Chapter 18: Kinetics

# 18.1 Chemical Reaction Rates

### Learning Outcomes

• Use the stoichiometry of a reaction to relate the rates at which the concentrations of its reactants and products change as the reaction proceeds
• Determine average and instantaneous reaction rates
• Differentiate between average and instantaneous reaction rates using a plot of concentration versus time

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. If we measure the concentration of hydrogen peroxide, $\ce{H2O2}$, in an aqueous solution, we find that it changes slowly over time as the $\ce{H2O2}$ decomposes, according to the equation:

$\ce{2H2O2}(aq)\rightarrow\ce{2H2O}(l)+\ce{O2}(g)$

The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown below:

$\begin{array}{cc}\hfill \text{rate of decomposition of }{\ce{H}}_{2}{\ce{O}}_{2}& =-\dfrac{\text{change in concentration of reactant}}{\text{time interval}}\hfill \\ & =-\dfrac{{\left[{\ce{H}}_{2}{\ce{O}}_{2}\right]}_{{t}_{2}}-{\left[{\ce{H}}_{2}{\ce{O}}_{2}\right]}_{{t}_{1}}}{{t}_{2}-{t}_{1}}\hfill \\ & =-\dfrac{\Delta\left[{\ce{H}}_{2}{\ce{O}}_{2}\right]}{\Delta t}\hfill \end{array}$

This mathematical representation of the change in species concentration over time is the rate expression for the reaction. The brackets indicate molar concentrations, and the symbol delta (Δ) indicates “change in.” Thus, $\ce{[H2O2]_{t1}}$ represents the molar concentration of hydrogen peroxide at some time t1; likewise, $\ce{[H2O2]_{t2}}$ represents the molar concentration of hydrogen peroxide at a later time t2; and Δ$\ce{H2O2}$ represents the change in molar concentration of hydrogen peroxide during the time interval Δt (that is, t2t1). Since the reactant concentration decreases as the reaction proceeds, Δ$\ce{H2O2}$ is a negative quantity; we place a negative sign in front of the expression because reaction rates are, by convention, positive quantities. Figure 18.1.1 provides an example of data collected during the decomposition of $\ce{H2O2}$.

Figure 18.1.1. The rate of decomposition of $\ce{H2O2}$ in an aqueous solution decreases as the concentration of $\ce{H2O2}$ decreases.

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 below for the first 6-hour period:

$\dfrac{-\Delta\left[{\ce{H}}_{2}{\ce{O}}_{2}\right]}{\Delta t}=\dfrac{-\left(\text{0.500 mol/L}-\text{1.000 mol/L}\right)}{\left(\text{6.00 h}-\text{0.00 h}\right)}=0.0833 \text{ mol}{\text{ L}}^{-1}{\text{h}}^{-1}$

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

$\dfrac{-\Delta\left[{\ce{H}}_{2}{\ce{O}}_{2}\right]}{\Delta t}=\dfrac{-\left(0.0625\text{ mol/L}-0.125\text{ mol/L}\right)}{\left(24.00\text{h}-18.00\text{h}\right)}=0.010\text{ mol}{\text{ L}}^{-1}{\text{h}}^{-1}$

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 above 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. If we plot the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of $\ce{H2O2}$ at any time t is given by the slope of a straight line that is tangent to the curve at that time (Figure 18.1.2). We can use calculus to evaluating the slopes of such tangent lines, but the procedure for doing so is beyond the scope of this chapter.

## 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

$\text{aA}\rightarrow{\text{bB}}$

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:

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

Consider the reaction represented by the following equation:

$\ce{2NH3}(g)\rightarrow\ce{N2}(g)+\ce{3H2}(g)$

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

$-\dfrac{{{\Delta \text{ mol } \ce{NH}}}_{3}}{\Delta t}\times \dfrac{\text{1 mol}{\ce{ N}}_{2}}{\text{2 mol}{\ce{ NH}}_{3}}=\dfrac{\Delta\text{mol}{\ce{ N}}_{2}}{\Delta t}$

This may be represented in an abbreviated format by omitting the units of the stoichiometric factor:

$-\dfrac{1}{2}\dfrac{{\Delta}{\text{ mol } \ce{NH}}_{3}}{\Delta t}=\dfrac{\Delta\text{mol}{\ce{N}}_{2}}{\Delta t}$

Note that a negative sign has been added to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). If the reactants and products are present in the same solution, the molar amounts may be replaced by concentrations:

$-\dfrac{1}{2}\dfrac{\Delta\left[{\ce{NH}}_{3}\right]}{\Delta t}=\dfrac{\Delta\left[{\ce{N}}_{\ce{2}}\right]}{\Delta t}$

Similarly, the rate of formation of $\ce{H2}$ is three times the rate of formation of $\ce{N2}$, because three moles of $\ce{H2}$ form during the time required for the formation of one mole of $\ce{N2}$:

$\dfrac{1}{3}\dfrac{\Delta\left[{\ce{H}}_{2}\right]}{\Delta t}=\dfrac{\Delta\left[{\ce{N}}_{\ce{2}}\right]}{\Delta t}$

Figure 18.1.4 illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen and hydrogen at 1100 °C. We can see from the slopes of the tangents drawn at t = 500 seconds that the instantaneous rates of change in the concentrations of the reactants and products 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:

$\dfrac{2.91\times {10}^{-6}M\text{/s}}{9.71\times {10}^{-6}M\text{/s}}\approx 3$

This graph shows the changes in concentrations of the reactants and products during the reaction $\ce{2NH3}\rightarrow \ce{3N2}+\ce{H2}.$ The rates of change of the three concentrations are related by their stoichiometric factors, as shown by the different slopes of the tangents at t = 500s.

### Example 18.1.1: Expressions for Relative Reaction Rates

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

$\ce{4NH3}(g)+\ce{5O2}(g)\rightarrow \ce{4NO}(g)+\ce{6H2O}(g)$

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

Show Solution

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

$-\dfrac{1}{4}\dfrac{\Delta\left[{\ce{NH}}_{3}\right]}{\Delta t}=-\dfrac{1}{5}\dfrac{\Delta\left[{\ce{O}}_{2}\right]}{\Delta t}=\dfrac{1}{4}\dfrac{\Delta\left[\ce{NO}\right]}{\Delta t}=\dfrac{1}{6}\dfrac{\Delta\left[{\ce{H}}_{2}\ce{O}\right]}{\Delta t}$

### Example 18.1.2: Reaction Rate Expressions for Decomposition of $\ce{H2O2}$

The graph in Figure 18.1.4 shows the rate of the decomposition of $\ce{H2O2}$ over time:

$\ce{2H2O2}\rightarrow\ce{2H2O}+\ce{O2}$

Based on these data, the instantaneous rate of decomposition of $\ce{H2O2}$ at t = 11.1 h is determined to be $3.20\times {10}^{-2}$ mol/L/h, that is:

$-\dfrac{\Delta\left[{\ce{H}}_{2}{\ce{O}}_{2}\right]}{\Delta t}=3.20\times {10}^{-2}{\text{mol L}}^{-1}{\text{h}}^{-1}$

What is the instantaneous rate of production of $\ce{H2O}$ and $\ce{O2}$?

Show Solution

Using the stoichiometry of the reaction, we may determine that $-\dfrac{1}{2}\text{ }\dfrac{\Delta\left[{\ce{H}}_{2}{\ce{O}}_{2}\right]}{\Delta t}=\dfrac{1}{2}\dfrac{\Delta\left[{\ce{H}}_{2}\ce{O}\right]}{\Delta t}=\dfrac{\Delta\left[{\ce{O}}_{2}\right]}{\Delta t}$.

Therefore, $\dfrac{1}{2}\times 3.20\times {10}^{-2}\text{mol}{\text{ L}}^{-1}{\text{h}}^{-1}=\dfrac{\Delta\left[{\text{O}}_{2}\right]}{\Delta t}$, and $\dfrac{\Delta\left[{\ce{O}}_{2}\right]}{\Delta t}=1.60\times {10}^{-2}\text{mol}{\text{ L}}^{-1}{\text{h}}^{-1}$

### 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

• relative reaction rates for:  $\text{aA}\rightarrow \text{bB}=-\left(\dfrac{1}{a}\right)\left(\dfrac{\Delta\text{A}}{\Delta t}\right)=\left(\dfrac{1}{b}\right)\left(\dfrac{\Delta\text{B}}{\Delta t}\right)$

### Try It

1. Ozone decomposes to oxygen according to the equation $\ce{2O3}(g)\rightarrow\ce{3O2}(g).$ Write the equation that relates the rate expressions for this reaction in terms of the disappearance of $\ce{O3}$ and the formation of oxygen.
2. 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 reaction $\ce{Cl2}(g)+\ce{3F2}(g)\rightarrow\ce{2ClF3}(g)$. Write the equation that relates the rate expressions for this reaction in terms of the disappearance of $\ce{Cl2}$ and $\ce{F2}$ and the formation of $\ce{ClF3}$.
3. Consider the following reaction in aqueous solution: $\ce{5Br-}(\mathit{\text{aq}})+\ce{BrO3-}(\mathit{\text{aq}})+\ce{6H+}(\mathit{\ce{aq}})\rightarrow\ce{3Br2}(\mathit{\text{aq}})+\ce{3H2O}(l)$ If the rate of disappearance of Br(aq) at a particular moment during the reaction is $3.5\times {10}^{-4}M{\text{s}}^{-1},$ what is the rate of appearance of Br2(aq) at that moment?
Show Selected Solutions
1. Write the rate of change with a negative sign for substances decreasing in concentration (reactants) and a positive sign for those substances being formed (products). Multiply each term by the reciprocal of its coefficient:
1. $\text{rate}=+\frac{1}{2}\frac{\Delta\left[{\ce{CIF}}_{3}\right]}{\Delta t}=-\frac{\Delta\left[{\ce{Cl}}_{2}\right]}{\Delta t}=-\frac{1}{3}\frac{\Delta\left[{\ce{F}}_{2}\right]}{\Delta t}$

## 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 relating reaction rate to changes in amount, concentration, or pressure of reactant or product species per unit time