Editing Chain reaction (section)

===Acetaldehyde pyrolysis and rate equation===

The [[pyrolysis]] (thermal decomposition) of [[acetaldehyde]], CH<sub>3</sub>CHO (g) → CH<sub>4</sub> (g) + CO (g), proceeds via the Rice-Herzfeld mechanism:<ref name=LM>{{cite book |last1=Laidler |first1=Keith J. |last2=Meiser |first2=John H. |title=Physical Chemistry |date=1982 |publisher=Benjamin/Cummings |isbn=0-8053-5682-7 |page=417}}</ref><ref>{{cite book |last1=Atkins |first1=Peter |last2=de Paula |first2=Julio |title=Atkins' Physical Chemistry |date=2006 |publisher=W. H. Freeman |isbn=0-7167-8759-8 |pages=830–1 |edition=8th}}</ref>

*Initiation (formation of [[Radical (chemistry)|free radicals]]):

: CH<sub>3</sub>CHO (g) → •CH<sub>3</sub> (g) + •CHO (g)	              k<sub>1</sub>

The methyl and CHO groups are [[Radical (chemistry)|free radicals]].

*Propagation (two steps):

: •CH<sub>3</sub> (g) + CH<sub>3</sub>CHO (g) →  CH<sub>4</sub> (g) + •CH<sub>3</sub>CO (g)            	k<sub>2</sub>

This reaction step provides [[methane]], which is one of the two main products.

: •CH<sub>3</sub>CO (g) → CO (g) + •CH<sub>3</sub> (g)		        k<sub>3</sub>

The product  •CH<sub>3</sub>CO (g) of the previous step gives rise to [[carbon monoxide]] (CO), which is the second main product.

The sum of the two propagation steps corresponds to the overall reaction CH<sub>3</sub>CHO (g) → CH<sub>4</sub> (g) + CO (g), [[Catalysis|catalyzed]] by a methyl radical •CH<sub>3</sub>.

*Termination:

: •CH<sub>3</sub> (g) + •CH<sub>3</sub> (g) → C<sub>2</sub>H<sub>6</sub> (g)              	k<sub>4</sub>

	This reaction is the only source of [[ethane]] (minor product) and it is concluded to be the main chain ending step.

Although this mechanism explains the principal products, there are others that are formed in a minor degree, such as [[acetone]] (CH<sub>3</sub>COCH<sub>3</sub>) and [[Propionaldehyde|propanal]] (CH<sub>3</sub>CH<sub>2</sub>CHO).

Applying the [[Steady state (chemistry)|Steady State Approximation]] for the intermediate species CH<sub>3</sub>(g) and CH<sub>3</sub>CO(g), the rate law for the formation of methane and the order of reaction are found:<ref name=LM/><ref name=Atkins/>

The rate of formation of the product methane is

<math chem>(1)... \frac{d\ce{[CH4]}}{dt} = k_2\ce{[CH3]} \ce{[CH3CHO]}</math>

For the intermediates

<math chem>(2)... \frac{d\ce{[CH_3]}}{dt} = k_1 \ce{[CH3CHO]} - k_2 \ce{[CH3]} \ce{[CH3CHO]} + k_3 \ce{[CH3CO]} - 2 k_4 \ce{[CH3]}^2 = 0</math> and

<math chem>(3)... \frac{d\ce{[CH3CO]}}{dt} = k_2 \ce{[CH3]} \ce{[CH3CHO]} - k_3 \ce{[CH3CO]} = 0</math>

Adding (2) and (3), we obtain <math chem>k_1 \ce{[CH3CHO]} - 2 k_4 \ce{[CH3]}^2 = 0</math>

so that <math chem>(4)...\ce{[CH3]} = \frac{k_1}{2k_4}\ce{[CH3CHO]}^{1/2}</math>

Using (4) in (1) gives the rate law <math chem>(5) \frac{d\ce{[CH4]}}{dt} = \frac{k_1}{2k_4} k_2 \ce{[CH3CHO]}^{3/2}</math>, which is order 3/2 in the reactant CH<sub>3</sub>CHO.