Extent of reaction

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In physical chemistry, the extent of reaction is a quantity that measures the extent in which the reaction proceeds. It is usually denoted by the Greek letter ξ. The extent of a reaction has units of amount (moles). It was introduced by the Belgian scientist Théophile de Donder.


Consider the reaction

A ⇌ B

Suppose an infinitesimal amount dξ of the reactant A that changes into B. The change of the amount of A can be represented by the equation dnA = – dξ and the change of B is dnB = dξ.[1] The extent of reaction is then defined as[2][3]

 d\xi= \frac{dn_i}{\nu_i}

where n_i denotes the amount of the i-th reactant and \nu_i is the stoichiometric number of the i-th reactant. In other words, it is the amount of substance that is being changed in an equilibrium reaction. Considering finite changes instead of infinitesimal changes, one can write the equation for the extent of a reaction as

\Delta \xi=\frac{\Delta n_i}{\nu_i}

The extent of a reaction is defined as zero at the beginning of the reaction. Thus the change of ξ is the extent itself.

\xi=\frac{\Delta n_i}{\nu_i}=\frac{n_{equilibrium}-n_{initial}}{\nu_i}


The relation between the change in Gibbs reaction energy and Gibbs energy can be defined as the slope of the Gibbs energy plotted against the extent of reaction at constant pressure and temperature.[1]

\Delta _r G=\left (\frac{\partial G}{\partial \xi}\right )_{p,T}

Analogously, the relation between the change in reaction enthalpy and enthalpy can be defined.[4]

\Delta _r H=\left (\frac{\partial H}{\partial \xi}\right )_{p,T}


The extent of reaction is a useful quantity in computations with equilibrium reactions. Let us consider the reaction

2A ⇌ B + 3 C

where the initial amounts are n_A = 2 mol , n_B=1mol , n_C=0 mol, and the equilibrium amount of A is 0.5 . We can calculate the extent of reaction from its definition

\xi=\frac{\Delta n_A}{\nu_A}=\frac{0.5-2}{-2}=0.75

Do not forget that the stoichiometric number of reactants is negative. Now when we know the extent, we can rearrange the equation and calculate the equilibrium amounts of B and C.

n_{equilibrium}=\xi \nu_i+n_{initial}
n_{B}=0.75*1+1=1.75 mol
n_{C}=0.75*3+0=2.25 mol


  1. ^ a b Atkins, Peter; de Paula, Julio (2006). Physical chemistry (8 ed.). p. 201. ISBN 0-7167-8759-8. 
  2. ^ Lisý, Ján Mikuláš; Valko, Ladislav (1979). Príklady a úlohy z fyzikálnej chémie. p. 593. 
  3. ^ Ulický, Ladislav (1983). Chemický náučný slovník. p. 313. 
  4. ^ Lisý, Ján Mikuláš; Valko, Ladislav (1979). Príklady a úlohy z fyzikálnej chémie. p. 593.