1827. Solution -
For a second-order reaction, the integrated rate law is written

$$
\frac{1}{[A]_{t}}=k t+\frac{1}{[A]_{0}}
$$

We know three variables in this equation: $[A]_{0}=0.200 \mathrm{~mol} / \mathrm{L}, k=5.76 \times 10^{-2} \mathrm{~L} / \mathrm{mol} / \mathrm{min}$, and $t=10.0 \mathrm{~min}$. Therefore, we can solve for $[A]$, the fourth variable:

$$
\begin{aligned}
\frac{1}{[A]_{t}} & =\left(5.76 \times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~min}^{-1}\right)(10 \mathrm{~min})+\frac{1}{0.200 \mathrm{~mol}^{-1}} \\
\frac{1}{[A]_{t}} & =\left(5.76 \times 10^{-1} \mathrm{~L} \mathrm{~mol}^{-1}\right)+5.00 \mathrm{~L} \mathrm{~mol}^{-1} \\
\frac{1}{[A]_{t}} & =5.58 \mathrm{~L} \mathrm{~mol}^{-1} \\
{[A]_{t} } & =1.79 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
$$

Therefore $0.179 \mathrm{~mol} / \mathrm{L}$ of butadiene remain at the end of 10.0 min , compared to the $0.200 \mathrm{~mol} / \mathrm{L}$ that was originally present.


1828. Check Your Learning -
If the initial concentration of butadiene is 0.0200 M , what is the concentration remaining after 20.0 min?


1829. Answer: -
$0.0195 \mathrm{~mol} / \mathrm{L}$

The integrated rate law for second-order reactions has the form of the equation of a straight line:

$$
\begin{aligned}
\frac{1}{[A]_{t}} & =k t+\frac{1}{[A]_{0}} \\
y & =m x+b
\end{aligned}
$$

A plot of $\frac{1}{[A]_{t}}$ versus $t$ for a second-order reaction is a straight line with a slope of $k$ and a $y$-intercept of $\frac{1}{[A]_{0}}$. If the plot is not a straight line, then the reaction is not second order.


1831. Graphical Determination of Reaction Order and Rate Constant -
The data below are for the same reaction described in Example 17.8. Prepare and compare two appropriate data plots to identify the reaction as being either first or second order. After identifying the reaction order, estimate a value for the rate constant.


1832. Solution -
| Trial | Time (s) | $\left\mathrm{C}_{4} \mathrm{H}_{6}\right$ |
| :--- | :--- | :--- |
| 1 | 0 | $1.00 \times 10^{-2}$ |
| 2 | 1600 | $5.04 \times 10^{-3}$ |
| 3 | 3200 | $3.37 \times 10^{-3}$ |
| 4 | 4800 | $2.53 \times 10^{-3}$ |
| 5 | 6200 | $2.08 \times 10^{-3}$ |

In order to distinguish a first-order reaction from a second-order reaction, prepare a plot of $\ln \left[\mathrm{C}_{4} \mathrm{H}_{6}\right]_{t}$ versus $t$ and compare it to a plot of $\frac{1}{\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]_{t}}$ versus $t$. The values needed for these plots follow.

| Time (s) | $\frac{1}{\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]}\left(M^{-1}\right)$ | $\ln \left[\mathrm{C}_{4} \mathrm{H}_{6}\right]$ |
| :--- | :--- | :--- |
| 0 | 100 | -4.605 |
| 1600 | 198 | -5.289 |
| 3200 | 296 | -5.692 |
| 4800 | 395 | -5.978 |
| 6200 | 481 | -6.175 |

The plots are shown in Figure 17.10, which clearly shows the plot of $\ln \left[\mathrm{C}_{4} \mathrm{H}_{6}\right]_{t}$ versus $t$ is not linear, therefore the reaction is not first order. The plot of $\frac{1}{\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]_{t}}$ versus $t$ is linear, indicating that the reaction is second order.


1832. Solution -
FIGURE 17.10 These two graphs show first- and second-order plots for the dimerization of $\mathrm{C}_{4} \mathrm{H}_{6}$. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.

According to the second-order integrated rate law, the rate constant is equal to the slope of the $\frac{1}{[A]_{t}}$ versus $t$ plot. Using the data for $t=0 \mathrm{~s}$ and $t=6200 \mathrm{~s}$, the rate constant is estimated as follows:

$$
k=\text { slope }=\frac{\left(481 M^{-1}-100 M^{-1}\right)}{(6200 \mathrm{~s}-0 \mathrm{~s})}=0.0614 \mathrm{M}^{-1} \mathrm{~s}^{-1}
$$


1833. Check Your Learning -
Do the following data fit a second-order rate law?

| Trial | Time (s) | $A$ |
| :--- | :--- | :--- |
| 1 | 5 | 0.952 |
| 2 | 10 | 0.625 |
| 3 | 15 | 0.465 |
| 4 | 20 | 0.370 |
| 5 | 25 | 0.308 |
| 6 | 35 | 0.230 |


1834. Answer: -
Yes. The plot of $\frac{1}{[A]_{t}}$ vs. $t$ is linear:


1835. Zero-Order Reactions -
For zero-order reactions, the differential rate law is:

$$
\text { rate }=k
$$

A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactant(s). This may seem counterintuitive, since the reaction rate certainly can't be finite when the reactant concentration is zero. For purposes of this introductory text, it will suffice to note that zero-order kinetics are observed for some reactions only under certain specific conditions. These same reactions exhibit different kinetic behaviors when the specific conditions aren't met, and for this reason the more prudent term pseudo-zero-order is sometimes used.

The integrated rate law for a zero-order reaction is a linear function:

$$
\begin{aligned}
{[A]_{t} } & =-k t+[A]_{0} \\
y & =m x+b
\end{aligned}
$$

A plot of $[A]$ versus $t$ for a zero-order reaction is a straight line with a slope of $-k$ and a $y$-intercept of $[A]_{0}$. Figure 17.11 shows a plot of $\left[\mathrm{NH}_{3}\right]$ versus $t$ for the thermal decomposition of ammonia at the surface of two different heated solids. The decomposition reaction exhibits first-order behavior at a quartz $\left(\mathrm{SiO}_{2}\right)$ surface, as suggested by the exponentially decaying plot of concentration versus time. On a tungsten surface, however, the plot is linear, indicating zero-order kinetics.


1837. Graphical Determination of Zero-Order Rate Constant -
Use the data plot in Figure 17.11 to graphically estimate the zero-order rate constant for ammonia decomposition at a tungsten surface.


1838. Solution -
The integrated rate law for zero-order kinetics describes a linear plot of reactant concentration, $[A]_{t}$, versus time, $t$, with a slope equal to the negative of the rate constant, $-k$. Following the mathematical approach of previous examples, the slope of the linear data plot (for decomposition on W ) is estimated from the graph. Using the ammonia concentrations at $t=0$ and $t=1000 \mathrm{~s}$ :

$$
k=- \text { slope }=-\frac{\left(0.0015 \mathrm{~mol} \mathrm{~L}^{-1}-0.0028 \mathrm{~mol} \mathrm{~L}^{-1}\right)}{(1000 \mathrm{~s}-0 \mathrm{~s})}=1.3 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
$$


1839. Check Your Learning -
The zero-order plot in Figure 17.11 shows an initial ammonia concentration of $0.0028 \mathrm{~mol} \mathrm{~L}^{-1}$ decreasing linearly with time for 1000 s . Assuming no change in this zero-order behavior, at what time (min) will the concentration reach $0.0001 \mathrm{~mol} \mathrm{~L}^{-1}$ ?


1840. Answer: -
35 min

FIGURE 17.11 The decomposition of $\mathrm{NH}_{3}$ on a tungsten $(\mathrm{W})$ surface is a zero-order reaction, whereas on a quartz $\left(\mathrm{SiO}_{2}\right)$ surface, the reaction is first order.

The Half-Life of a Reaction
The half-life of a reaction ( $\boldsymbol{t}_{\mathbf{1} / \mathbf{2}}$ ) is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide (Figure 17.2) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of $\mathrm{H}_{2} \mathrm{O}_{2}$ decreases from 1.000 M to 0.500 M . During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M ; during the third half-life, it decreases from 0.250 M to 0.125 M . The concentration of $\mathrm{H}_{2} \mathrm{O}_{2}$ decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.


1841. First-Order Reactions -
An equation relating the half-life of a first-order reaction to its rate constant may be derived from the integrated rate law as follows:

$$
\begin{aligned}
\ln \frac{[A]_{0}}{[A]_{t}} & =k t \\
t & =\ln \frac{[A]_{0}}{[A]_{t}} \times \frac{1}{k}
\end{aligned}
$$

Invoking the definition of half-life, symbolized $t_{1 / 2}$, requires that the concentration of $A$ at this point is one-half its initial concentration: $t=t_{1 / 2},[A]_{t}=\frac{1}{2}[A]_{0}$.

Substituting these terms into the rearranged integrated rate law and simplifying yields the equation for halflife:

$$
\begin{aligned}
t_{1 / 2} & =\ln \frac{[A]_{0}}{\frac{1}{2}[A]_{0}} \times \frac{1}{k} \\
& =\ln 2 \times \frac{1}{k}=0.693 \times \frac{1}{k} \\
t_{1 / 2} & =\frac{0.693}{k}
\end{aligned}
$$

This equation describes an expected inverse relation between the half-life of the reaction and its rate constant, k. Faster reactions exhibit larger rate constants and correspondingly shorter half-lives. Slower reactions exhibit smaller rate constants and longer half-lives.


1843. Calculation of a First-order Rate Constant using Half-Life -
Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at $40^{\circ} \mathrm{C}$, using the data given in Figure 17.12.

FIGURE 17.12 The decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}\left(2 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\right)$ at $40^{\circ} \mathrm{C}$ is illustrated. The intensity of the color symbolizes the concentration of $\mathrm{H}_{2} \mathrm{O}_{2}$ at the indicated times; $\mathrm{H}_{2} \mathrm{O}_{2}$ is actually colorless.


1844. Solution -
Inspecting the concentration/time data in Figure 17.12 shows the half-life for the decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}$ is $2.16 \times 10^{4} \mathrm{~s}$ :

$$
\begin{aligned}
t_{1 / 2} & =\frac{0.693}{k} \\
k & =\frac{0.693}{t_{1 / 2}}=\frac{0.693}{2.16 \times 10^{4} \mathrm{~s}}=3.21 \times 10^{-5} \mathrm{~s}^{-1}
\end{aligned}
$$


1845. Check Your Learning -
The first-order radioactive decay of iodine-131 exhibits a rate constant of $0.138 \mathrm{~d}^{-1}$. What is the half-life for this decay?

Answer:
5.02 d.


1846. Second-Order Reactions -
Following the same approach as used for first-order reactions, an equation relating the half-life of a secondorder reaction to its rate constant and initial concentration may be derived from its integrated rate law:

$$
\frac{1}{[A]_{t}}=k t+\frac{1}{[A]_{0}}
$$

or

$$
\frac{1}{[A]}-\frac{1}{[A]_{0}}=k t
$$

Restrict $t$ to $t_{1 / 2}$

$$
t=t_{1 / 2}
$$

define $[A]_{t}$ as one-half $[A]_{0}$

$$
[A]_{t}=\frac{1}{2}[A]_{0}
$$

and then substitute into the integrated rate law and simplify:

$$
\begin{aligned}
\frac{1}{\frac{1}{2}[A]_{0}}-\frac{1}{[A]_{0}} & =k t_{1 / 2} \\
\frac{2}{[A]_{0}}-\frac{1}{[A]_{0}} & =k t_{1 / 2} \\
\frac{1}{[A]_{0}} & =k t_{1 / 2} \\
t_{1 / 2} & =\frac{1}{k[A]_{0}}
\end{aligned}
$$

For a second-order reaction, $t_{1 / 2}$ is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.


1847. Zero-Order Reactions -
As for other reaction orders, an equation for zero-order half-life may be derived from the integrated rate law:

$$
[A]=-k t+[A]_{0}
$$

Restricting the time and concentrations to those defined by half-life: $t=t_{1 / 2}$ and $[A]=\frac{[A]_{0}}{2}$. Substituting these terms into the zero-order integrated rate law yields:

$$
\begin{aligned}
\frac{[\mathrm{A}]_{0}}{2} & =-k t_{1 / 2}+[\mathrm{A}]_{0} \\
k t_{1 / 2} & =\frac{[\mathrm{A}]_{0}}{2} \\
t_{1 / 2} & =\frac{[A]_{0}}{2 k}
\end{aligned}
$$

As for all reaction orders, the half-life for a zero-order reaction is inversely proportional to its rate constant. However, the half-life of a zero-order reaction increases as the initial concentration increases.

Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table 17.2.

Summary of Rate Laws for Zero-, First-, and Second-Order Reactions

| Zero-Order |  | First-Order | Second-Order |
| :--- | :--- | :--- | :--- |
| rate law | rate $=k$ | rate $=k[A]$ | rate $=k[A]^{2}$ |
| units of rate constant | $M s^{-1}$ | $\mathrm{~s}^{-1}$ | $M^{-1} \mathrm{~s}^{-1}$ |
| integrated rate law | $[A]=-k t+[A]_{0}$ | $\ln [A]=-k t+\ln [A]_{0}$ | $\frac{1}{[A]}=k t+\left(\frac{1}{[A]_{0}}\right)$ |

TABLE 17.2


1847. Zero-Order Reactions -
TABLE 17.2

|  | Zero-Order | First-Order | Second-Order |
| :--- | :--- | :--- | :--- |
| plot needed for linear fit of rate data | $[A]$ vs. $t$ | $\ln [A]$ vs. $t$ | $\frac{1}{[A]}$ vs. $t$ |
| relationship between slope of linear <br> plot and rate constant | $k=-$ slope | $k=-$ slope | $k=$ slope |
| half-life | $t_{1 / 2}=\frac{[A]_{0}}{2 k}$ | $t_{1 / 2}=\frac{0.693}{k}$ | $t_{1 / 2}=\frac{1}{[A]_{0} k}$ |

TABLE 17.2


1849. Half-Life for Zero-Order and Second-Order Reactions -
What is the half-life for the butadiene dimerization reaction described in Example 17.8?


1850. Solution -
The reaction in question is second order, is initiated with a $0.200 \mathrm{~mol} \mathrm{~L}^{-1}$ reactant solution, and exhibits a rate constant of $0.0576 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~min}^{-1}$. Substituting these quantities into the second-order half-life equation:

$$
t_{1 / 2}=\frac{1}{\left[\left(0.0576 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~min}^{-1}\right)\left(0.200 \mathrm{~mol} \mathrm{~L}^{-1}\right)\right]}=18 \mathrm{~min}
$$


1851. Check Your Learning -
What is the half-life (min) for the thermal decomposition of ammonia on tungsten (see Figure 17.11)?


1852. Answer: -
87 min


1853. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates
- Define the concepts of activation energy and transition state
- Use the Arrhenius equation in calculations relating rate constants to temperature

We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates.

Collision theory is based on the following postulates:

1. The rate of a reaction is proportional to the rate of reactant collisions:

$$
\text { reaction rate } \propto \frac{\# \text { collisions }}{\text { time }}
$$

2. The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.
3. The collision must occur with adequate energy to permit mutual penetration of the reacting species' valence shells so that the electrons can rearrange and form new bonds (and new chemical species).

We can see the importance of the two physical factors noted in postulates 2 and 3 , the orientation and energy
of collisions, when we consider the reaction of carbon monoxide with oxygen:

$$
2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)
$$

Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient amounts, the reaction will occur at high temperature and pressure.

The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules:

$$
\mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{O}(g)
$$


1853. LEARNING OBJECTIVES -
$$
\mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{O}(g)
$$

Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure 17.13. In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms $(\mathrm{O}=\mathrm{C}=\mathrm{O})$. This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction.

FIGURE 17.13 Illustrated are two collisions that might take place between carbon monoxide and oxygen molecules. The orientation of the colliding molecules partially determines whether a reaction between the two molecules will occur.

If the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. In addition to a proper orientation, the collision must also occur with sufficient energy to result in product formation. When reactant species collide with both proper orientation and adequate energy, they combine to form an unstable species called an activated complex or a transition state. These species are very short lived and usually undetectable by most analytical instruments. In some cases, sophisticated spectral measurements have been used to observe transition states.

Collision theory explains why most reaction rates increase as concentrations increase. With an increase in the concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. More collisions mean a faster reaction rate, assuming the energy of the collisions is adequate.


1854. Activation Energy and the Arrhenius Equation -
The minimum energy necessary to form a product during a collision between reactants is called the activation energy $\left(\boldsymbol{E}_{\mathrm{a}}\right)$. How this energy compares to the kinetic energy provided by colliding reactant molecules is a primary factor affecting the rate of a chemical reaction. If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly since only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, a large fraction of molecules will be adequately energetic and the reaction will proceed rapidly.

Figure 17.14 shows how the energy of a chemical system changes as it undergoes a reaction converting reactants to products according to the equation

$$
A+B \longrightarrow C+D
$$

These reaction diagrams are widely used in chemical kinetics to illustrate various properties of the reaction of interest. Viewing the diagram from left to right, the system initially comprises reactants only, $A+B$. Reactant molecules with sufficient energy can collide to form a high-energy activated complex or transition state. The unstable transition state can then subsequently decay to yield stable products, $C+D$. The diagram depicts the reaction's activation energy, $E_{a}$, as the energy difference between the reactants and the transition state. Using a specific energy, the enthalpy (see chapter on thermochemistry), the enthalpy change of the reaction, $\Delta H$, is estimated as the energy difference between the reactants and products. In this case, the reaction is exothermic $(\Delta H<0)$ since it yields a decrease in system enthalpy.

FIGURE 17.14 Reaction diagram for the exothermic reaction $A+B \longrightarrow C+D$.
The Arrhenius equation relates the activation energy and the rate constant, $k$, for many chemical reactions:

$$
k=A e^{-E_{\mathrm{a}} / R T}
$$


1854. Activation Energy and the Arrhenius Equation -
$$
k=A e^{-E_{\mathrm{a}} / R T}
$$

In this equation, $R$ is the ideal gas constant, which has a value $8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K}, \mathrm{T}$ is temperature on the Kelvin scale, $E_{\mathrm{a}}$ is the activation energy in joules per mole, $e$ is the constant 2.7183 , and $A$ is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.

Postulates of collision theory are nicely accommodated by the Arrhenius equation. The frequency factor, $A$, reflects how well the reaction conditions favor properly oriented collisions between reactant molecules. An increased probability of effectively oriented collisions results in larger values for $A$ and faster reaction rates.
The exponential term, $e^{-E a / R T}$, describes the effect of activation energy on reaction rate. According to kinetic molecular theory (see chapter on gases), the temperature of matter is a measure of the average kinetic energy of its constituent atoms or molecules. The distribution of energies among the molecules composing a sample of matter at any given temperature is described by the plot shown in Figure 17.15(a). Two shaded areas under the curve represent the numbers of molecules possessing adequate energy $(R T)$ to overcome the activation barriers $\left(E_{a}\right)$. A lower activation energy results in a greater fraction of adequately energized molecules and a faster reaction.

The exponential term also describes the effect of temperature on reaction rate. A higher temperature represents a correspondingly greater fraction of molecules possessing sufficient energy ( $R T$ ) to overcome the activation barrier $\left(E_{a}\right)$, as shown in Figure $17.15(\mathbf{b})$. This yields a greater value for the rate constant and a correspondingly faster reaction rate.

FIGURE 17.15 Molecular energy distributions showing numbers of molecules with energies exceeding (a) two different activation energies at a given temperature, and (b) a given activation energy at two different temperatures.

A convenient approach for determining $E_{\mathrm{a}}$ for a reaction involves the measurement of $k$ at two or more different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation


1854. Activation Energy and the Arrhenius Equation -
A convenient approach for determining $E_{\mathrm{a}}$ for a reaction involves the measurement of $k$ at two or more different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation

$$
\begin{aligned}
\ln k & =\left(\frac{-E_{\mathrm{a}}}{R}\right)\left(\frac{1}{T}\right)+\ln A \\
y & =m x+b
\end{aligned}
$$

A plot of $\ln k$ versus $\frac{1}{T}$ is linear with a slope equal to $\frac{-E_{\mathrm{a}}}{R}$ and a $y$-intercept equal to $\ln A$.


1855. Determination of $E_{\mathrm{a}}$ -
The variation of the rate constant with temperature for the decomposition of $\mathrm{HI}(g)$ to $\mathrm{H}_{2}(g)$ and $\mathrm{I}_{2}(g)$ is given here. What is the activation energy for the reaction?

$$
2 \mathrm{HI}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)
$$

| $\boldsymbol{T}(\mathrm{K})$ | $k(\mathrm{~L} / \mathrm{mol} / \mathrm{s})$ |
| :---: | :---: |
| 555 | $3.52 \times 10^{-7}$ |
| 575 | $1.22 \times 10^{-6}$ |
| 645 | $8.59 \times 10^{-5}$ |
| 700 | $1.16 \times 10^{-3}$ |
| 781 | $3.95 \times 10^{-2}$ |


1856. Solution -
Use the provided data to derive values of $\frac{1}{T}$ and $\ln k$ :

| $\frac{1}{\mathrm{~T}}\left(\mathrm{~K}^{-1}\right)$ | $\ln \boldsymbol{k}$ |
| :--- | :--- |
| $1.80 \times 10^{-3}$ | -14.860 |
| $1.74 \times 10^{-3}$ | -13.617 |

| $\frac{1}{\mathrm{~T}}\left(\mathrm{~K}^{-1}\right)$ | $\ln k$ |
| :--- | :--- |
| $1.55 \times 10^{-3}$ | -9.362 |
| $1.43 \times 10^{-3}$ | -6.759 |
| $1.28 \times 10^{-3}$ | -3.231 |

Figure 17.16 is a graph of $\ln k$ versus $\frac{1}{T}$. In practice, the equation of the line (slope and $y$-intercept) that best fits these plotted data points would be derived using a statistical process called regression. This is helpful for most experimental data because a perfect fit of each data point with the line is rarely encountered. For the data here, the fit is nearly perfect and the slope may be estimated using any two of the provided data pairs. Using the first and last data points permits estimation of the slope.

FIGURE 17.16 This graph shows the linear relationship between $\ln k$ and $\frac{1}{T}$ for the reaction $2 \mathrm{HI} \longrightarrow \mathrm{H}_{2}+\mathrm{I}_{2}$ according to the Arrhenius equation.


1856. Solution -
$$
\begin{aligned}
\text { Slope }= & \frac{\Delta(\ln k)}{\Delta\left(\frac{1}{T}\right)} \\
= & \frac{(-14.860)-(-3.231)}{\left(1.80 \times 10^{-3} \mathrm{~K}^{-1}\right)-\left(1.28 \times 10^{-3} \mathrm{~K}^{-1}\right)} \\
= & \frac{-11.629}{0.52 \times 10^{-3} \mathrm{~K}^{-1}}=-2.2 \times 10^{4} \mathrm{~K} \\
= & -\frac{E_{\mathrm{a}}}{R} \\
E_{\mathrm{a}}= & - \text { slope } \times R=-\left(-2.2 \times 10^{4} \mathrm{~K} \times 8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right) \\
& 1.8 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1} \text { or } 180 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$

Alternative approach: A more expedient approach involves deriving activation energy from measurements of the rate constant at just two temperatures. In this approach, the Arrhenius equation is rearranged to a convenient two-point form:

$$
\ln \frac{k_{1}}{k_{2}}=\frac{E_{\mathrm{a}}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)
$$

Rearranging this equation to isolate activation energy yields:

$$
E_{\mathrm{a}}=-R\left(\frac{\ln k_{2}-\ln k_{1}}{\left(\frac{1}{T_{2}}\right)-\left(\frac{1}{T_{1}}\right)}\right)
$$

Any two data pairs may be substituted into this equation-for example, the first and last entries from the above data table:


1856. Solution -
Any two data pairs may be substituted into this equation-for example, the first and last entries from the above data table:

$$
E_{\mathrm{a}}=-8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\left(\frac{-3.231-(-14.860)}{1.28 \times 10^{-3} \mathrm{~K}^{-1}-1.80 \times 10^{-3} \mathrm{~K}^{-1}}\right)
$$

and the result is $E_{\mathrm{a}}=1.8 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}$ or $180 \mathrm{~kJ} \mathrm{~mol}^{-1}$
This approach yields the same result as the more rigorous graphical approach used above, as expected. In practice, the graphical approach typically provides more reliable results when working with actual experimental data.


1857. Check Your Learning -
The rate constant for the rate of decomposition of $\mathrm{N}_{2} \mathrm{O}_{5}$ to NO and $\mathrm{O}_{2}$ in the gas phase is $1.66 \mathrm{~L} / \mathrm{mol} / \mathrm{s}$ at 650 K and $7.39 \mathrm{~L} / \mathrm{mol} / \mathrm{s}$ at 700 K :

$$
2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}(g)+3 \mathrm{O}_{2}(g)
$$

Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition.


1858. Answer: -
$1.1 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}$ or $110 \mathrm{~kJ} \mathrm{~mol}^{-1}$


1859. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Distinguish net reactions from elementary reactions (steps)
- Identify the molecularity of elementary reactions
- Write a balanced chemical equation for a process given its reaction mechanism
- Derive the rate law consistent with a given reaction mechanism

Chemical reactions very often occur in a step-wise fashion, involving two or more distinct reactions taking place in sequence. A balanced equation indicates what is reacting and what is produced, but it reveals no details about how the reaction actually takes place. The reaction mechanism (or reaction path) provides details regarding the precise, step-by-step process by which a reaction occurs.

The decomposition of ozone, for example, appears to follow a mechanism with two steps:

$$
\begin{aligned}
& \mathrm{O}_{3}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{O} \\
& \mathrm{O}+\mathrm{O}_{3}(g) \longrightarrow 2 \mathrm{O}_{2}(g)
\end{aligned}
$$

Each of the steps in a reaction mechanism is an elementary reaction. These elementary reactions occur precisely as represented in the step equations, and they must sum to yield the balanced chemical equation representing the overall reaction:

$$
2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)
$$

Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediates.

While the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react
to give three molecules of oxygen, the mechanism of the reaction does not involve the direct collision and reaction of two ozone molecules. Instead, one $\mathrm{O}_{3}$ decomposes to yield $\mathrm{O}_{2}$ and an oxygen atom, and a second $\mathrm{O}_{3}$ molecule subsequently reacts with the oxygen atom to yield two additional $\mathrm{O}_{2}$ molecules.


1859. LEARNING OBJECTIVES -
Unlike balanced equations representing an overall reaction, the equations for elementary reactions are explicit representations of the chemical change taking place. The reactant(s) in an elementary reaction's equation undergo only the bond-breaking and/or making events depicted to yield the product(s). For this reason, the rate law for an elementary reaction may be derived directly from the balanced chemical equation describing the reaction. This is not the case for typical chemical reactions, for which rate laws may be reliably determined only via experimentation.


1860. Unimolecular Elementary Reactions -
The molecularity of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a unimolecular reaction involves the reaction of a single reactant species to produce one or more molecules of product:

$$
A \longrightarrow \text { products }
$$

The rate law for a unimolecular reaction is first order:

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

A unimolecular reaction may be one of several elementary reactions in a complex mechanism. For example, the reaction:

$$
\mathrm{O}_{3} \longrightarrow \mathrm{O}_{2}+\mathrm{O}
$$

illustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism as described above. However, some unimolecular reactions may be the only step of a single-step reaction mechanism. (In other words, an "overall" reaction may also be an elementary reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, $\mathrm{C}_{4} \mathrm{H}_{8}$, to ethylene, $\mathrm{C}_{2} \mathrm{H}_{4}$, is represented by the following chemical equation:

This equation represents the overall reaction observed, and it might also represent a legitimate unimolecular elementary reaction. The rate law predicted from this equation, assuming it is an elementary reaction, turns out to be the same as the rate law derived experimentally for the overall reaction, namely, one showing firstorder behavior:

$$
\text { rate }=-\frac{\Delta\left[\mathrm{C}_{4} \mathrm{H}_{8}\right]}{\Delta t}=k\left[\mathrm{C}_{4} \mathrm{H}_{8}\right]
$$

This agreement between observed and predicted rate laws is interpreted to mean that the proposed unimolecular, single-step process is a reasonable mechanism for the butadiene reaction.


1861. Bimolecular Elementary Reactions -
A bimolecular reaction involves two reactant species, for example:

$$
\begin{gathered}
A+B \longrightarrow \text { products } \\
\text { and } \\
2 A \longrightarrow \text { products }
\end{gathered}
$$

For the first type, in which the two reactant molecules are different, the rate law is first-order in $A$ and first order in $B$ (second-order overall):

$$
\text { rate }=kA
$$

For the second type, in which two identical molecules collide and react, the rate law is second order in $A$ :

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

Some chemical reactions occur by mechanisms that consist of a single bimolecular elementary reaction. One example is the reaction of nitrogen dioxide with carbon monoxide:

$$
\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)
$$

(see Figure 17.17)

FIGURE 17.17 The probable mechanism for the reaction between $\mathrm{NO}_{2}$ and CO to yield NO and $\mathrm{CO}_{2}$.
Bimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is the second step of the two-step ozone decomposition mechanism discussed earlier in this section:

$$
\mathrm{O}(\mathrm{~g})+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g})
$$


1862. Termolecular Elementary Reactions -
An elementary termolecular reaction involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:

$$
\begin{aligned}
& 2 \mathrm{NO}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} \\
& \text { rate }=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]
\end{aligned}
$$

Likewise, the reaction of nitric oxide with chlorine appears to involve termolecular steps:

$$
\begin{aligned}
& 2 \mathrm{NO}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NOCl} \\
& \text { rate }=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right]
\end{aligned}
$$


1863. Relating Reaction Mechanisms to Rate Laws -
It's often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the rate-limiting step (or rate-determining step) of the reaction Figure 17.18.

FIGURE 17.18 A cattle chute is a nonchemical example of a rate-determining step. Cattle can only be moved from one holding pen to another as quickly as one animal can make its way through the chute. (credit: Loren Kerns)

As described earlier, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, the rate law must be determined from experimental data and the reaction mechanism subsequently deduced from the rate law (and sometimes from other data). The reaction of $\mathrm{NO}_{2}$ and CO provides an illustrative example:

$$
\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g)
$$

For temperatures above $225^{\circ} \mathrm{C}$, the rate law has been found to be:

$$
\text { rate }=k\left\mathrm{NO}_{2}\right
$$

The reaction is first order with respect to $\mathrm{NO}_{2}$ and first-order with respect to CO . This is consistent with a single-step bimolecular mechanism and it is possible that this is the mechanism for this reaction at high temperatures.

At temperatures below $225^{\circ} \mathrm{C}$, the reaction is described by a rate law that is second order with respect to $\mathrm{NO}_{2}$ :

$$
\text { rate }=k\left[\mathrm{NO}_{2}\right]^{2}
$$

This rate law is not consistent with the single-step mechanism, but is consistent with the following two-step mechanism:


1863. Relating Reaction Mechanisms to Rate Laws -
$$
\text { rate }=k\left[\mathrm{NO}_{2}\right]^{2}
$$

This rate law is not consistent with the single-step mechanism, but is consistent with the following two-step mechanism:

$$
\begin{aligned}
& \mathrm{NO}_{2}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}_{3}(g)+\mathrm{NO}(g) \text { (slow) } \\
& \mathrm{NO}_{3}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}_{2}(g) \text { (fast) }
\end{aligned}
$$

The rate-determining (slower) step gives a rate law showing second-order dependence on the $\mathrm{NO}_{2}$ concentration, and the sum of the two equations gives the net overall reaction.

In general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving a rapidly reversible reaction the rate law for the overall reaction may be more difficult to derive.

As discussed in several chapters of this text, a reversible reaction is at equilibrium when the rates of the forward and reverse processes are equal. Consider the reversible elementary reaction in which NO dimerizes to yield an intermediate species $\mathrm{N}_{2} \mathrm{O}_{2}$. When this reaction is at equilibrium:

$$
\begin{aligned}
& \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{~N}_{2} \mathrm{O}_{2} \\
& \mathrm{rate}_{\text {forward }}=\text { rate }_{\text {reverse }} \\
& k_{1}[\mathrm{NO}]^{2}=k_{-1}\left[\mathrm{~N}_{2} \mathrm{O}_{2}\right]
\end{aligned}
$$

This expression may be rearranged to express the concentration of the intermediate in terms of the reactant NO:


1863. Relating Reaction Mechanisms to Rate Laws -
This expression may be rearranged to express the concentration of the intermediate in terms of the reactant NO:

$$
\left(\frac{\mathrm{k}_{1}[\mathrm{NO}]^{2}}{\mathrm{k}_{-1}}\right)=\left[\mathrm{N}_{2} \mathrm{O}_{2}\right]
$$

Since intermediate species concentrations are not used in formulating rate laws for overall reactions, this approach is sometimes necessary, as illustrated in the following example exercise.


1865. Deriving a Rate Law from a Reaction Mechanism -
The two-step mechanism below has been proposed for a reaction between nitrogen monoxide and molecular chlorine:

$$
\begin{array}{ll}
\text { Step 1: } \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{NOCl}_{2}(g) & \text { fast } \\
\text { Step 2: } \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) \longrightarrow 2 \mathrm{NOCl}(g) & \text { slow }
\end{array}
$$

Use this mechanism to derive the equation and predicted rate law for the overall reaction.


1866. Solution -
The equation for the overall reaction is obtained by adding the two elementary reactions:

$$
2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)
$$

To derive a rate law from this mechanism, first write rates laws for each of the two steps.

$$
\begin{aligned}
& \operatorname{rate}_{1}=k_{1}[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right] \text { for the forward reaction of step } 1 \\
& \text { rate }_{-1}=k_{-1}\left[\mathrm{NOCl}_{2}\right] \text { for the reverse reaction of step } 1 \\
& \operatorname{rate}_{2}=k_{2}\left\mathrm{NOCl}_{2}\right \text { for step } 2
\end{aligned}
$$

Step 2 is the rate-determining step, and so the rate law for the overall reaction should be the same as for this step. However, the step 2 rate law, as written, contains an intermediate species concentration, [ $\mathrm{NOCl}_{2}$ ]. To remedy this, use the first step's rate laws to derive an expression for the intermediate concentration in terms of the reactant concentrations.

Assuming step 1 is at equilibrium:

$$
\begin{aligned}
\text { rate }_{1} & =\text { rate }_{-1} \\
k_{1}[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right] & =k_{-1}\left[\mathrm{NOCl}_{2}\right] \\
{\left[\mathrm{NOCl}_{2}\right] } & =\left(\frac{k_{1}}{k_{-1}}\right)[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]
\end{aligned}
$$

Substituting this expression into the rate law for step 2 yields:

$$
\operatorname{rate}_{2}=\operatorname{rate}_{\text {overall }}=\left(\frac{k_{2} k_{1}}{k_{-1}}\right)[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right]
$$


1867. Check Your Learning -
The first step of a proposed multistep mechanism is:

$$
\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{~F}(g) \text { fast }
$$

Derive the equation relating atomic fluorine concentration to molecular fluorine concentration.


1868. Answer: -
$[\mathrm{F}]=\left(\frac{k_{1}\left[\mathrm{~F}_{2}\right]}{k_{-1}}\right)^{1 / 2}$


1868. Answer: - 1868.1. Catalysis
LEARNING OBJECTIVES
By the end of this section, you will be able to:

- Explain the function of a catalyst in terms of reaction mechanisms and potential energy diagrams
- List examples of catalysis in natural and industrial processes


1869. Catalysts Do Not Affect Equilibrium -
A catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations.

The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation

$$
\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)
$$

A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year.

Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry.


1871. Fritz Haber -
In the early 20th century, German chemist Fritz Haber (Figure 17.19) developed a practical process for converting diatomic nitrogen, which cannot be used by plants as a nutrient, to ammonia, a form of nitrogen that is easiest for plants to absorb.

$$
\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)
$$

The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for $78 \%$ of air, diatomic nitrogen $\left(\mathrm{N}_{2}\right)$ is nutritionally unavailable due the tremendous stability of the nitrogennitrogen triple bond. For plants to use atmospheric nitrogen, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation).

Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements. The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate. Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008.

FIGURE 17.19 The work of Nobel Prize recipient Fritz Haber revolutionized agricultural practices in the early 20th century. His work also affected wartime strategies, adding chemical weapons to the artillery.


1871. Fritz Haber -
FIGURE 17.19 The work of Nobel Prize recipient Fritz Haber revolutionized agricultural practices in the early 20th century. His work also affected wartime strategies, adding chemical weapons to the artillery.

In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, "During peace time a scientist belongs to the World, but during war time he belongs to his country." ${ }^{1}$ Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science.

Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995.

It has long been known that nitrogen and hydrogen react to form ammonia. However, it became possible to manufacture ammonia in useful quantities by the reaction of nitrogen and hydrogen only in the early 20th century after the factors that influence its equilibrium were understood.

To be practical, an industrial process must give a large yield of product relatively quickly. One way to increase the yield of ammonia is to increase the pressure on the system in which $\mathrm{N}_{2}, \mathrm{H}_{2}$, and $\mathrm{NH}_{3}$ are at equilibrium or are coming to equilibrium.

$$
\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)
$$

The formation of additional amounts of ammonia reduces the total pressure exerted by the system and somewhat reduces the stress of the increased pressure.


1871. Fritz Haber -
The formation of additional amounts of ammonia reduces the total pressure exerted by the system and somewhat reduces the stress of the increased pressure.

Although increasing the pressure of a mixture of $\mathrm{N}_{2}, \mathrm{H}_{2}$, and $\mathrm{NH}_{3}$ will increase the yield of ammonia, at low temperatures, the rate of formation of ammonia is slow. At room temperature, for example, the reaction is so slow that if we prepared a mixture of $\mathrm{N}_{2}$ and $\mathrm{H}_{2}$, no detectable amount of ammonia would form during our lifetime. The formation of ammonia from hydrogen and nitrogen is an exothermic process:

$$
\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \quad \Delta H=-92.2 \mathrm{~kJ}
$$

Thus, increasing the temperature to increase the rate lowers the yield. If we lower the temperature to shift the equilibrium to favor the formation of more ammonia, equilibrium is reached more slowly because of the large decrease of reaction rate with decreasing temperature.

Part of the rate of formation lost by operating at lower temperatures can be recovered by using a catalyst. The net effect of the catalyst on the reaction is to cause equilibrium to be reached more rapidly.

[^10]In the commercial production of ammonia, conditions of about $500^{\circ} \mathrm{C}, 150-900 \mathrm{~atm}$, and the presence of a catalyst are used to give the best compromise among rate, yield, and the cost of the equipment necessary to produce and contain high-pressure gases at high temperatures (Figure 17.20).

FIGURE 17.20 Commercial production of ammonia requires heavy equipment to handle the high temperatures and pressures required. This schematic outlines the design of an ammonia plant.

Among the factors affecting chemical reaction rates discussed earlier in this chapter was the presence of a catalyst, a substance that can increase the reaction rate without being consumed in the reaction. The concepts introduced in the previous section on reaction mechanisms provide the basis for understanding how catalysts are able to accomplish this very important function.


1871. Fritz Haber -
Figure 17.21 shows reaction diagrams for a chemical process in the absence and presence of a catalyst. Inspection of the diagrams reveals several traits of these reactions. Consistent with the fact that the two diagrams represent the same overall reaction, both curves begin and end at the same energies (in this case, because products are more energetic than reactants, the reaction is endothermic). The reaction mechanisms, however, are clearly different. The uncatalyzed reaction proceeds via a one-step mechanism (one transition state observed), whereas the catalyzed reaction follows a two-step mechanism (two transition states observed) with a notably lesser activation energy. This difference illustrates the means by which a catalyst functions to accelerate reactions, namely, by providing an alternative reaction mechanism with a lower activation energy. Although the catalyzed reaction mechanism for a reaction needn't necessarily involve a different number of steps than the uncatalyzed mechanism, it must provide a reaction path whose rate determining step is faster (lower $E_{\mathrm{a}}$ ).

FIGURE 17.21 Reaction diagrams for an endothermic process in the absence (red curve) and presence (blue curve) of a catalyst. The catalyzed pathway involves a two-step mechanism (note the presence of two transition states) and an intermediate species (represented by the valley between the two transitions states).


1873. Reaction Diagrams for Catalyzed Reactions -
The two reaction diagrams here represent the same reaction: one without a catalyst and one with a catalyst. Estimate the activation energy for each process, and identify which one involves a catalyst.

(a)

(b)


1874. Solution -
Activation energies are calculated by subtracting the reactant energy from the transition state energy.

$$
\begin{aligned}
& \operatorname{diagram}(\mathrm{a}): E_{\mathrm{a}}=32 \mathrm{~kJ}-6 \mathrm{~kJ}=26 \mathrm{~kJ} \\
& \operatorname{diagram}(\mathrm{~b}): E_{\mathrm{a}}=20 \mathrm{~kJ}-6 \mathrm{~kJ}=14 \mathrm{~kJ}
\end{aligned}
$$

The catalyzed reaction is the one with lesser activation energy, in this case represented by diagram (b).


1875. Check Your Learning -
Reaction diagrams for a chemical process with and without a catalyst are shown below. Both reactions involve a two-step mechanism with a rate-determining first step. Compute activation energies for the first step of each mechanism, and identify which corresponds to the catalyzed reaction. How do the second steps of these two mechanisms compare?


1876. Answer: -
For the first step, $E_{\mathrm{a}}=80 \mathrm{~kJ}$ for (a) and 70 kJ for (b), so diagram (b) depicts the catalyzed reaction. Activation energies for the second steps of both mechanisms are the same, 20 kJ .


1877. Homogeneous Catalysts -
A homogeneous catalyst is present in the same phase as the reactants. It interacts with a reactant to form an intermediate substance, which then decomposes or reacts with another reactant in one or more steps to regenerate the original catalyst and form product.

As an important illustration of homogeneous catalysis, consider the earth's ozone layer. Ozone in the upper atmosphere, which protects the earth from ultraviolet radiation, is formed when oxygen molecules absorb ultraviolet light and undergo the reaction:

$$
3 \mathrm{O}_{2}(\mathrm{~g}) \xrightarrow{h v} 2 \mathrm{O}_{3}(\mathrm{~g})
$$

Ozone is a relatively unstable molecule that decomposes to yield diatomic oxygen by the reverse of this equation. This decomposition reaction is consistent with the following two-step mechanism:

$$
\begin{aligned}
& \mathrm{O}_{3} \longrightarrow \mathrm{O}_{2}+\mathrm{O} \\
& \mathrm{O}+\mathrm{O}_{3} \longrightarrow 2 \mathrm{O}_{2}
\end{aligned}
$$

A number of substances can catalyze the decomposition of ozone. For example, the nitric oxide -catalyzed decomposition of ozone is believed to occur via the following three-step mechanism:

$$
\begin{aligned}
& \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \\
& \mathrm{O}_{3}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g) \\
& \mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g)
\end{aligned}
$$

As required, the overall reaction is the same for both the two-step uncatalyzed mechanism and the three-step NO-catalyzed mechanism:

$$
2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)
$$

Notice that NO is a reactant in the first step of the mechanism and a product in the last step. This is another characteristic trait of a catalyst: Though it participates in the chemical reaction, it is not consumed by the reaction.


1879. Mario J. Molina -
The 1995 Nobel Prize in Chemistry was shared by Paul J. Crutzen, Mario J. Molina (Figure 17.22), and F. Sherwood Rowland "for their work in atmospheric chemistry, particularly concerning the formation and decomposition of ozone." ${ }^{2}$ Molina, a Mexican citizen, carried out the majority of his work at the Massachusetts Institute of Technology (MIT).

FIGURE 17.22 (a) Mexican chemist Mario Molina (1943 -) shared the Nobel Prize in Chemistry in 1995 for his research on (b) the Antarctic ozone hole. (credit a: courtesy of Mario Molina; credit b: modification of work by NASA)

In 1974, Molina and Rowland published a paper in the journal Nature detailing the threat of chlorofluorocarbon gases to the stability of the ozone layer in earth's upper atmosphere. The ozone layer protects earth from solar radiation by absorbing ultraviolet light. As chemical reactions deplete the amount of ozone in the upper atmosphere, a measurable "hole" forms above Antarctica, and an increase in the amount of solar ultraviolet radiation - strongly linked to the prevalence of skin cancers-reaches earth's surface. The work of Molina and Rowland was instrumental in the adoption of the Montreal Protocol, an international treaty signed in 1987 that successfully began phasing out production of chemicals linked to ozone destruction.

Molina and Rowland demonstrated that chlorine atoms from human-made chemicals can catalyze ozone destruction in a process similar to that by which NO accelerates the depletion of ozone. Chlorine atoms are generated when chlorocarbons or chlorofluorocarbons-once widely used as refrigerants and propellants-are photochemically decomposed by ultraviolet light or react with hydroxyl radicals. A sample mechanism is shown here using methyl chloride:

$$
\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{OH} \longrightarrow \mathrm{Cl}+\text { other products }
$$

Chlorine radicals break down ozone and are regenerated by the following catalytic cycle:


1879. Mario J. Molina -
Chlorine radicals break down ozone and are regenerated by the following catalytic cycle:

$$
\begin{aligned}
& \mathrm{Cl}+\mathrm{O}_{3} \longrightarrow \mathrm{ClO}+\mathrm{O}_{2} \\
& \mathrm{ClO}+\mathrm{O} \longrightarrow \mathrm{Cl}+\mathrm{O}_{2} \\
& \text { overall Reaction: } \mathrm{O}_{3}+\mathrm{O} \longrightarrow 2 \mathrm{O}_{2}
\end{aligned}
$$

A single monatomic chlorine can break down thousands of ozone molecules. Luckily, the majority of atmospheric chlorine exists as the catalytically inactive forms $\mathrm{Cl}_{2}$ and $\mathrm{ClONO}_{2}$.

[^11]Since receiving his portion of the Nobel Prize, Molina has continued his work in atmospheric chemistry at MIT.


1881. Glucose-6-Phosphate Dehydrogenase Deficiency -
Enzymes in the human body act as catalysts for important chemical reactions in cellular metabolism. As such, a deficiency of a particular enzyme can translate to a life-threatening disease. G6PD (glucose-6-phosphate dehydrogenase) deficiency, a genetic condition that results in a shortage of the enzyme glucose-6-phosphate dehydrogenase, is the most common enzyme deficiency in humans. This enzyme, shown in Figure 17.23, is the rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells (Figure 17.24).

FIGURE 17.23 Glucose-6-phosphate dehydrogenase is a rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells.

A disruption in this pathway can lead to reduced glutathione in red blood cells; once all glutathione is consumed, enzymes and other proteins such as hemoglobin are susceptible to damage. For example, hemoglobin can be metabolized to bilirubin, which leads to jaundice, a condition that can become severe. People who suffer from G6PD deficiency must avoid certain foods and medicines containing chemicals that can trigger damage their glutathione-deficient red blood cells.

FIGURE 17.24 In the mechanism for the pentose phosphate pathway, G6PD catalyzes the reaction that regulates NADPH, a co-enzyme that regulates glutathione, an antioxidant that protects red blood cells and other cells from oxidative damage.


1882. Heterogeneous Catalysts -
A heterogeneous catalyst is a catalyst that is present in a different phase (usually a solid) than the reactants. Such catalysts generally function by furnishing an active surface upon which a reaction can occur. Gas and liquid phase reactions catalyzed by heterogeneous catalysts occur on the surface of the catalyst rather than within the gas or liquid phase.

Heterogeneous catalysis typically involves the following processes:

1. Adsorption of the reactant(s) onto the surface of the catalyst
2. Activation of the adsorbed reactant(s)
3. Reaction of the adsorbed reactant(s)
4. Desorption of product(s) from the surface of the catalyst

Figure 17.25 illustrates the steps of a mechanism for the reaction of compounds containing a carbon-carbon double bond with hydrogen on a nickel catalyst. Nickel is the catalyst used in the hydrogenation of polyunsaturated fats and oils (which contain several carbon-carbon double bonds) to produce saturated fats and oils (which contain only carbon-carbon single bonds).

FIGURE 17.25 Mechanism for the Ni-catalyzed reaction $\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}$. (a) Hydrogen is adsorbed on the surface, breaking the $\mathrm{H}-\mathrm{H}$ bonds and forming $\mathrm{Ni}-\mathrm{H}$ bonds. (b) Ethylene is adsorbed on the surface, breaking the $\mathrm{C}-\mathrm{C}$ $\pi$-bond and forming $\mathrm{Ni}-\mathrm{C}$ bonds. (c) Atoms diffuse across the surface and form new $\mathrm{C}-\mathrm{H}$ bonds when they collide.
(d) $\mathrm{C}_{2} \mathrm{H}_{6}$ molecules desorb from the Ni surface.

Many important chemical products are prepared via industrial processes that use heterogeneous catalysts, including ammonia, nitric acid, sulfuric acid, and methanol. Heterogeneous catalysts are also used in the catalytic converters found on most gasoline-powered automobiles (Figure 17.26).


1884. Automobile Catalytic Converters -
Scientists developed catalytic converters to reduce the amount of toxic emissions produced by burning gasoline in internal combustion engines. By utilizing a carefully selected blend of catalytically active metals, it is possible to effect complete combustion of all carbon-containing compounds to carbon dioxide while also reducing the output of nitrogen oxides. This is particularly impressive when we consider that one step involves adding more oxygen to the molecule and the other involves removing the oxygen (Figure 17.26).

FIGURE 17.26 A catalytic converter allows for the combustion of all carbon-containing compounds to carbon dioxide, while at the same time reducing the output of nitrogen oxide and other pollutants in emissions from gasoline-burning engines.

Most modern, three-way catalytic converters possess a surface impregnated with a platinum-rhodium catalyst, which catalyzes the conversion of nitric oxide into dinitrogen and oxygen as well as the conversion of carbon monoxide and hydrocarbons such as octane into carbon dioxide and water vapor:

$$
\begin{aligned}
& 2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \\
& 2 \mathrm{CO}(g)+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{~g}) \\
& 2 \mathrm{C}_{8} \mathrm{H}_{18}(g)+25 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 16 \mathrm{CO}_{2}(\mathrm{~g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g})
\end{aligned}
$$

In order to be as efficient as possible, most catalytic converters are preheated by an electric heater. This ensures that the metals in the catalyst are fully active even before the automobile exhaust is hot enough to maintain appropriate reaction temperatures.


1885. LINK TO LEARNING -
The University of California at Davis' "ChemWiki" provides a thorough explanation (http://openstax.org/l/ 16 catconvert) of how catalytic converters work.


1887. Enzyme Structure and Function -
The study of enzymes is an important interconnection between biology and chemistry. Enzymes are usually proteins (polypeptides) that help to control the rate of chemical reactions between biologically important compounds, particularly those that are involved in cellular metabolism. Different classes of enzymes perform a variety of functions, as shown in Table 17.3.

Classes of Enzymes and Their Functions

| Class | Function |
| :--- | :--- |
| oxidoreductases | redox reactions |
| transferases | transfer of functional groups |
| hydrolases | hydrolysis reactions |
| lyases | group elimination to form double bonds |
| isomerases | isomerization |
| ligases | bond formation with ATP hydrolysis |

TABLE 17.3

Enzyme molecules possess an active site, a part of the molecule with a shape that allows it to bond to a specific substrate (a reactant molecule), forming an enzyme-substrate complex as a reaction intermediate. There are two models that attempt to explain how this active site works. The most simplistic model is referred to as the lock-and-key hypothesis, which suggests that the molecular shapes of the active site and substrate are complementary, fitting together like a key in a lock. The induced fit hypothesis, on the other hand, suggests that the enzyme molecule is flexible and changes shape to accommodate a bond with the substrate. This is not to suggest that an enzyme's active site is completely malleable, however. Both the lock-and-key model and the induced fit model account for the fact that enzymes can only bind with specific substrates, since in general a particular enzyme only catalyzes a particular reaction (Figure 17.27).

FIGURE 17.27 (a) According to the lock-and-key model, the shape of an enzyme's active site is a perfect fit for the substrate. (b) According to the induced fit model, the active site is somewhat flexible, and can change shape in order to bond with the substrate.


1888. LINK TO LEARNING -
The Royal Society of Chemistry (http://openstax.org/l/16enzymes) provides an excellent introduction to enzymes for students and teachers.

The connection between the rate of a reaction and its equilibrium constant is one we can easily determine with just a bit of algebraic substitution. For a reaction where substance A forms B (and the reverse)

$$
A \rightleftharpoons B
$$

The rate of the forward reaction is

$$
\operatorname{Rate}(f)=k(f)[A]
$$

And the rate of the reverse reaction is

$$
\operatorname{Rate}(r)=k(r)[B]
$$

Once equilibrium is established, the rates of the forward and reverse reactions are equal:

$$
\operatorname{Rate}(f)=\operatorname{Rate}(r)=k(f)[A]=k(r)[B]
$$

Rearranging a bit, we get

$$
\operatorname{Rate}_{f}=\operatorname{Rate}_{r} \text { so } k(f)[A]=k(r)[B]
$$

Also recall that the equilibrium constant is simply the ratio of product to reactant concentration at equilibrium:

$$
\begin{gathered}
\frac{k(f)}{k(r)}=\frac{[B]}{[A]} \\
K=\frac{[B]}{[A]}
\end{gathered}
$$

So the equilibrium constant turns out to be the ratio of the forward to the reverse rate constants. This relationship also helps cement our understanding of the nature of a catalyst. That is, a catalyst does not change the fundamental equilibrium (or the underlying thermodynamics) of a reaction. Rather, what it does is alter the rate constant for the reaction - that is, both rate constants, forward and reverse, equally. In doing so, catalysts usually speed up the rate at which reactions attain equilibrium (though they can be used to slow down the rate of reaction as well!).


1889. Key Terms -
activated complex (also, transition state) unstable combination of reactant species formed during a chemical reaction
activation energy $\left(\boldsymbol{E}_{\mathrm{a}}\right)$ minimum energy necessary in order for a reaction to take place
Arrhenius equation mathematical relationship between a reaction's rate constant, activation energy, and temperature
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
bimolecular reaction elementary reaction involving two reactant species
catalyst substance that increases the rate of a reaction without itself being consumed by the reaction
collision theory model that emphasizes the energy and orientation of molecular collisions to explain and predict reaction kinetics
elementary reaction reaction that takes place in a single step, precisely as depicted in its chemical equation
frequency factor (A) proportionality constant in the Arrhenius equation, related to the relative number of collisions having an orientation capable of leading to product formation
half-life of a reaction ( $\boldsymbol{t}_{\mathbf{l} / \mathbf{2}}$ ) time required for half of a given amount of reactant to be consumed
heterogeneous catalyst catalyst present in a different phase from the reactants, furnishing a surface at which a reaction can occur
homogeneous catalyst catalyst present in the same phase as the reactants
initial rate instantaneous rate of a chemical reaction at $t=0 \mathrm{~s}$ (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
integrated rate law equation that relates the
concentration of a reactant to elapsed time of reaction
intermediate species produced in one step of a reaction mechanism and consumed in a subsequent step
method of initial rates common experimental approach to determining rate laws that involves measuring reaction rates at varying initial reactant concentrations
molecularity number of reactant species involved in an elementary reaction
overall reaction order sum of the reaction orders for each substance represented in the rate law
rate constant (k) proportionality constant in a rate law
rate expression mathematical representation defining reaction rate as change in amount, concentration, or pressure of reactant or product species per unit time


1889. Key Terms -
rate constant (k) proportionality constant in a rate law
rate expression mathematical representation defining reaction rate as change in amount, concentration, or pressure of reactant or product species per unit time
rate law (also, rate equation) (also, differential rate laws) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants
rate of reaction measure of the speed at which a chemical reaction takes place
rate-determining step (also, rate-limiting step) slowest elementary reaction in a reaction mechanism; determines the rate of the overall reaction
reaction diagram used in chemical kinetics to illustrate various properties of a reaction
reaction mechanism stepwise sequence of elementary reactions by which a chemical change takes place
reaction order value of an exponent in a rate law (for example, zero order for 0 , first order for 1 , second order for 2 , and so on)
termolecular reaction elementary reaction involving three reactant species
unimolecular reaction elementary reaction involving a single reactant species


1890. Key Equations -
relative reaction rates for $a \mathrm{~A} \longrightarrow b \mathrm{~B}=-\frac{1}{a} \frac{\Delta[\mathrm{~A}]}{\Delta t}=\frac{1}{b} \frac{\Delta[\mathrm{~B}]}{\Delta t}$
integrated rate law for zero-order reactions: $[A]_{t}=-k t+[A]_{0}$,
half-life for a zero-order reaction $t_{1 / 2}=\frac{[A]_{0}}{2 k}$
integrated rate law for first-order reactions: $\ln [A]_{t}=-k t+\ln [A]_{0}$,
half-life for a first-order reaction $t_{1 / 2}=\frac{0.693}{k}$
integrated rate law for second-order reactions: $\frac{1}{[A]_{t}}=k t+\frac{1}{[A]_{0}}$,
half-life for a second-order reaction $t_{1 / 2}=\frac{1}{[A]_{0} k}$
$k=A e^{-E_{\mathrm{a}} / R T}$
$\ln k=\left(\frac{-E_{\mathrm{a}}}{R}\right)\left(\frac{1}{T}\right)+\ln A$
$\ln \frac{k_{1}}{k_{2}}=\frac{E_{\mathrm{a}}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$


1891. Summary - 1891.1. Chemical Reaction Rates
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.


1891. Summary - 1891.2. Factors Affecting Reaction Rates
The rate of a chemical reaction is affected by several parameters. Reactions involving two phases proceed more rapidly when there is greater surface area contact. If temperature or reactant concentration is increased, the rate of a given reaction generally increases as well. A catalyst can increase the rate of a reaction by providing an alternative pathway with a lower activation energy.


1891. Summary - 1891.3. Rate Laws
Rate laws (differential rate laws) provide a mathematical description of how changes in the concentration of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the concentration of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.


1891. Summary - 1891.4. Integrated Rate Laws
Integrated rate laws are mathematically derived from differential rate laws, and they describe the time dependence of reactant and product concentrations.

The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. A reaction's half-life varies with rate constant and,
for some reaction orders, reactant concentration. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases.


1891. Summary - 1891.5. Collision Theory
Chemical reactions typically require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction's rate constant, activation energy, temperature, and dependence on collision orientation.


1891. Summary - 1891.6. Reaction Mechanisms
The sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The molecularity of an elementary reaction is the number of reactant species involved, typically one (unimolecular), two (bimolecular), or, less commonly, three (termolecular). The overall rate of a reaction is determined by the rate of the slowest in its mechanism, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible.


1891. Summary - 1891.7. Catalysis
Catalysts affect the rate of a chemical reaction by altering its mechanism to provide a lower activation energy, but they do not affect equilibrium. Catalysts
can be homogenous (in the same phase as the reactants) or heterogeneous (a different phase than
the reactants).


1892. Exercises - 1892.1. Chemical Reaction Rates
1. What is the difference between average rate, initial rate, and instantaneous rate?
2. Ozone decomposes to oxygen according to the equation $2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)$. Write the equation that relates the rate expressions for this reaction in terms of the disappearance of $\mathrm{O}_{3}$ 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 reaction $\mathrm{Cl}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{ClF}_{3}(g)$. Write the equation that relates the rate expressions for this reaction in terms of the disappearance of $\mathrm{Cl}_{2}$ and $\mathrm{F}_{2}$ and the formation of $\mathrm{ClF}_{3}$.
4. A study of the rate of dimerization of $\mathrm{C}_{4} \mathrm{H}_{6}$ gave the data shown in the table:
$2 \mathrm{C}_{4} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{8} \mathrm{H}_{12}$

| Time (s) | 0 | 1600 | 3200 | 4800 | 6200 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\left\mathrm{C}_{4} \mathrm{H}_{6}\right$ | $1.00 \times 10^{-2}$ | $5.04 \times 10^{-3}$ | $3.37 \times 10^{-3}$ | $2.53 \times 10^{-3}$ | $2.08 \times 10^{-3}$ |


1892. Exercises - 1892.1. Chemical Reaction Rates
(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 $\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]$. What are the units of this rate?
(c) Determine the average rate of formation of $\mathrm{C}_{8} \mathrm{H}_{12}$ 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 $2 A \longrightarrow B$ gave the following data:

| Time (s) | 0.0 | 5.0 | 10.0 | 15.0 | 20.0 | 25.0 | 35.0 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $A$ | 1.00 | 0.775 | 0.625 | 0.465 | 0.360 | 0.285 | 0.230 |


1892. Exercises - 1892.1. Chemical Reaction Rates
(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 .
6. Consider the following reaction in aqueous solution:
$5 \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}{ }^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{Br}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)$
If the rate of disappearance of $\mathrm{Br}^{-}(\mathrm{aq})$ at a particular moment during the reaction is $3.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$, what is the rate of appearance of $\mathrm{Br}_{2}(a q)$ at that moment?


1892. Exercises - 1892.2. Factors Affecting Reaction Rates
7. Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.
8. Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)
9. Go to the PhET Reactions \& Rates (http://openstax.org/l/16PHETreaction) interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen ( O ) and carbon monoxide ( CO ) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on "Reload Launcher" and change to "Angled shot" to see the difference.
(a) What happens when the angle of the collision is changed?
(b) Explain how this is relevant to rate of reaction.
10. In the PhET Reactions \& Rates (http://openstax.org/l/16PHETreaction) interactive, use the "Many Collisions" tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select "Show bonds" under Options. How is the rate of the reaction affected by concentration and temperature?
11. In the PhET Reactions \& Rates (http://openstax.org/l/16PHETreaction) interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select "Show Bonds" under Options.
(a) Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?
(b) Click "Pause" and then "Reset All," and then enter 15 molecules of A and 10 molecules of BC once again. Select "Show Bonds" under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction.


1892. Exercises - 1892.3. Rate Laws
12. How do the rate of a reaction and its rate constant differ?
13. Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:
(a) What is the order of the reaction with respect to that reactant?
(b) Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?
14. Tripling the concentration of a reactant increases the rate of a reaction nine-fold. With this knowledge, answer the following questions:
(a) What is the order of the reaction with respect to that reactant?
(b) Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four-fold.


1892. Exercises - 1892.3. Rate Laws
What is the order of the reaction with respect to that reactant?
15. How will the rate of reaction change for the process: $\mathrm{CO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g)$ if the rate law for the reaction is rate $=k\left[\mathrm{NO}_{2}\right]^{2}$ ?
(a) Decreasing the pressure of $\mathrm{NO}_{2}$ from 0.50 atm to 0.250 atm .
(b) Increasing the concentration of CO from 0.01 M to 0.03 M .
16. How will each of the following affect the rate of the reaction: $\mathrm{CO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g)$ if the rate law for the reaction is rate $=k\left\mathrm{NO}_{2}\right$ ?
(a) Increasing the pressure of $\mathrm{NO}_{2}$ from 0.1 atm to 0.3 atm
(b) Increasing the concentration of CO from 0.02 M to 0.06 M .
17. Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction $\mathrm{NO}+\mathrm{O}_{3} \longrightarrow \mathrm{NO}_{2}+\mathrm{O}_{2}$ is first order with respect to both NO and $\mathrm{O}_{3}$ with a rate constant of $2.20 \times 10^{7}$ $\mathrm{L} / \mathrm{mol} / \mathrm{s}$. What is the instantaneous rate of disappearance of NO when $[\mathrm{NO}]=3.3 \times 10^{-6} \mathrm{M}$ and $\left[\mathrm{O}_{3}\right]=5.9 \times$ $10^{-7} \mathrm{M}$ ?


1892. Exercises - 1892.3. Rate Laws
18. Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:
${ }_{15}^{32} \mathrm{P} \longrightarrow{ }_{16}^{32} \mathrm{~S}+\mathrm{e}^{-}$
rate $=4.85 \times 10^{-2}$ day $^{-1}\left[{ }^{32} \mathrm{P}\right]$
What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M ?
19. The rate constant for the radioactive decay of ${ }^{14} \mathrm{C}$ is $1.21 \times 10^{-4}$ year ${ }^{-1}$. The products of the decay are nitrogen atoms and electrons (beta particles):
${ }_{6}^{14} \mathrm{C} \longrightarrow{ }_{7}^{14} \mathrm{~N}+\mathrm{e}^{-}$
rate $=k\left[{ }_{6}^{14} \mathrm{C}\right]$
What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of $6.5 \times$ $10^{-9} \mathrm{M}$ ?
20. The decomposition of acetaldehyde is a second order reaction with a rate constant of $4.71 \times 10^{-8} \mathrm{~L} \mathrm{~mol}^{-1}$ $\mathrm{s}^{-1}$. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of $5.55 \times 10^{-4} \mathrm{M}$ ?
21. Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by $25-30 \%$. Women metabolize alcohol a little more slowly than men:


1892. Exercises - 1892.3. Rate Laws
| $\left\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right$ | $4.4 \times 10^{-2}$ | $3.3 \times 10^{-2}$ | $2.2 \times 10^{-2}$ |
| :---: | :---: | :---: | :---: |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~h}^{-1}\right)$ | $2.0 \times 10^{-2}$ | $2.0 \times 10^{-2}$ | $2.0 \times 10^{-2}$ |

Determine the rate law, the rate constant, and the overall order for this reaction.
22. Under certain conditions the decomposition of ammonia on a metal surface gives the following data:

| $\left\mathrm{NH}_{3}\right$ | $1.0 \times 10^{-3}$ | $2.0 \times 10^{-3}$ | $3.0 \times 10^{-3}$ |
| :---: | :---: | :---: | :---: |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~h}^{-1}\right)$ | $1.5 \times 10^{-6}$ | $1.5 \times 10^{-6}$ | $1.5 \times 10^{-6}$ |

Determine the rate law, the rate constant, and the overall order for this reaction.
23. Nitrosyl chloride, NOCl , decomposes to NO and $\mathrm{Cl}_{2}$.
$2 \mathrm{NOCl}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$
Determine the rate law, the rate constant, and the overall order for this reaction from the following data:


1892. Exercises - 1892.3. Rate Laws
| $\mathrm{NOCl}$ | 0.10 | 0.20 | 0.30 |
| :---: | :---: | :---: | :---: |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~h}^{-1}\right)$ | $8.0 \times 10^{-10}$ | $3.2 \times 10^{-9}$ | $7.2 \times 10^{-9}$ |

24. From the following data, determine the rate law, the rate constant, and the order with respect to $A$ for the reaction $A \longrightarrow 2 C$.

| $A$ | $1.33 \times 10^{-2}$ | $2.66 \times 10^{-2}$ | $3.99 \times 10^{-2}$ |
| :---: | :---: | :---: | :---: |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~h}^{-1}\right)$ | $3.80 \times 10^{-7}$ | $1.52 \times 10^{-6}$ | $3.42 \times 10^{-6}$ |

25. Nitrogen monoxide reacts with chlorine according to the equation:
$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)$
The following initial rates of reaction have been observed for certain reactant concentrations:

| [NO] (mol/L) | $\left\mathrm{Cl}_{2}\right$ |  |
| :---: | :---: | :---: |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~h}^{-1}\right)$ |  |  |
| 0.50 | 0.50 | 1.14 |
| 1.00 | 0.50 | 4.56 |
| 1.00 | 1.00 | 9.12 |


1892. Exercises - 1892.3. Rate Laws
What is the rate law that describes the rate's dependence on the concentrations of NO and $\mathrm{Cl}_{2}$ ? What is the rate constant? What are the orders with respect to each reactant?
26. Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: $\mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)$
Determine the rate law, the rate constant, and the orders with respect to each reactant from the following data:

| $\mathrm{NO}$ | 0.30 | 0.60 | 0.60 |
| :---: | :---: | :---: | :---: |
| $\left\mathrm{H}_{2}\right$ | 0.35 | 0.35 | 0.70 |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)$ | $2.835 \times 10^{-3}$ | $1.134 \times 10^{-2}$ | $2.268 \times 10^{-2}$ |

27. For the reaction $A \longrightarrow B+C$, the following data were obtained at $30^{\circ} \mathrm{C}$ :

| $A$ | 0.230 | 0.356 | 0.557 |
| :---: | :---: | :---: | :---: |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)$ | $4.17 \times 10^{-4}$ | $9.99 \times 10^{-4}$ | $2.44 \times 10^{-3}$ |

(a) What is the order of the reaction with respect to [ $A$ ], and what is the rate law?
(b) What is the rate constant?
28. For the reaction $Q \longrightarrow W+X$, the following data were obtained at $30^{\circ} \mathrm{C}$ :


1892. Exercises - 1892.3. Rate Laws
| $[Q]_{\text {initial }}(M)$ | 0.170 | 0.212 | 0.357 |
| :---: | :---: | :---: | :---: |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)$ | $6.68 \times 10^{-3}$ | $1.04 \times 10^{-2}$ | $2.94 \times 10^{-2}$ |

(a) What is the order of the reaction with respect to [ $Q$ ], and what is the rate law?
(b) What is the rate constant?
29. The rate constant for the first-order decomposition at $45^{\circ} \mathrm{C}$ of dinitrogen pentoxide, $\mathrm{N}_{2} \mathrm{O}_{5}$, dissolved in chloroform, $\mathrm{CHCl}_{3}$, is $6.2 \times 10^{-4} \mathrm{~min}^{-1}$.
$2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$
What is the rate of the reaction when $\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.40 \mathrm{M}$ ?
30. The annual production of $\mathrm{HNO}_{3}$ in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.
(a) $4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$
(b) $2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$
(c) $3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$


1892. Exercises - 1892.3. Rate Laws
The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in $\mathrm{O}_{2}$, what is the rate of formation of $\mathrm{NO}_{2}$ when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M ? The rate constant for the reaction is $5.8 \times 10^{-6} \mathrm{~L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}$.
31. The following data have been determined for the reaction:
$\mathrm{I}^{-}+\mathrm{OCl}^{-} \longrightarrow \mathrm{IO}^{-}+\mathrm{Cl}^{-}$


1893. 1 -
2
3

| $\left[\mathrm{I}^{-}\right]_{\text {initial }}(M)$ | 0.10 | 0.20 | 0.30 |
| :---: | :---: | :---: | :---: |
| $\left[\mathrm{OCl}^{-}\right]_{\text {initial }}(M)$ | 0.050 | 0.050 | 0.010 |
| Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)$ | $3.05 \times 10^{-4}$ | $6.20 \times 10^{-4}$ | $1.83 \times 10^{-4}$ |

Determine the rate law and the rate constant for this reaction.


1893. 1 - 1893.1. Integrated Rate Laws
32. Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of $A$ at varying times.
33. Use the data provided to graphically determine the order and rate constant of the following reaction:
$\mathrm{SO}_{2} \mathrm{Cl}_{2} \longrightarrow \mathrm{SO}_{2}+\mathrm{Cl}_{2}$

| Time (s) | 0 | $5.00 \times 10^{3}$ | $1.00 \times 10^{4}$ | $1.50 \times 10^{4}$ |
| :---: | :---: | :---: | :---: | :---: |
| $\left\mathrm{SO}_{2} \mathrm{Cl}_{2}\right$ | 0.100 | 0.0896 | 0.0802 | 0.0719 |
| Time (s) | $2.50 \times 10^{4}$ | $3.00 \times 10^{4}$ | $4.00 \times 10^{4}$ |  |
| $\left\mathrm{SO}_{2} \mathrm{Cl}_{2}\right$ | 0.0577 | 0.0517 | 0.0415 |  |

34. Pure ozone decomposes slowly to oxygen, $2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)$. Use the data provided in a graphical method and determine the order and rate constant of the reaction.


1893. 1 - 1893.1. Integrated Rate Laws
| Time (h) | 0 | $2.0 \times 10^{3}$ | $7.6 \times 10^{3}$ | $1.00 \times 10^{4}$ |
| :---: | :---: | :---: | :---: | :---: |
| $\left\mathrm{O}_{3}\right$ | $1.00 \times 10^{-5}$ | $4.98 \times 10^{-6}$ | $2.07 \times 10^{-6}$ | $1.66 \times 10^{-6}$ |
| Time (h) | $1.23 \times 10^{4}$ | $1.43 \times 10^{4}$ | $1.70 \times 10^{4}$ |  |
| $\left\mathrm{O}_{3}\right$ | $1.39 \times 10^{-6}$ | $1.22 \times 10^{-6}$ | $1.05 \times 10^{-6}$ |  |

35. From the given data, use a graphical method to determine the order and rate constant of the following reaction:
$2 X \longrightarrow Y+Z$

| $2 X \longrightarrow Y+$ Time (s) | 5.0 | 10.0 | 15.0 | 20.0 | 25.0 | 30.0 | 35.0 | 40.0 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $X$ | 0.0990 | 0.0497 | 0.0332 | 0.0249 | 0.0200 | 0.0166 | 0.0143 | 0.0125 |


1893. 1 - 1893.1. Integrated Rate Laws
36. What is the half-life for the first-order decay of phosphorus- 32 ? $\left({ }_{15}^{32} \mathrm{P} \longrightarrow{ }_{16}^{32} \mathrm{~S}+\mathrm{e}^{-}\right)$The rate constant for the decay is $4.85 \times 10^{-2}$ day $^{-1}$.
37. What is the half-life for the first-order decay of carbon-14? $\left({ }_{6}^{14} \mathrm{C} \longrightarrow{ }_{7}^{14} \mathrm{~N}+\mathrm{e}^{-}\right)$The rate constant for the decay is $1.21 \times 10^{-4}$ year $^{-1}$.
38. What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M ? The rate constant for this second-order reaction is $8.0 \times 10^{-8} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$.
39. What is the half-life for the decomposition of $\mathrm{O}_{3}$ when the concentration of $\mathrm{O}_{3}$ is $2.35 \times 10^{-6} \mathrm{M}$ ? The rate constant for this second-order reaction is $50.4 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~h}^{-1}$.
40. The reaction of compound $A$ to give compounds $C$ and $D$ was found to be second-order in $A$. The rate constant for the reaction was determined to be $2.42 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$. If the initial concentration is $0.500 \mathrm{~mol} / \mathrm{L}$, what is the value of $t_{1 / 2}$ ?


1893. 1 - 1893.1. Integrated Rate Laws
41. The half-life of a reaction of compound $A$ to give compounds $D$ and $E$ is 8.50 min when the initial concentration of $A$ is 0.150 M . How long will it take for the concentration to drop to 0.0300 M if the reaction is (a) first order with respect to $A$ or (b) second order with respect to $A$ ?
42. Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of $3 \times 10^{4} \mathrm{~g} / \mathrm{mol}$ that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate law that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing $0.15 \mu \mathrm{~g}\left(0.15 \times 10^{-6} \mathrm{~g}\right)$ of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.


1893. 1 - 1893.1. Integrated Rate Laws
| [Penicillin] (M) | Rate $\left(\mathrm{mol} \mathrm{L}^{-1} \mathbf{m i n}^{-1}\right)$ |  |
| :--- | :--- | :--- |
| $2.0 \times 10^{-6}$ | $1.0 \times 10^{-10}$ |  |
| $3.0 \times 10^{-6}$ | $1.5 \times 10^{-10}$ |  |
| $4.0 \times 10^{-6}$ | $2.0 \times 10^{-10}$ |  |

43. Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h , respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days ( 48 h )?
44. There are two molecules with the formula $\mathrm{C}_{3} \mathrm{H}_{6}$. Propene, $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}$, is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:


1893. 1 - 1893.1. Integrated Rate Laws
When heated to $499^{\circ} \mathrm{C}$, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of $5.95 \times 10^{-4} \mathrm{~s}^{-1}$. What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at $499^{\circ} \mathrm{C}$ ?
45. Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min . (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is ${ }_{9}^{18} \mathrm{~F} \longrightarrow{ }_{8}^{18} \mathrm{O}+{ }_{+1}^{0} \mathrm{e}$ ) Physicians use ${ }^{18} \mathrm{~F}$ to study the brain by injecting a quantity of fluorosubstituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.
(a) What is the rate constant for the decomposition of fluorine-18?
(b) If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h ?
(c) How long does it take for $99.99 \%$ of the ${ }^{18} \mathrm{~F}$ to decay?
46. Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for $\frac{1}{64}$ of the initial dose to remain in the athlete's body?
47. Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about $93.79 \%$ of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.


1893. 1 - 1893.1. Integrated Rate Laws
48. Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to $160^{\circ} \mathrm{C}$ and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:


1893. 1 - 1893.1. Integrated Rate Laws
|  |  |  |  |  |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Initial
 $\left[\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right]$
 $(M)$ | 4.88 | 3.52 | 2.29 | 1.81 | 5.33 | 4.05 | 2.95 | 1.72 |
| $t(\mathrm{~s})$ | 300 | 300 | 300 | 300 | 180 | 180 | 180 | 180 |
| $\%$
 Decomposed | 52.0 | 52.9 | 53.2 | 53.9 | 34.6 | 35.9 | 36.0 | 35.4 |

49. For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene $\left(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\right)$ has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:

The isomerization of cyclobutene to butadiene is first-order and the rate constant has been measured as $2.0 \times 10^{-4} \mathrm{~s}^{-1}$ at $150^{\circ} \mathrm{C}$ in a 0.53 -L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at $150^{\circ} \mathrm{C}$ with an initial pressure of 55 torr.


1893. 1 - 1893.2. Collision Theory
50. Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?
51. When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?
52. What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?
53. Account for the relationship between the rate of a reaction and its activation energy.
54. Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.
55. How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate.
56. The rate of a certain reaction doubles for every $10^{\circ} \mathrm{C}$ rise in temperature.
(a) How much faster does the reaction proceed at $45^{\circ} \mathrm{C}$ than at $25^{\circ} \mathrm{C}$ ?
(b) How much faster does the reaction proceed at $95^{\circ} \mathrm{C}$ than at $25^{\circ} \mathrm{C}$ ?
57. In an experiment, a sample of $\mathrm{NaClO}_{3}$ was $90 \%$ decomposed in 48 min . Approximately how long would this decomposition have taken if the sample had been heated $20^{\circ} \mathrm{C}$ higher? (Hint: Assume the rate doubles for each $10^{\circ} \mathrm{C}$ rise in temperature.)
58. The rate constant at $325^{\circ} \mathrm{C}$ for the decomposition reaction $\mathrm{C}_{4} \mathrm{H}_{8} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}$ is $6.1 \times 10^{-8} \mathrm{~s}^{-1}$, and the activation energy is 261 kJ per mole of $\mathrm{C}_{4} \mathrm{H}_{8}$. Determine the frequency factor for the reaction.


1893. 1 - 1893.2. Collision Theory
59. The rate constant for the decomposition of acetaldehyde, $\mathrm{CH}_{3} \mathrm{CHO}$, to methane, $\mathrm{CH}_{4}$, and carbon monoxide, CO , in the gas phase is $1.1 \times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$ at 703 K and $4.95 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$ at 865 K . Determine the activation energy for this decomposition.
60. An elevated level of the enzyme alkaline phosphatase (ALP) in human serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from $30^{\circ} \mathrm{C}$ to $37^{\circ} \mathrm{C}$. What is the activation energy for the ALP-catalyzed conversion of PNPP to PNP and phosphate?
61. In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?
(a) the change in free energy per second
(b) the change in temperature per second
(c) the number of collisions per second
(d) the number of product molecules
62. Hydrogen iodide, HI , decomposes in the gas phase to produce hydrogen, $\mathrm{H}_{2}$, and iodine, $\mathrm{I}_{2}$. The value of the rate constant, $k$, for the reaction was measured at several different temperatures and the data are shown here: