If you like us, please share us on social media.
The latest UCD Hyperlibrary newsletter is now complete, check it out.

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

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 Representation

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

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

The "rate" is the reaction rate (in units of molar/time) and \(k\) 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 as demonstrated below.

The Integral Representation

First, write the differential form of the rate law.

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

Rearrange to give:

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

Second, integrate both sides of the equation.

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

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

Recall from calculus that:

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

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 

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

where [A] is the concentration at time \(t\) and \([A]_o\) is the concentration at time 0, and \(k\) is  the first-order rate constant.



Figure 1: Decay profiles for first-order reactions with large and small rate constants. 


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 \(y=e^x\) 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.

\[ [A] = \dfrac{1}{2} [A]_o \]

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.



Figure 2: Half lives graphically desmonstrated for first-order reaction. Notice the the half-life is indenpen of initial concentration. This is not te case with other reaction orders.


Example 1: Estimated Rate Constants

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?


From the above equation

\[k = \dfrac{–0.693}{(600 s} = 0.00115 s^{–1}\]

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 2: 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?


There are two ways to approach this problems:

  • Approach #1: Recognize that the final concentration of A is \(\dfrac{1}{8}\) of the final concentration and hence three half lives (\(\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}\)) have elapsed during this reaction. \[t_{1/2} = \dfrac{36\, \text{min}}{3}= 12 \; \text{min}\]
  • 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.

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

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

\[t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{0.0578 \, \text{min}^{-1}} \approx 12\, \text{min}\]

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.


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​ \[\dfrac{[H_2O_2]_t}{[H_2O_2]_0} = e^{-kt}\] This is a simple plug and play application once you have identified this equation. \[\dfrac{[H_2O_2]_{t=600\;s}}{[H_2O_2]_0} = e^{-(6.40 \times 10^{-5} s^{-1}) (600 \, s)}\] \[\dfrac{[H_2O_2]_{t=600\;s}}{[H_2O_2]_0} = 0.9629\] 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​ \[\dfrac{[H_2O_2]_t}{[H_2O_2]_0} = e^{-kt}\] This is a simple plug and play application once you have identified this equation. \[\dfrac{[H_2O_2]_{t=450\;s}}{[H_2O_2]_0} = e^{-(6.40 \times 10^{-5} s^{-1}) (450 \, s)}\] \[\dfrac{[H_2O_2]_{t=450\;s}}{[H_2O_2]_0} = 0.9720\] So 100-96.3=2.8% of the hydrogen peroxide has decayed by 450 s.


  1. Petrucci, Ralph H. General Chemistry: Principles and Modern Applications 9th Ed. New Jersey: Pearson Education Inc. 2007.


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

You must to post a comment.
Last modified
07:17, 3 Apr 2015


This page has no custom tags.


Upper Divisional

Creative Commons License Unless otherwise noted, content in the UC Davis ChemWiki is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License. Permissions beyond the scope of this license may be available at copyright@ucdavis.edu. Questions and concerns can be directed toward Prof. Delmar Larsen (dlarsen@ucdavis.edu), Founder and Director. Terms of Use