posted by jan .
I refer to the following question. Is my explanation correct, that step two
C6H14(g)--> 6CO2(g) + 7H2O(g)
can be considered combustion reaction?
Posted by jan on Tuesday, March 31, 2009 at 10:28pm.
C6H14(l)->C6H14(g) ->6CO2(g) + 7H2O(g) ->6CO2(g) + 7H2O(l)
Hexane is a liquid at room temp. It can be converted into carbon dioxide and water through the above processes. There are 3 steps. Steps 2 & 3 are exothermic, right? I think step 2
C6H14(g)->6CO2(g) + 7H2O(g) is something like combustion so it is exothermic.Please help.
Chemistry pls help - bobpursley, Tuesday, March 31, 2009 at 10:50pm
Chemistry pls help - DrBob222, Tuesday, March 31, 2009 at 10:56pm
But step 1 is endothermic.
You have omitted the oxygen that is responsible for combustion. If you will copy this as three steps and not run them together you will be less likely to omit reacts you need PLUS you will not confuse products with reactants.
C6H14(l) + heat ->C6H14(g) (step 1)
2C6H14(g) + 19O2 ->12CO2(g) + 14H2O(g)+ heat (step 2)
14H2O(g) ==> 14H2O(l) + heat (step 3)
Steps 2 and 3 are exothermic as you suggest. Step 2 is a combustion reaction. Step 1 is endothermic.