# 5.5: The Change of Concentration with Time (Integrated Rate Laws)

- Last updated

- Save as PDF

- Page ID
- 81883

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)##### Learning Objectives

- To apply rate laws to zeroth, first and second order reactions.

Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data. Often, the exponents in the rate law are the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth, first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data. The learning objective of this Module is to know how to determine the reaction order from experimental data.

## Zeroth-Order Reactions

A **zeroth-order reaction** is one whose rate is independent of concentration; its differential rate law is

\[\text{rate} = k. \nonumber \]

We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:

\[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \label{14.4.1} \]

Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of \(−k\). The value of \(k\) is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of \(k\), a positive value.

The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form

\[[A] = [A]_0 − kt \label{14.4.2} \]

where \([A]_0\) is the initial concentration of reactant \(A\). Equation \ref{14.4.2} has the form of the algebraic equation for a straight line,

\[y = mx + b, \nonumber \]

with \(y = [A]\), \(mx = −kt\), and \(b = [A]_0\).)

##### Units

In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second.

Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N_{2}O on a platinum (Pt) surface to produce N_{2} and O_{2}, which occurs at temperatures ranging from 200°C to 400°C:

\[\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \label{14.4.3} \]

Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N_{2}O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N_{2}O to react with the entire Pt surface, doubling or quadrupling the N_{2}O concentration will have no effect on the reaction rate. At very low concentrations of N_{2}O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N_{2}O concentration. The reaction rate is as follows:

\[\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \label{14.4.4} \]

Thus the rate at which N_{2}O is consumed and the rates at which N_{2} and O_{2} are produced are independent of concentration. As shown in Figure \(\PageIndex{2}\), the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N_{2}O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally.

A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the **enzyme** alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is

where \ce{NAD^{^{+}}\)} (nicotinamide adenine dinucleotide) and \(\ce{NADH}\) (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (Figure \(\PageIndex{3a}\)). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in Figure \(\PageIndex{3}\)).

These examples illustrate two important points:

- In a zeroth-order reaction, the reaction rate does not depend on the reactant concentration.
- A linear change in concentration with time is a clear indication of a zeroth-order reaction.

## First-Order Reactions

In a **first-order reaction**, the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:

\[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \label{14.4.5} \]

If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s^{−1}).

The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:

\[[A] = [A]_0e^{−kt} \label{14.4.6} \]

where \([A]_0\) is the initial concentration of reactant \(A\) at \(t = 0\); \(k\) is the rate constant; and *e* is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation \(\ref{14.4.6}\) predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation \(\ref{14.4.6}\) and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of \(A\) and \(t\):

\[\ln[A] = \ln[A]_0 − kt \label{14.4.7} \]

Because Equation \(\ref{14.4.7}\) has the form of the algebraic equation for a straight line,

\[y = mx + b, \nonumber \]

with \(y = \ln[A]\) and \(b = \ln[A]_0\), a plot of \(\ln[A]\) versus \(t\) for a first-order reaction should give a straight line with a slope of \(−k\) and an intercept of \(\ln[A]_0\). Either the differential rate law (Equation \(\ref{14.4.5}\)) or the integrated rate law (Equation \(\ref{14.4.7}\)) can be used to determine whether a particular reaction is first order.

First-order reactions are very common. One reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin. Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows:

Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in Figure \(\PageIndex{5}\) is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in Figure \(\PageIndex{4}\) have been studied extensively to find ways of maximizing the concentration of the active species.

If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order.

The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Table \(\PageIndex{1}\). The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin.

Experiment | [Cisplatin]_{0} (M) |
Initial Rate (M/min) |
---|---|---|

1 | 0.0060 | 9.0 × 10^{−6} |

2 | 0.012 | 1.8 × 10^{−5} |

3 | 0.024 | 3.6 × 10^{−5} |

4 | 0.030 | 4.5 × 10^{−5} |

Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in Table \(\PageIndex{1}\) shows that the reaction rate doubles [(1.8 × 10^{−5} M/min) ÷ (9.0 × 10^{−6} M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10^{−5} M/min) ÷ (9.0 × 10^{−6} M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = *k*[cisplatin]^{1}. Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table \(\PageIndex{1}\). For example, substituting the values for Experiment 3 into Equation \(\ref{14.4.5}\),

3.6 × 10^{−5} M/min = *k*(0.024 M)

1.5 × 10^{−3} min^{−1} = *k*

Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.

##### Example \(\PageIndex{1}\)

At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction:

\[\ce{CH_3CH_2Cl(g) ->[\Delta] HCl(g) + C_2H_4(g)} \nonumber \]

Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction.

Experiment |
[CH_{3}CH_{2}Cl]_{0} (M) |
Initial Rate (M/s) |
---|---|---|

1 | 0.010 | 1.6 × 10^{−8} |

2 | 0.015 | 2.4 × 10^{−8} |

3 | 0.030 | 4.8 × 10^{−8} |

4 | 0.040 | 6.4 × 10^{−8} |

**Given:** balanced chemical equation, initial concentrations of reactant, and initial rates of reaction

**Asked for:** reaction order and rate constant

###### Strategy:

- Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species.
- Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.

**C** Use measured concentrations and rate data from any of the experiments to find the rate constant.

###### Solution

The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.

**A** Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH_{3}CH_{2}Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH_{3}CH_{2}Cl].

**B** This behavior is characteristic of a first-order reaction, for which the rate law is rate =* k*[CH_{3}CH_{2}Cl].

**C** We can calculate the rate constant (*k*) using any row in the table. Selecting Experiment 1 gives the following:

1.60 × 10^{−8} M/s = *k*(0.010 M)

1.6 × 10^{−6} s^{−1} = *k*

##### Exercise \(\PageIndex{1}\)

Sulfuryl chloride (SO_{2}Cl_{2}) decomposes to SO_{2} and Cl_{2} by the following reaction:

\[SO_2Cl_2(g) → SO_2(g) + Cl_2(g) \nonumber \]

Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.

Experiment |
[SO_{2}Cl_{2}]_{0} (M) |
Initial Rate (M/s) |
---|---|---|

1 | 0.0050 | 1.10 × 10^{−7} |

2 | 0.0075 | 1.65 × 10^{−7} |

3 | 0.0100 | 2.20 × 10^{−7} |

4 | 0.0125 | 2.75 × 10^{−7} |

**Answer**- first order;
*k*= 2.2 × 10^{−5}s^{−1}

We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Figure \(\PageIndex{6a}\) shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C.

The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in Figure \(\PageIndex{6}\). The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in Figure \(\PageIndex{6}\) for *t* = 100 min ([cisplatin] = 0.0086 M) and *t* = 1000 min ([cisplatin] = 0.0022 M),

\[\begin{align*}\textrm{slope}&=\dfrac{\ln [\textrm{cisplatin}]_{1000}-\ln [\textrm{cisplatin}]_{100}}{\mathrm{1000\;min-100\;min}}

\\[4pt] -k&=\dfrac{\ln 0.0022-\ln 0.0086}{\mathrm{1000\;min-100\;min}}=\dfrac{-6.12-(-4.76)}{\mathrm{900\;min}}=-1.51\times10^{-3}\;\mathrm{min^{-1}}

\\[4pt] k&=1.5\times10^{-3}\;\mathrm{min^{-1}}\end{align*} \nonumber \]

The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min^{−1} because the times plotted on the horizontal axes in parts (a) and (b) in Figure \(\PageIndex{6}\) are in minutes rather than seconds.

The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions.

Video Example Using the First-Order Integrated Rate Law Equation:

Example Using the First-Order Integrated Rate Law Equation(opens in new window) [youtu.be]

##### Example \(\PageIndex{2}\)

If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M (*k* = 1.6 × 10^{−6} s^{−1}) ?

**Given:** initial concentration, rate constant, and time interval

**Asked for:** concentration at specified time and time required to obtain particular concentration

###### Strategy:

- Substitute values for the initial concentration ([A]
_{0}) and the calculated rate constant for the reaction (*k*) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time*t*. - Given a concentration [A], solve the integrated rate law for time
*t*.

###### Solution

The exponential form of the integrated rate law for a first-order reaction (Equation \(\ref{14.4.6}\)) is [A] = [A]_{0}*e*^{−kt}.

**A** Having been given the initial concentration of ethyl chloride ([A]_{0}) and having the rate constant of *k* = 1.6 × 10^{−6} s^{−1}, we can use the rate law to calculate the concentration of the reactant at a given time *t*. Substituting the known values into the integrated rate law,

\[\begin{align*}[\mathrm{CH_3CH_2Cl}]_{\mathrm{10\;h}}&=[\mathrm{CH_3CH_2Cl}]_0e^{-kt}

\\[4pt] &=\textrm{0.0200 M}(e^{\large{-(1.6\times10^{-6}\textrm{ s}^{-1})[(10\textrm{ h})(60\textrm{ min/h})(60\textrm{ s/min})]}})

\\[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber \]

We could also have used the logarithmic form of the integrated rate law (Equation \(\ref{14.4.7}\)):

\[\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=\ln [\mathrm{CH_3CH_2Cl}]_0-kt

\\[4pt] &=\ln 0.0200-(1.6\times10^{-6}\textrm{ s}^{-1})[(\textrm{10 h})(\textrm{60 min/h})(\textrm{60 s/min})]

\\[4pt] &=-3.912-0.0576=-3.970 \nonumber

\\[4pt] [\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=e^{-3.970}\textrm{ M} \nonumber

\\[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber \]

**B** To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for \(t\). Equation \(\ref{14.4.7}\) gives the following:

\[\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_t &=\ln[\mathrm{CH_3CH_2Cl}]_0-kt

\\[4pt] kt &=\ln[\mathrm{CH_3CH_2Cl}]_0-\ln[\mathrm{CH_3CH_2Cl}]_t=\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t}

\\[4pt] t &=\dfrac{1}{k}\left (\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \right )=\dfrac{1}{1.6\times10^{-6}\textrm{ s}^{-1}}\left(\ln \dfrac{0.0200\textrm{ M}}{0.0050\textrm{ M}}\right)

\\[4pt] &=\dfrac{\ln 4.0}{1.6\times10^{-6}\textrm{ s}^{-1}}=8.7\times10^5\textrm{ s}=240\textrm{ h}=2.4\times10^2\textrm{ h} \nonumber \end{align*} \nonumber \]

##### Exercise \(\PageIndex{2}\)

In the exercise in Example \(\PageIndex{1}\), you found that the decomposition of sulfuryl chloride (\(\ce{SO2Cl2}\)) is first order, and you calculated the rate constant at 320°C.

- Use the form(s) of the integrated rate law to find the amount of \(\ce{SO2Cl2}\) that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C.
- How long would it take for 90% of the SO
_{2}Cl_{2}to decompose?

**Answer a**- 0.0252 M
**Answer b**- 29 h

## Second-Order Reactions

The simplest kind of **second-order reaction** is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form

\[\ce{2A → products.}\nonumber \]

A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).

The differential rate law for the simplest second-order reaction in which 2A → products is as follows:

\[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \label{14.4.8} \]

Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M^{−1}·s^{−1}). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).

For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:

\[\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \label{14.4.9} \]

Because Equation \(\ref{14.4.9}\) has the form of an algebraic equation for a straight line, *y* = *mx* + *b*, with *y* = 1/[A] and *b* = 1/[A]_{0}, a plot of 1/[A] versus *t* for a simple second-order reaction is a straight line with a slope of *k* and an intercept of 1/[A]_{0}.

Second-order reactions generally have the form 2A → products or A + B → products.

Video Discussing the Second-Order Integrated Rate Law Equation: Second-Order Integrated Rate Law Equation(opens in new window) [youtu.be]

Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO_{2} to NO and O_{2} and the decomposition of HI to I_{2} and H_{2}. Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.

Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:

Figure \(\PageIndex{7}\)

For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law (Equation \(\ref{14.4.8}\)) or the integrated rate law (Equation \(\ref{14.4.9}\)).

Time (min) |
[Monomer] (M) |
Instantaneous Rate (M/min) |
---|---|---|

10 | 0.0044 | 8.0 × 10^{−5} |

26 | 0.0034 | 5.0 × 10^{−5} |

44 | 0.0027 | 3.1 × 10^{−5} |

70 | 0.0020 | 1.8 × 10^{−5} |

120 | 0.0014 | 8.0 × 10^{−6} |

To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in Table \(\PageIndex{2}\). From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:

\[\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7 \nonumber \]

Because (1.7)^{2} = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.

rate ∝ [monomer]^{2}

This means that the reaction is second order in the monomer. Using Equation \(\ref{14.4.8}\) and the data from any row in Table \(\PageIndex{2}\), we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:

\[\begin{align}\textrm{rate}&=k[\textrm A]^2 \\8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \\4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align} \nonumber \]

We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in Figure \(\PageIndex{8a}\). The measurements show that the concentration of the monomer (initially 5.4 × 10^{−3} M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus *t* should be a straight line, as shown in Figure \(\PageIndex{8b}\). Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, *k* = 4.1 M^{−1}·min^{−1}, which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.

For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders.

##### Example \(\PageIndex{3}\)

At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.

\[\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \nonumber \]

Experimental data for the reaction at 300°C and four initial concentrations of NO_{2} are listed in the following table:

Experiment |
[NO_{2}]_{0} (M) |
Initial Rate (M/s) |
---|---|---|

1 | 0.015 | 1.22 × 10^{−4} |

2 | 0.010 | 5.40 × 10^{−5} |

3 | 0.0080 | 3.46 × 10^{−5} |

4 | 0.0050 | 1.35 × 10^{−5} |

Determine the reaction order and the rate constant.

**Given: **balanced chemical equation, initial concentrations, and initial rates

**Asked for:** reaction order and rate constant

###### Strategy:

- From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.
- Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant (
*k*).

###### Solution

**A** We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO_{2} concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10^{−5}) ÷ (1.35 × 10^{−5}) = 4.0], which means that the reaction rate is proportional to [NO_{2}]^{2}. Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO_{2}]^{2}. This behavior is characteristic of a second-order reaction.

**B** We have rate = *k*[NO_{2}]^{2}. We can calculate the rate constant (*k*) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:

\[\begin{align*}\textrm{rate}&=k[\mathrm{NO_2}]^2

\\5.40\times10^{-5}\textrm{ M/s}&=k(\mathrm{\mathrm{0.010\;M}})^2

\\0.54\mathrm{\;M^{-1}\cdot s^{-1}}&=k\end{align*} \nonumber \]

##### Exercise \(\PageIndex{3}\)

When the highly reactive species HO_{2} forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:

\[2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber \]

The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:

Experiment |
[HO_{2}]_{0} (M) |
Initial Rate (M/s) |
---|---|---|

1 | 1.1 × 10^{−8} |
1.7 × 10^{−7} |

2 | 2.5 × 10^{−8} |
8.8 × 10^{−7} |

3 | 3.4 × 10^{−8} |
1.6 × 10^{−6} |

4 | 5.0 × 10^{−8} |
3.5 × 10^{−6} |

Determine the reaction order and the rate constant.

**Answer**- second order in HO
_{2};*k*= 1.4 × 10^{9}M^{−1}·s^{−1}

If a plot of reactant concentration versus time is **not **linear, but a plot of 1/(reactant concentration) versus time is linear, then the reaction is second order.

##### Example \(\PageIndex{4}\)

If a flask that initially contains 0.056 M NO_{2} is heated at 300°C, what will be the concentration of NO_{2} after 1.0 h? How long will it take for the concentration of NO_{2} to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction (Equation \ref{14.4.9}) and the rate constant calculated above.

**Given:** balanced chemical equation, rate constant, time interval, and initial concentration

**Asked for:** final concentration and time required to reach specified concentration

###### Strategy:

- Given
*k*,*t*, and [A]_{0}, use the integrated rate law for a second-order reaction to calculate [A]. - Setting [A] equal to 1/10 of [A]
_{0}, use the same equation to solve for \(t\).

###### Solution

**A** We know *k *and [NO_{2}]_{0}, and we are asked to determine [NO_{2}] at *t* = 1 h (3600 s). Substituting the appropriate values into Equation \ref{14.4.9},

\[\begin{align*}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt \\[4pt] &=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \\[4pt] &=2.0\times10^3\textrm{ M}^{-1}\end{align*} \nonumber \]

Thus [NO_{2}]_{3600} = 5.1 × 10^{−4} M.

**B** In this case, we know *k* and [NO_{2}]_{0}, and we are asked to calculate at what time [NO_{2}] = 0.1[NO_{2}]_{0} = 0.1(0.056 M) = 0.0056 M. To do this, we solve Equation \(\ref{14.4.9}\) for *t*, using the concentrations given.

\[ \begin{align*} t &=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k} \\[4pt] &=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}} \\[4pt] &=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \end{align*} \nonumber \]

NO_{2} decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.

##### Exercise \(\PageIndex{4}\)

In the previous exercise, you calculated the rate constant for the decomposition of HO_{2} as *k* = 1.4 × 10^{9} M^{−1}·s^{−1}. This high rate constant means that HO_{2} decomposes rapidly under the reaction conditions given in the problem. In fact, the HO_{2} molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO_{2}, calculate the concentration of HO_{2} that remains after 1.0 h at 25°C. How long will it take for 90% of the HO_{2} to decompose? Use the integrated rate law for a second-order reaction (Equation \(\ref{14.4.9}\)) and the rate constant calculated in the exercise in Example \(\PageIndex{3}\).

**Answer**- 2.0 × 10
^{−13}M; 6.4 × 10^{−6}s

In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form \(A + B \rightarrow products\), in which the reaction is first order in \(A\) and first order in \(B\). The differential rate law for this reaction is as follows:

\[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A][\textrm B] \nonumber \]

Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant.

## Summary

The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.

**zeroth-order reaction:**\[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k \nonumber \] \[[A] = [A]_0 − kt \nonumber \]**first-order reaction**: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \nonumber \] \[[A] = [A]_0e^{−kt} \nonumber \] \[\ln[A] = \ln[A]_0 − kt \nonumber \]**second-order reaction:**\[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^2 \nonumber \] \[\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \nonumber \]