Calculate the standard enthalpy of combustion.

united stars

Using the standard enthalpies of formation below,calculate the standard enthalpy of combustion of hexane C6H14 = -4109 kjmolOL,C02=-394KJMOL, H20 -286Kjmol.

I have not done chemistry before. Can someone explain in laymans term what is happening here and how I do this equation or any useful links that will help me.Thank You.

gerardduff

united stars wrote: »

Using the standard enthalpies of formation below,calculate the standard enthalpy of combustion of hexane C6H14 = -4109 kjmolOL,C02=-394KJMOL, H20 -286Kjmol.

I have not done chemistry before. Can someone explain in laymans term what is happening here and how I do this equation or any useful links that will help me.Thank You.

Neither have I done chemistry before. I have a just a basic understanding of chemistry.

But, I suppose you have to balance the equation. So it looks something like this...

[latex]C_6H_{14}--^{^{Heat}}_{_{oxygen}}-->6CO_2 + 7H_2O[/latex]

So you need to balance

-4109 = x - 6*394 - 7*286

so,

x=+157 kJoules per mol

and the reaction in exothermal....which is pretty handy if you had a few mashmellows you needed toasting.

I just thought about what the simplest reaction I could write down which involved all three compounds in the question. That was easy for me as the easy ones are the only ones I knows.:)

What's happening is you are setting fire to some hydro carbon in the form of Hexane gas. It, like all matter, has an associated internal energy which can be described in a myriad of ways, one of which is as enthalpy. You should think of it as 'potential energy'. Enthalpy is a thermodynamic property which plays the role of internal energy under certain conditions (to do with the speed of the reaction and whether it is confined/constant temp/constant pressure). It is best to read about this say here.

Hope you found this of some help, and if I left something out I'm sorry.

craggles

You're going to need to take oxygen into account on the left side of the arrow as well.