Here is a test problem a little larger than the example given in the
distribution for REACT.
============================= problem.m===================================
 .2500e-05	no2m		->	no + om
 .5556e+10	om		->	oh + ohm
 .1000e+10	oh + ohm	->	om
 .1000e+10	oh + no2m	->	no2 + ohm
 .2500e+09	om + no2m	->	no2 + ohm + ohm
 .2500e+10	om + o2		->	o3m
 .5500e+04	o3m 		->	o2 + om
 .1000e+09	no2 + no2	->	n2o4
 .1300e+04	n2o4		->	no2 + no2
 .1200e+04	n2o4		->	no2m + no3m
 .3000e+03	no2		->	x
 .1000e+09	no2 + x		->	no2m + no3m
 .1000e+09	no2 + no	->	n2o3
 .2000e+04	n2o3		->	no2 + no
 .2000e+04	n2o3		->	no2m + no2m
 .1000e-00	x + no		->	no2m + no2m
============================= problem.m===================================
and here are the parameters.
============================= problem.p===================================
0 .36e+04 .06e+04 1.0e-03
no2m	0.1e-0	1.0e-8	1.0e-03
o2	.3e-03	1.0e-8	1.0e-03
ohm	.1e-06	1.0e-8 1.0e-03
============================= problem.p===================================
The solution is found by
...............................................................
%react problem
1)  .2500e-05	no2m		->	no + om
2)  .5556e+10	om		->	oh + ohm
3)  .1000e+10	oh + ohm	->	om
4)  .1000e+10	oh + no2m	->	no2 + ohm
5)  .2500e+09	om + no2m	->	no2 + ohm + ohm
6)  .2500e+10	om + o2		->	o3m
7)  .5500e+04	o3m 		->	o2 + om
8)  .1000e+09	no2 + no2	->	n2o4
9)  .1300e+04	n2o4		->	no2 + no2
10)  .1200e+04	n2o4		->	no2m + no3m
11)  .3000e+03	no2		->	x
12)  .1000e+09	no2 + x		->	no2m + no3m
13)  .1000e+09	no2 + no	->	n2o3
14)  .2000e+04	n2o3		->	no2 + no
15)  .2000e+04	n2o3		->	no2m + no2m
16)  .1000e-00	x + no		->	no2m + no2m
...............................................................
6.340u 0.340s 0:12.26 54% 0+22k 0+2io 0pf+0w
(6.34 seconds run time)

I can look at the results using select (here I have a file problem.e
with only the words 
no2 no no3m
rather than everything to reduce the amount of output)
...............................................................
% select no2aq
16 reacions involving 12 species
no2m  no  om  oh  ohm  no2  o2  o3m  n2o4  no3m  x  n2o3 
       time            no2             no           no3m
     0.0000e+00      0.0000e+00      0.0000e+00      0.0000e+00 
     6.0000e+02      8.5578e-11      4.5160e-05      2.1075e-05 
     1.2000e+03      6.7257e-11      6.5957e-05      3.1552e-05 
     1.8000e+03      5.2981e-11      8.3997e-05      4.0644e-05 
     2.4000e+03      4.6293e-11      9.9261e-05      4.8192e-05 
     3.0000e+03      4.1795e-11      1.1373e-04      5.5295e-05 
     3.6000e+03      3.8639e-11      1.2502e-04      6.0946e-05 
...............................................................

Now how do I know that the tolerance values are ok?  I could rerun
with smaller values and look for an effect. But in this case I was
carefull to write reactions conserving all 'n' containing species (not
so with o an h, in this sample I make OH- readily (ohm) but not H+
where I should!). If I add up all the 'n's in the result I should get
what I started with (0.10 == no2m initial, see problem.p).  Here is
the problem.e file to extract all the species with 'n's:

=============================problem.e===================================
 no2m  no  no2  n2o4 n2o4  no3m  x  n2o3 n2o3
=============================problem.e===================================

notice n2o4 and n2o3 are in the list twice so they get counted twice!
I just pipe the output to Gary Pearlman's dm to add them up!
(recall that the first column is time)

%extract problem | dm ' x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 '

to get:
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1

You could also use an awk script

% extract problem | awk '{a= $2+$3+$4+$5+$6+$7+$8+$9+$10; print a}'
.1
.1
.1
.1
.1
.1
.1
...............................................................


You may have noticed that since I use '-' as a token in the reaction
arrow I designate charge with a lowercase m or p appended to the name.

Tokens parsed as word delimeters in the mechanism file are:
	"<blank>$<TAB>,+;:(){}[]"  and "->"

That is 
	[CH3] + (CH2) -> {stuff}
is equivalent to
	CH3 + CH2 -> stuff

In a future release I may remove the "(){}[]" from the token list after
I decide what if anything I want to do with automated invocation of
the steady state approximation (I was thinking about enclosing
variables in {} if they could be approximated by ss).

___________________________________________________________
|Mike Whitbeck             | whitbeck@unssun.unr.edu      |
|Desert Research Inst.     | whitbeck@wheeler.wrc.unr.edu |
|POB 60220                 | whitbeck@sanjuan.UUCP        |
|RENO, NV 89506            | 702-673-7348                 |
|__________________________|______________________________|

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