In a first order reaction, the rate of reaction is directly proportional to the concentration of a single $A species and is independent of the concentrations of all others. Therefore, the reaction rate equation will be:

$$-\ \frac{d[A]}{dt} = k \cdot [A]$$

where k is the specific velocity coefficient. There are many reactions that follow this type of behavior. Among them are frequent decompositions and isomerizations.

To obtain the equation that allows us to know how the concentration of $A$ varies with time, we have to integrate the initial expression. We will have to separate variables first:

$$-\ \frac{d[A]}{dt} = k \cdot [A]\ \ \Rightarrow\ \ -\ \frac{d[A]}{[A]} = k \cdot dt$$

and now we integrate:

$$-\ \int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = k \cdot \int_0^t dt$$

$$-\ \left ( {\ln [A] - \ln [A]_0} \right ) = k \cdot t$$

$$\ln {\frac{[A]}{[A]_0}} = -\ k \cdot t$$

$$[A] = [A]_0 \cdot e^{-\ k\cdot t}$$

where $[A]_0$ represents the concentration of the reactant $A$ at the start of the reaction, that is, when $t = 0$, and $[A]$ is the concentration of the same at any instant $t$.

The solution of the previous equation is done algebraically when we know the value of the specific speed constant. Otherwise, it can be done graphically by plotting $[A]/[A]_0$ as a function of time.

But the simplest is to draw the line of least squares that fits the experimental values, representing the logarithm of the concentration as a function of time. That gives us a line whose slope is – k.

Let's see a practical example of this.

As we have said, many of the first-order reactions are decompositions or isomerizations of a chemical species, usually by the action of heat, some type of radiation (for example, light),... etc. Azomethane, with the empirical formula $C2H6N2, is a yellow, easily explosive gas with a boiling point t _{eb} = 1.5 ºC, which at temperatures between 60 and 100 ºC breaks down into methyl radicals and nitrogen gas (#1). The methyl groups formed quickly join to give ethane. The reaction is the following:

$$CH3-N=N-CH_3 \longrightarrow CH3-CH3 + N_2$$

If the concentration of azomethane is measured over time, under certain conditions of stable pressure and temperature, values such as those in the following table can be obtained:

*azomethane decomposition*

Time / min |
[CH _{3} -N=N-CH _{3} ] / 10 ^{-2} mol · dm ^{-3} |

0 | 1.21 |

10 | 1.04 |

twenty | 0.880 |

33 | 0.706 |

46 | 0.575 |

65 | 0.434 |

If we take the logarithm of the azomethane concentrations and plot it against time, we get the following:

Time / min |
[CH3-N=N-CH3] / 10-2 mol dm-3 |
ln([CH3-N=N-CH3]) |

0 | 1.21 | -4.41 |

10 | 1.04 | -4.57 |

twenty | 0.880 | - 4.73 |

33 | 0.706 | - 4.95 |

46 | 0.575 | -5.16 |

65 | 0.434 | - 5.44 |

*Graphical representation of the logarithm of the concentration*

*azomethane versus time*

Graphically, we can see that the logarithm of the azomethane concentration decreases linearly with time. Therefore, it is confirmed that the decomposition reaction of azomethane that we have described before is first order with respect to it. That is, we have to:

$$v = v_{C_{2}H_{6}N_2} = - \frac{d[C_{2}H_{6}N_{2}]}{dt} = k \cdot [C_{2}H_ {6}N_{2}]$$

As we have seen before, the integration of this type of differential equation leads us to the following expression:

$$[C_{2}H_{6}N_{2}] = [C_{2}H_{6}N_{2}]_0 \cdot e^{- k \cdot t} \Leftrightarrow \ln ([C_ {2}H_{6}N_{2}]) = \ln ([C_{2}H_{6}N_{2}]_0) - k \cdot t$$

So the slope of the line observed in the graph will be the value of $-\ k$. There are many ways to calculate this slope, although the most widely used and well-known is to calculate the least squares line that fits the data in the table, a procedure widely used in statistics and data processing. We will not explain how it is done here because there is abundant bibliography (#3) on it and it is a type of calculation that is studied in any statistical mathematics course. The value obtained for the specific rate constant of the reaction is:

$$k = 0.0157 min^{-\ 1}$$

Therefore, the rate of decomposition of azomethane, according to the data handled in this example, responds to the equation:

$$v = 0.0157 .[C_{2}H_{6}N_{2}]$$

Expressing time in minutes.

Many times, when we are studying reactions in a gaseous medium, instead of using the concentration of a species we can use the partial pressure of said component, since, by the gas law:

$$p_A \cdot V = n_A \cdot R \cdot T \Leftrightarrow p_A = \frac{n_A}{V} \cdot R \cdot T \Leftrightarrow p_A = [A] \cdot R \cdot T$$

so we will have:

$$\frac{[A]}{[A]_0} = e^{- k \cdot t} \Leftrightarrow \frac{\frac{p_A}{R \cdot T}}{\frac{(p_A)_0 }{R \cdot T}} = e^{- k \cdot t} \Leftrightarrow \frac{p_A}{(p_A)_0} = e^{- k \cdot t}$$

And therefore, we can also write the logarithmic equation of the first-order reactions using partial pressures:

$$\ln p_A = \ln (p_A)_0 - k \cdot t$$

equivalent to the one we used in the example to graphically determine the value of $k$. Of course, care must be taken to express the rate of reaction in the appropriate units. That is, in this case, units of pressure divided by units of time. Although they are rarely used, in SI we would have $Pa \cdot s^{-1}$.

**Bibliography :**

# 1 - H.Beyer, W.Walter – *Manual of Organic Chemistry* – 1st ed., 1980 (p. 187) – Ed. Reverté.

#2 - HCRamsperger - *J. Am. Chem. Soc.* , **49** , 912 (1927).

# 3 - There is a good article on this subject in Wikipedia: http://es.wikipedia.org/wiki/Regresi%C3%B3n_lineal