From: zenadsl6186@zen.co.uk   
      
   iain-3@truecircuits.com wrote:   
      
   > If I do the simple heat of formation at 0 K equation for a   
   > hydrogen-oxygen rocket, I get:   
   >   
   > H2 + 0.5*O2 -> H20 + 238.9 kJ/mol  (= 13.27 kJ/g)   
   >   
   > If I do the same equation for a carbon monoxide rocket, I get:   
   >   
   > C + 0.5*O2 -> CO + 825.0 kJ/mol    (= 29.45 kJ/g)   
      
   Wrong ballpark. The standard enthalpy of formation at ntp of CO is -110.53   
   kJ/mol (water vapour is -241.82 kJ/mol).   
      
      
   --   
   Peter Fairbrother   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)