac{\Delta\left[{\text{F}}_{2}\right]}{\Delta t}$

4. A study of the rate of dimerization of C₄H₆ gave the data shown in the table:
2 C₄H₆ → C₈H₁₂

Time (s)	0	1600	3200	4800	6200
[C₄H₆] (M)	1.00 x 10⁻²	5.04 x10⁻³	3.37 x 10⁻³	2.53 x 10⁻³	2.08 x 10⁻³

(a) Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s.

(b) Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus [C₄H₆]. What are the units of this rate?

(c) Determine the average rate of formation of C₈H₁₂ at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b).

5. A study of the rate of the reaction represented as 2A → B gave the following data:

Time (s)	0.0	5.0	10.0	15.0	20.0	25.0	35.0
[A] (M)	1.00	0.775	0.625	0.465	0.360	0.285	0.230

(a) Determine the average rate of disappearance of A between 0.0 s and 10.0 s, and between 10.0 s and 20.0 s.

(b) Estimate the instantaneous rate of disappearance of A at 15.0 s from a graph of time versus [A]. What are the units of this rate?

(c) Use the rates found in parts (a) and (b) to determine the average rate of formation of B between 0.00 s and 10.0 s, and the instantaneous rate of formation of B at 15.0 s.

Answer(s): (a) average rate, 0 − 10 s = 0.0375 mol L⁻¹ s⁻¹; average rate, 10 − 20 s = 0.0265 mol L⁻¹ s⁻¹; (b) instantaneous rate, 15 s = 0.023 mol L⁻¹ s⁻¹; (c) average rate for B formation = 0.0188 mol L⁻¹ s⁻¹; instantaneous rate for B formation = 0.012 mol L⁻¹ s⁻¹

6. Consider the following reaction in aqueous solution:

5 Br^–(aq) + BrO₃^–(aq) + 6 H⁺(aq) → 3 Br₂(aq) + 3 H₂O(l)

If the rate of disappearance of Br^–(aq) at a particular moment during the reaction is 3.5 x 10⁻⁴ mol L⁻¹s⁻¹, what is the rate of appearance of Br₂(aq) at that moment?

Glossary

average rate: rate of a chemical reaction computed as the ratio of a measured change in amount or concentration of substance to the time interval over which the change occurred

initial rate: instantaneous rate of a chemical reaction at t = 0 s (immediately after the reaction has begun)

instantaneous rate: rate of a chemical reaction at any instant in time, determined by the slope of the line tangential to a graph of concentration as a function of time

rate of reaction: measure of the speed at which a chemical reaction takes place

rate expression: mathematical representation defining reaction rate as change in amount, concentration, or pressure of reactant or product species per unit time

License

Icon for the Creative Commons Attribution 4.0 International License

Relative Rates of Reaction

Expressions for Relative Reaction Rates

Check Your Learning

Reaction Rate Expressions for Decomposition of H2O2

Check Your Learning

Glossary

License

Share This Book

Reaction Rate Expressions for Decomposition of H₂O₂