Hi, I am doing a practice test for my chemistry final and I do not understand one of the answer they gave for this thermochemical equation question. I do not understand how they got 400kJ.
The question is: Butane has a deltaH of -2871 kJ/mol when it undergoes complete combustion into carbon dioxide and liquid water. Given the following deltaHf values:
Carbon dioxide = -393.5kJ/mol
Oxygen gas = 0kJ/mol
liquid water = -285.8 kJ/mol
a) write the balanced thermochemical equation for the complete combustion of 1 mol of butane.
The answer given is: C4H10 =6.5O2 ---> 4Co2 + 5H20 + 400kJ.
I just don't understand how they calculated the kJ to be 400. Any help is greatly appreciated.
You need the delta Hformation of C4H10.
Then
dHcomb = [dHproducts]-[dHreactants]
dHcomb = [(4*-393.5) + (5*-285.8)]-[(dHf C4H10)] = ?
Thank you, so when I did that I got -137kJ. so is their answer wrong or mine. They had 400kJ.
To understand how the answer of 400 kJ was obtained, let's break down the process step by step.
First, we need to write the balanced thermochemical equation for the complete combustion of 1 mol of butane. The chemical formula for butane is C4H10. The products of combustion are carbon dioxide (CO2) and water (H2O).
The balanced equation looks like this:
C4H10 + O2 -> 4CO2 + 5H2O
Now, let's consider the enthalpies of formation (deltaHf) of the given substances:
DeltaHf of CO2 = -393.5 kJ/mol
DeltaHf of O2 = 0 kJ/mol (since it is in its elemental form)
DeltaHf of H2O = -285.8 kJ/mol
We can use these values to calculate the deltaH of the reaction, which represents the change in enthalpy for the reaction.
DeltaH = (4 * DeltaHf of CO2) + (5 * DeltaHf of H2O) - (DeltaHf of C4H10) - (6.5 * DeltaHf of O2)
Plugging in the values we have:
DeltaH = (4 * -393.5 kJ/mol) + (5 * -285.8 kJ/mol) - (-2871 kJ/mol) - (6.5 * 0 kJ/mol)
Simplifying the equation gives:
DeltaH = -1574 kJ/mol + (-1429 kJ/mol) + 2871 kJ/mol
DeltaH = -132 kJ/mol
So, the calculated deltaH is -132 kJ/mol.
However, the given thermochemical equation in the answer suggests that the deltaH is positive (400 kJ), indicating that energy is released (exothermic) during the combustion reaction. It is likely that the given deltaH value of -2871 kJ/mol for butane is actually an enthalpy of combustion, which is the negative of the enthalpy of formation.
To obtain the positive value, we need to multiply the deltaH value by -1:
DeltaH = -(-2871 kJ/mol)
DeltaH = 2871 kJ/mol
Therefore, the correct balanced thermochemical equation for the complete combustion of 1 mol of butane is:
C4H10 + 13/2 O2 -> 4CO2 + 5H2O + 2871 kJ