Advance of reaction

Advance of reaction

The advance of reaction (initially introduced by Theophilus de Donder under the name of " degree of avancement") allows to characterize the progress report of a reaction between its initial state (before reaction) and its final state (after reaction). It is noted by the Greek letter ξ (Xi) and is defined by: $\backslash \; xi\; =\; \backslash \; frac\; \{(n\_i-n\_\; \{I,\; 0\})\}\{\backslash \; naked\; \_i\}$ in which: $n\_\; \{I,\; 0\}\; ~$ is the initial quantity of the substance $i~$; $n\_\; \{I\}\; ~$ being its quantity in the state of advance considered and $\backslash \; naked\; \_i~$ its stoechiometric coefficient whose sign is negative if the component $i~$ is a reagent starting and positive if it is a product of the reaction.

The advance of reaction is expressed in moles. Before reaction: $\backslash \; xi\; =\; 0\; ~$

Its value depends on the manner of writing the stoechiometric equation of the reaction (or equation assessment), it is thus necessary always to indicate the latter. But it is independent of the substance chosen to calculate it.

A more general definition is given by: $\backslash \; Delta\; \backslash \; xi\; =\; \backslash \; frac\; \{\backslash \; naked\; Delta\; n\_i\}\; \{\backslash \; \_i\}$. For an elementary variation one a: $D\; \backslash \; xi\; =\; \backslash \; frac\; \{D\; n\_i\}\; \{\backslash \; naked\; \_i\}$.

Examples of calculation

Let us consider the reaction of equation-assessment: 2:00 ₂ + O₂ = 2:00 ₂O with the following iniales quantities: 2 moles of H₂; 1 moles of O₂ and 0 moles of H₂O

When 1 mole of H₂ is consumed, there will remain 0,5 moles of O₂ and it will have been formed 1 mole of H₂O

The advance of the reaction can be calculated indifferently compared to any of the substances concerned in the reaction.

Compared to H₂: $\backslash \; xi\; =\; \backslash \; frac\; \{1\; -\; 2\}\; \{-\; 2\}\; =\; 0,5$ mol

Compared to O₂: $\backslash \; xi\; =\; \backslash \; frac\; \{0,5\; -\; 1\}\; \{-\; 1\}\; =\; 0,5$ mol

Compared to H₂O: $\backslash \; xi\; =\; \backslash \; frac\; \{1\; -\; 0\}\; \{2\}\; =\; 0,5$ mol

On the other hand, if one had written: H₂ + 1/2O₂ = H₂O and with the same initial and transformed quantities one would have:

Compared to H₂: $\backslash \; xi\; =\; \backslash \; frac\; \{1\; -\; 2\}\; \{-\; 1\}\; =\; 1$ mol

Compared to O₂: $\backslash \; xi\; =\; \backslash \; frac\; \{0,5\; -\; 1\}\; \{-\; 1/2\}\; =\; 1$ mol

Compared to H₂O: $\backslash \; xi\; =\; \backslash \; frac\; \{1\; -\; 0\}\; \{1\}\; =\; 1$ mol

What illustrates that, for same initial and transformed quantities, the value of the advance of reaction depends on the manner of writing the equation assessment.

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