12.5: Integrated Rate Laws
 Page ID
 1897
Skills to Develop
 Explain the form and function of an integrated rate law
 Perform integrated rate law calculations for zero, first, and secondorder reactions
 Define halflife and carry out related calculations
 Identify the order of a reaction from concentration/time data
The rate laws we have seen thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.
Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first, second, and zeroorder reactions.
FirstOrder Reactions
An equation relating the rate constant k to the initial concentration [A]_{0} and the concentration [A]_{t} present after any given time t can be derived for a firstorder reaction and shown to be:
or
or
\[[A]=[A]_0e^{−kt}\]
Example \(\PageIndex{1}\): The Integrated Rate Law for a FirstOrder Reaction
The rate constant for the firstorder decomposition of cyclobutane, C_{4}H_{8} at 500 °C is 9.2
\[\ce{C4H8⟶2C2H4}\]
How long will it take for 80.0% of a sample of C_{4}H_{8} to decompose?
Solution
We use the integrated form of the rate law to answer questions regarding time:
\[\ln\left(\dfrac{[A]_0}{[A]}\right)=kt\]
There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [A]_{0}, [A], and k, and need to find t.
The initial concentration of C_{4}H_{8}, [A]_{0}, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of x or 0.200x. Rearranging the rate law to isolate t and substituting the provided quantities yields:
\(\begin{align*}
t&=\ln\dfrac{[x]}{[0.200x]}×\dfrac{1}{k}\\
&=\mathrm{\ln\dfrac{0.100\:mol\: L^{−1}}{0.020\:mol\: L^{−1}}×\dfrac{1}{9.2×10^{−3}\:s^{−1}}}\\
&=\mathrm{1.609×\dfrac{1}{9.2×10^{−3}\:s^{−1}}}\\
&=\mathrm{1.7×10^2\:s}
\end{align*}\)
Exercise \(\PageIndex{1}\)
Iodine131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine131 decays to xenon131 according to the equation:
\[\textrm{I131 ⟶ Xe131 + electron}\]
The decay is firstorder with a rate constant of 0.138 d^{−1}. All radioactive decay is first order. How many days will it take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe131?
 Answer

16.7 days
We can use integrated rate laws with experimental data that consist of time and concentration information to determine the order and rate constant of a reaction. The integrated rate law can be rearranged to a standard linear equation format:
\[\begin{align*}
\ln[A]&=(−k)(t)+\ln[A]_0\\
y&=mx+b
\end{align*}\]
A plot of ln[A] versus t for a firstorder reaction is a straight line with a slope of −k and an intercept of ln[A]_{0}. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A.
Example \(\PageIndex{2}\): Determination of Reaction Order by Graphing
Show that the data in this Figure can be represented by a firstorder rate law by graphing ln[H_{2}O_{2}] versus time. Determine the rate constant for the rate of decomposition of H_{2}O_{2} from this data.
Solution
The data from this Figure with the addition of values of ln[H_{2}O_{2}] are given in Figure \(\PageIndex{1}\).
Figure \(\PageIndex{1}\): The linear relationship between the ln[H_{2}O_{2}] and time shows that the decomposition of hydrogen peroxide is a firstorder reaction.
Trial  Time (h)  [H_{2}O_{2}] (M)  ln[H_{2}O_{2}] 

1  0  1.000  0.0 
2  6.00  0.500  −0.693 
3  12.00  0.250  −1.386 
4  18.00  0.125  −2.079 
5  24.00  0.0625  −2.772 
The plot of ln[H_{2}O_{2}] versus time is linear, thus we have verified that the reaction may be described by a firstorder rate law.
The rate constant for a firstorder reaction is equal to the negative of the slope of the plot of ln[H_{2}O_{2}] versus time where:
\[\ce{slope}=\dfrac{\textrm{change in }y}{\textrm{change in }x}=\dfrac{Δy}{Δx}=\dfrac{Δ\ln[\ce{H2O2}]}{Δt}\]
In order to determine the slope of the line, we need two values of ln[H_{2}O_{2}] at different values of t (one near each end of the line is preferable). For example, the value of ln[H_{2}O_{2}] when t is 6.00 h is −0.693; the value when t = 12.00 h is −1.386:
\[\begin{align*}
\ce{slope}&=\mathrm{\dfrac{−1.386−(−0.693)}{12.00\: h−6.00\: h}}\\
&=\mathrm{\dfrac{−0.693}{6.00\: h}}\\
&=\mathrm{−1.155×10^{−2}\:h^{−1}}\\
k&=\mathrm{−slope=−(−1.155×10^{−1}\:h^{−1})=1.155×10^{−1}\:h^{−1}}
\end{align*}\]
Exercise \(\PageIndex{2}\)
Graph the following data to determine whether the reaction
Trial  Time (s)  [A] 

1  4.0  0.220 
2  8.0  0.144 
3  12.0  0.110 
4  16.0  0.088 
5  20.0  0.074 
 Answer

The plot of ln[A] vs. t is not a straight line. The equation is not first order:
SecondOrder Reactions
The equations that relate the concentrations of reactants and the rate constant of secondorder reactions are fairly complicated. We will limit ourselves to the simplest secondorder reactions, namely, those with rates that are dependent upon just one reactant’s concentration and described by the differential rate law:
\[\ce{Rate}=k[A]^2\]
For these secondorder reactions, the integrated rate law is:
\[\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0}\]
where the terms in the equation have their usual meanings as defined earlier.
Example \(\PageIndex{3}\): The Integrated Rate Law for a SecondOrder Reaction
The reaction of butadiene gas (C_{4}H_{6}) with itself produces C_{8}H_{12} gas as follows:
\[\ce{2C4H6}(g)⟶\ce{C8H12}(g)\]
The reaction is second order with a rate constant equal to 5.76
Solution
We use the integrated form of the rate law to answer questions regarding time. For a secondorder reaction, we have:
We know three variables in this equation: [A]_{0} = 0.200 mol/L, k = 5.76
\[\begin{align*}
\dfrac{1}{[A]}&=\mathrm{(5.76×10^{−2}\:L\: mol^{−1}\:min^{−1})(10\:min)+\dfrac{1}{0.200\:mol^{−1}}}\\
\dfrac{1}{[A]}&=\mathrm{(5.76×10^{−1}\:L\: mol^{−1})+5.00\:L\: mol^{−1}}\\
\dfrac{1}{[A]}&=\mathrm{5.58\:L\: mol^{−1}}\\
[A]&=\mathrm{1.79×10^{−1}\:mol\: L^{−1}}
\end{align*}\]
Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present.
Exercise \(\PageIndex{3}\)
If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min?
 Answer

0.0196 mol/L
The integrated rate law for our secondorder reactions has the form of the equation of a straight line:
\[\begin{align*}
\dfrac{1}{[A]}&=kt+\dfrac{1}{[A]_0}\\
y&=mx+b
\end{align*}\]
A plot of \(\dfrac{1}{[A]}\) versus t for a secondorder reaction is a straight line with a slope of k and an intercept of \(\dfrac{1}{[A]_0}\). If the plot is not a straight line, then the reaction is not second order.
Example \(\PageIndex{4}\): Determination of Reaction Order by Graphing
Test the data given to show whether the dimerization of C_{4}H_{6} is a first or a secondorder reaction.
Solution
Trial  Time (s)  [C_{4}H_{6}] (M) 

1  0  1.00 
2  1600  5.04 
3  3200  3.37 
4  4800  2.53 
5  6200  2.08 
In order to distinguish a firstorder reaction from a secondorder reaction, we plot ln[C_{4}H_{6}] versus t and compare it with a plot of
Time (s)  \(\dfrac{1}{[\ce{C4H6}]}\:(M^{−1})\)  ln[C_{4}H_{6}] 

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 \(\PageIndex{2}\). As you can see, the plot of ln[C_{4}H_{6}] versus t is not linear, therefore the reaction is not first order. The plot of
Figure \(\PageIndex{2}\): These two graphs show first and secondorder plots for the dimerization of C_{4}H_{6}. Since the firstorder plot (left) is not linear, we know that the reaction is not first order. The linear trend in the secondorder plot (right) indicates that the reaction follows secondorder kinetics.
Exercise \(\PageIndex{4}\)
Does the following data fit a secondorder rate law?
Trial  Time (s)  [A] (M) 

1  5  0.952 
2  10  0.625 
3  15  0.465 
4  20  0.370 
5  25  0.308 
6  35  0.230 
 Answer:

Yes. The plot of
\(\dfrac{1}{[A]}\) vs. t is linear:
ZeroOrder Reactions
For zeroorder reactions, the differential rate law is:
\[\ce{Rate}=k[A]^0=k\]
A zeroorder reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactants.
The integrated rate law for a zeroorder reaction also has the form of the equation of a straight line:
\(\begin{align*}
[A]&=−kt+[A]_0\\
y&=mx+b
\end{align*}\)
A plot of [A] versus t for a zeroorder reaction is a straight line with a slope of −k and an intercept of [A]_{0}. Figure \(\PageIndex{3}\) shows a plot of [NH_{3}] versus t for the decomposition of ammonia on a hot tungsten wire and for the decomposition of ammonia on hot quartz (SiO_{2}). The decomposition of NH_{3} on hot tungsten is zero order; the plot is a straight line. The decomposition of NH_{3} on hot quartz is not zero order (it is first order). From the slope of the line for the zeroorder decomposition, we can determine the rate constant:
\[\ce{slope}=−k=\mathrm{1.3110^{−6}\:mol/L/s}\]
Figure \(\PageIndex{3}\): The decomposition of NH_{3} on a tungsten (W) surface is a zeroorder reaction, whereas on a quartz (SiO_{2}) surface, the reaction is first order.
The HalfLife of a Reaction
The halflife of a reaction (t_{1/2}) is the time required for onehalf of a given amount of reactant to be consumed. In each succeeding halflife, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide as an example, we find that during the first halflife (from 0.00 hours to 6.00 hours), the concentration of H_{2}O_{2} decreases from 1.000 M to 0.500 M. During the second halflife (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M; during the third halflife, it decreases from 0.250 M to 0.125 M. The concentration of H_{2}O_{2} decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a firstorder reaction, and, as can be shown, the halflife of a firstorder reaction is independent of the concentration of the reactant. However, halflives of reactions with other orders depend on the concentrations of the reactants.
FirstOrder Reactions
We can derive an equation for determining the halflife of a firstorder reaction from the alternate form of the integrated rate law as follows:
\[\begin{align*}
\ln\dfrac{[A]_0}{[A]}&=kt\\
t&=\ln\dfrac{[A]_0}{[A]}×\dfrac{1}{k}
\end{align*}\]
If we set the time t equal to the halflife,
Therefore:
\[\begin{align*}
t_{1/2}&=\ln\dfrac{[A]_0}{\dfrac{1}{2}[A]_0}×\dfrac{1}{k}\\
&=\ln 2×\dfrac{1}{k}=0.693×\dfrac{1}{k}
\end{align*}\]
Thus:
We can see that the halflife of a firstorder reaction is inversely proportional to the rate constant k. A fast reaction (shorter halflife) will have a larger k; a slow reaction (longer halflife) will have a smaller k.
Example \(\PageIndex{5}\): Calculation of a Firstorder Rate Constant using HalfLife
Calculate the rate constant for the firstorder decomposition of hydrogen peroxide in water at 40 °C, using the data given in Figure \(\PageIndex{4}\).
Figure \(\PageIndex{4}\): The decomposition of H_{2}O_{2} \(\ce{(2H2O2⟶2H2O + O2)}\)
Solution The halflife for the decomposition of H_{2}O_{2} is 2.16
\[\begin{align*}
t_{1/2}&=\dfrac{0.693}{k}\\
k&=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{2.16×10^4\:\ce s}=3.21×10^{−5}\:\ce s^{−1}
\end{align*}\]
Exercise \(\PageIndex{1}\)
The firstorder radioactive decay of iodine131 exhibits a rate constant of 0.138 d^{−1}. What is the halflife for this decay?
Answer
5.02 d.
SecondOrder Reactions
We can derive the equation for calculating the halflife of a second order as follows:
\[\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0}\]
or
\[\dfrac{1}{[A]}−\dfrac{1}{[A]_0}=kt\]
If
\[t=t_{1/2}\]
then
\[[A]=\dfrac{1}{2}[A]_0\]
and we can write:
\dfrac{1}{\dfrac{1}{2}[A]_0}−\dfrac{1}{[A]_0}&=kt_{1/2}\\
2[A]_0−\dfrac{1}{[A]_0}&=kt_{1/2}\\
\dfrac{1}{[A]_0}&=kt_{1/2}
\end{align*}\)
Thus:
\[t_{1/2}=\dfrac{1}{k[A]_0}\]
For a secondorder reaction, \(t_{1/2}\)
ZeroOrder Reactions
We can derive an equation for calculating the halflife of a zero order reaction as follows:
\[[A]=−kt+[A]_0\]
When half of the initial amount of reactant has been consumed \(t=t_{1/2}\)
\dfrac{[A]_0}{2}&=−kt_{1/2}+[A]_0\\
kt_{1/2}&=\dfrac{[A]_0}{2}
\end{align*}\]
and
\[t_{1/2}=\dfrac{[A]_0}{2k}\]
The halflife of a zeroorder reaction increases as the initial concentration increases. Equations for both differential and integrated rate laws and the corresponding halflives for zero, first, and secondorder reactions are summarized in Table \(\PageIndex{1}\).
ZeroOrder  FirstOrder  SecondOrder  

rate law  rate = k  rate = k[A]  rate = k[A]^{2} 
units of rate constant  M s^{−1}  s^{−1}  M^{−1} s^{−1} 
integrated rate law  [A] = −kt + [A]_{0}  ln[A] = −kt + ln[A]_{0}  \(\dfrac{1}{[A]}=kt+\left(\dfrac{1}{[A]_0}\right)\) 
plot needed for linear fit of rate data  [A] vs. t  ln[A] vs. t  \(\dfrac{1}{[A]}\) vs. t 
relationship between slope of linear plot and rate constant  k = −slope  k = −slope  k = +slope 
halflife  \(t_{1/2}=\dfrac{[A]_0}{2k}\)  \(t_{1/2}=\dfrac{0.693}{k}\)  \(t_{1/2}=\dfrac{1}{[A]_0k}\) 
Summary
Differential rate laws can be determined by the method of initial rates or other methods. We measure values for the initial rates of a reaction at different concentrations of the reactants. From these measurements, we determine the order of the reaction in each reactant. Integrated rate laws are determined by integration of the corresponding differential rate laws. Rate constants for those rate laws are determined from measurements of concentration at various times during a reaction.
The halflife of a reaction is the time required to decrease the amount of a given reactant by onehalf. The halflife of a zeroorder reaction decreases as the initial concentration of the reactant in the reaction decreases. The halflife of a firstorder reaction is independent of concentration, and the halflife of a secondorder reaction decreases as the concentration increases.
Key Equations
 integrated rate law for zeroorder reactions: \([A]=−kt+[A]_0\), \(t_{1/2}=\dfrac{[A]_0}{2k}\)
 integrated rate law for firstorder reactions: \(\ln[A]=−kt+[A]_0\), \(t_{1/2}=\dfrac{0.693}{k}\)
 integrated rate law for secondorder reactions: \(\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0}\), \(t_{1/2}=\dfrac{1}{[A]_0k}\)
Glossary
 halflife of a reaction (t_{l/2})
 time required for half of a given amount of reactant to be consumed
 integrated rate law
 equation that relates the concentration of a reactant to elapsed time of reaction
Contributors
Paul Flowers (University of North Carolina  Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf1932bd...a7ac8df6@9.110).