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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Kinetics > Reaction Rates > First-Order Reactions

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First-Order Reactions

A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.

The differential rate law

The differential equation describing first-order kinetics is given below:

\[ Rate = - \frac{d[A]}{dt} = k[A]^1 = k[A] \tag{1}\]

The "rate" is the reaction rate (in units of molar/time) and is the reaction rate coefficient (in units of 1/time). However, the units of k vary for non-first-order reactions. These differential equations are separable, which simplifies the solutions.

The integral rate law

First, write the differential form of the rate law.

\[ Rate = - \frac{d[A]}{dt} = k[A] \tag{2}\]

Rearrange to give:

\[ \frac{d[A]}{[A]} = - k\,dt \tag{3}\]

Second, integrate both sides of the equation.

\[ \int_{[A]_o}^{[A]} \frac{d[A]}{[A]} = -\int_{t_o}^{t} k\, dt \tag{4a}\]

\[ \int_{[A]_{o}}^{[A]} \frac{1}{[A]} d[A] = -\int_{t_o}^{t} k\, dt \tag{4b}\]

Recall from calculus that:

\[ \int  \frac{1}{x} = \ln(x) \tag{5}\]

Upon integration,

\[ \ln[A] - \ln[A]_o = -kt \tag{6}\]

Rearrange to solve for [A] to obtain one form of the rate law:

\[ \ln[A] = \ln[A]_o - kt \tag{7}\]

This can be rearranged to:

\[ \ln [A] = -kt + \ln [A]_o \tag{8}\]

This can further be arranged into y=mx +b form:

\[ \ln [A] = -kt + \ln [A]_o \tag{9}\]

The equation is a straight line with slope m:

\[mx=-kt \tag{10}\]

and y-intercept b:

\[b=\ln [A]_o \tag{11}\]

Now, recall from the laws of logarithms that 

\[ \ln {\left(\frac{[A]_t}{ [A]_o}\right)}= -kt \tag{12}\]

where [A] is the concentration at time t and [A]o is the concentration at time 0, and k is  the rate constant.

   

Because the logarithms of numbers do not have any units, the product -kt also lacks units. This concludes that unit of k in a first order of reaction must be time-1. Examples of time-1 include s-1 or min-1. Thus, the equation of a straight line is applicable:

\[ \ln [A] = -kt + \ln [A]_o.\tag{15}\]

To test if it the reaction is a first-order reaction, plot the natural logarithm of a reactant concentration versus time and see whether the graph is linear. If the graph is linear and has a negative slope, the reaction must be a first-order reaction.

To create another form of the rate law, raise each side of the previous equation to the exponent, e:

\[ \large e^{\ln[A]} = e^{\ln[A]_o - kt} \tag{16}\]

Simplifying gives the second form of the rate law:

\[ [A] = [A]_{o}e^{- kt}\tag{17}\]

The integrated forms of the rate law can be used to find the population of reactant at any time after the start of the reaction. Plotting ln[A] with respect to time for a first-order reaction gives a straight line with the slope of the line equal to -k. More information can be found in the article on rate laws.

This general relationship, in which a quantity changes at a rate that depends on its instantaneous value, is said to follow an exponential law. Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms. The reason that the exponential function y=ex so efficiently describes such changes is that dy/dx = ex; that is, ex is its own derivative, making the rate of change of y identical to its value at any point.

Graphing first-order reactions

The following graphs represents concentration of reactants versus time for a first-order reaction.

Image04.jpg         Image05.jpg

Plotting ln[A] with respect to time for a first-order reaction gives a straight line with the slope of the line equal to -k.

First order.jpg

Half-lives of first order reactions

The half-life (t1/2) is a timescale on which the initial population is decreased by half of its original value, represented by the following equation.

\[ [A] = \frac{1}{2} [A]_o \]

After a period of one half-life, t = t and we can write

\[\dfrac{[A]_{1/2}}{[A]_o} = \dfrac{1}{2}=e^{-k\,t_{1/2}} \tag{18}\]

Taking logarithms of both sides (remember that \(\ln e^x = x\)) yields

\[ \ln 0.5 = -kt\tag{19}\]

Solving for the half-life, we obtain the simple relation

\[ t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{k}\tag{20}\]

This indicates that the half-life of a first-order reaction is a constant.

 

 

Example 5

The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant in reciprocal seconds?

SOLUTION

From the above equation, k = –0.693/(600 s) = 0.00115 s–1

The decay of radioactive nuclei is always a first-order process.

Notice that, for first-order reactions, the half-life is independent of the initial concentration of reactant, which is a unique aspect to first-order reactions. The practical implication of this is that it takes as much time for [A] to decrease from 1 M to 0.5 M as it takes for [A] to decrease from 0.1 M to 0.05 M. In addition, the rate constant and the half life of a first-order process are inversely related.

Example: Determining Half life

If 3.0 g of substance A decomposes for 36 minutes the mass of unreacted A remaining is found to be 0.375 g. What is the half life of this reaction if it follows first-order kinetics?

SOLUTION

There are two ways to approach this problems:

  • Approach #1: Recognize that the final concentration of A is \(\frac{1}{8}\) of the final concentration and hence three half lives (\(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\)) have elapsed during this reaction. \[t_{1/2} = \frac{36\, \text{min}}{3}= 12 \; \text{min}\]
  • Approach #2: "The brute force approach" involves solving for k from the integral rate law equation (eq. 12 or 17) and then relating k to the t1/2 via equation 20.

\[\frac{[A]_t}{[A]_o} =e^{-k\,t}\]

\[k= -\frac{\ln \frac{[A]_t}{[A]_o}}{t} = -\frac{\ln \frac{0.375\, g}{3\, g}}{36\, \text{min}} = 0.0578 \, \text{min}^{-1}\]

\[t_{1/2}=\dfrac{\ln{2}}{k} \approx \frac{0.693}{0.1268 \, \text{min}^{-1}} \approx 12\, \text{min}\]

The first approach is considerably faster (if the number of half lives evolved is apparent).

Practice Problems

Calculate the half-life of the reactions below:

  1. If 4.00 g A are allowed to decompose for 40 min, the mass of A remaining undecomposed is found to be 0.80 g. 
  2. If 8.00 g A are allowed to decompose for 34 min, the mass of A remaining undecomposed is found to be 0.70 g. 
  3. If 9.00 g A are allowed to decompose for 24 min, the mass of A remaining undecomposed is found to be 0.50 g.

Determine the percent H2O2 that decomposes in the time using \(k=6.40 \times 10^{-5} s^{-1}\)

  1. The time for the concentration to decompose is 600.0 s after the reaction begins. Use the value of k above. 
  2. The time for the concentration to decompose is 450 s after the reaction begins. Use the value of k above.

Solutions

Use the half life reaction that contains initial concentration and final concentration. Plug in the appropriate variables and solve to obtain:

  1. 17.2 min
  2. 9.67 min
  3. 5.75 min
  4. Rearranging Eq. 17 to solve for the  \([H_2O_2]_t/[H_2O_2]_0\) ratio​ \[\frac{[H_2O_2]_t}{[H_2O_2]_0} = e^{-kt}\] This is a simple plug and play application once you have identified this equation. \[\frac{[H_2O_2]_{t=600\;s}}{[H_2O_2]_0} = e^{-(6.40 \times 10^{-5} s^{-1}) (600 \, s)}\] \[\frac{[H_2O_2]_{t=600\;s}}{[H_2O_2]_0} = 0.9629\] So 100-96.3=3.71% of the hydrogen peroxide has decayed by 600 s.
  5. Rearranging Eq. 17 to solve for the  \([H_2O_2]_t/[H_2O_2]_0\) ratio​ \[\frac{[H_2O_2]_t}{[H_2O_2]_0} = e^{-kt}\] This is a simple plug and play application once you have identified this equation. \[\frac{[H_2O_2]_{t=450\;s}}{[H_2O_2]_0} = e^{-(6.40 \times 10^{-5} s^{-1}) (450 \, s)}\] \[\frac{[H_2O_2]_{t=450\;s}}{[H_2O_2]_0} = 0.9720\] So 100-96.3=2.8% of the hydrogen peroxide has decayed by 450 s.

Summary

The kinetics of any reaction depend on the reaction mechanism, or rate law, and the initial conditions. Assuming that for the reaction A -> products, there is an initial concentration of reactant of [A]0 at time t=0, and the rate law is an integral order in A, then the kinetics of the the first order reaction can be summarized as follows:

1st.jpg

References

  1. Petrucci, Ralph H. General Chemistry: Principles and Modern Applications 9th Ed. New Jersey: Pearson Education Inc. 2007.
  2. http://en.wikipedia.org/wiki/Half-life

Contributors

  • Rachael Curtis (UCD), Cathy Nguyen (UCD)

Last modified
14:29, 13 Jan 2015

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