chemical equilibrium reaction for burned gases in engine I do not understand how equation(3) occurs.(taken from the book Internal Combustion Engines by John B.Heywood)
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Consider a reactive mixture of ideal gases. The reactant species $M_a,M_b$ etc.and the product species $M_l,M_m$ etc. are related by the general reaction whose stoichiometry is given by:
$$\upsilon _aM_a+\upsilon _bM_b+...=\upsilon _lM_l+\upsilon _mM_m+...$$
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This can be otherwise written as:
$$ \sum_{i}\upsilon _iM_i=0$$
where the $\upsilon _i$ are the stoichiometric coefficients and by convention are positive for the product species and negative for the reactant species.

Let $\delta n_a$ of $M_a$ react with $\delta n_b$ of $M_b$,etc. and produce $\delta n_l$ of $M_l$,$\delta n_m$ of $M_m$,etc.. 

These amounts are in proportion,given by the equation (3):

(3) $$\delta n_i=\upsilon _i\delta n$$              

1.Does $\delta n$ here signify 'extent of reaction'?

2.I would eventually like to use the number of moles of each species in expressing the chemical potential,with the gibbs free energy already known. But,the part I don't understant is equation 3. What does the author mean by proportional? Could you give a example?

Gibbs free energy is given as:
$$(\Delta G)_(pressure,temper_)=\sum_{i}\mu _i\delta n_i$$

which by equation(3) can be re-written as(WHY??):

$$(\Delta G)_(pressure,temper_)=\sum_{i}\mu _i\nu _i\delta n$$
It's simpler than it might look. As a random example:

$$\ce{3A + 2B-&gt;2C + 4D}$$

If I were to ask you how much of $\ce{D}$ would be produced after $5\  mol$ of $\ce{B}$ were consumed, then you would immediately answer $10\  mol$ "due to the stoichiometry". In more elaborate terms, for a $\Delta n_B=n_B^{after}-n_B^{before}=-5\ mol$, one has $\Delta n_D=n_D^{after}-n_D^{before}=+10\ mol$. Any infinitesimal change in quantity of any species $i$ ($\delta n_i$) is accompanied by a shift in the quantities of all other species precisely determined by the stoichiometry of the reaction (the stoichiometric coefficients, $v_i$), such that $\frac{\delta n_i}{v_i}$ is constant for all species $i$. For the above reaction, $-\frac{\delta n_A}{3}=-\frac{\delta n_B}{2}=\frac{\delta n_C}{2}=\frac{\delta n_D}{4}$. The ratio $\frac{\delta n_i}{v_i}$ is defined as $\delta n$, and you can think of it as a counter for "how many times the reaction happened" ($\delta n&gt;0$ if left to right, $\delta n&lt;0$ if right to left).

$\delta n$ is a general change in amount of substances, but the particular case where changes are calculated relative to the equilibrium condition ($\Delta n=n^{equilibrium}-n^{before}$), then it represents the extent of reaction.