In an experiment liquid hexane, C6H14 (l) is combusted completely to produce CO2(g) and H2O (l) according to the equation:
2 C6H14 (l) + 19 O2(g) -> 12 CO2(g) + 14H2O (l)
The change in H degrees for the reaction is -8.39 x 10^3 KJ. Calculate change in H degrees for C6H14 (l).
AAARRRGGGG. H with a little circle at the top right is NOT degrees.
I will shift to more conventional symbols
Hr (heat of reaction,or reaction enthalpy)= -8.39E3 KJ
Hcombustion= -4.25 kJ/mole
Notice the units for heat of combustion. It is commonly on a per mole basis, but can be on a per mass, or per volume basis. Heats of reaction are usually in heat units, based on the reaction, but it is not unusual to find them in a per mole or per gram basis. In this instance, I assumed from the wording it was based on the amounts in moles of the coefficents. Better wording would have been desired.
To calculate the change in enthalpy for C6H14 (l) in this reaction, we will use the principle of Hess's Law. Hess's Law states that the change in enthalpy of a reaction can be calculated by taking the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants.
In this case, we need to find the enthalpy change for the combustion of C6H14 (l). We are given the enthalpy change for the overall reaction, which is -8.39 x 10^3 KJ.
To find the change in enthalpy for C6H14 (l), we need to consider the stoichiometry of the reaction. From the balanced equation, we can see that the coefficient of C6H14 (l) is 2.
Therefore, the change in enthalpy for C6H14 (l) can be calculated as follows:
Change in H degrees for C6H14 (l) = (Change in H degrees for the overall reaction) / (Coefficient of C6H14 (l))
= (-8.39 x 10^3 KJ) / (2)
= -4.195 x 10^3 KJ
So, the change in enthalpy for C6H14 (l) in this reaction is -4.195 x 10^3 KJ.