Enthalpy of Combustion -- Alkanes Write a balanced equation for the combustion o

Question

Enthalpy of Combustion -- Alkanes

Write a balanced equation for the combustion of CH3C8H17(l) (methyl octane)
-- i.e. its reaction with O2(g) forming the products CO2(g) and H2O(l).

I got the balance equation CH3C8H17 + 14O2 -> 9 CO2 + 10H2O

Question: Calculate the difference, H-E=(PV) for the combustion reaction of 1 mole of methyl octane.
(Assume standard state conditions and 298 K for all reactants and products.)

My Answer: n = 9 + 10 -14 = 5

And use (PV) = nRT = 5RT = 5 * 8.3145 * 298 = 12338J

But when I submit my answer, it says 12338 is wrong. Could someone tell me why my method is wrong?

Explanation / Answer

Combustion of alkanes gives CO2 (g) + H2O(l) whether it is liquid or gas.

But not H2O (l).

So delta n = 9-14= - 5 since we will consider only gaseous reactants and products.

So difference = - 5 x 8.3145 x 298 = - 12338 J

Important thing is that combustion of alkanes is an exothermic reaction, means that it releases heat. Negative sign indicates that the release of energy.

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