The enthalpy of combustion of butane C4H10 is described by the reaction:
C4H10(g) + (13/2) O2(g) -> 4CO2(g) + 5H2O(g) ΔH°rxn = –2613 kJ/mol
Given the following enthalpies of formation:
ΔH°f[CO2(g)] = -393.5 kJ/mol
ΔH°f[H2O(g)] = -241.8 kJ/mol
Calculate the value of enthalpy of formation of butane from the data given.
dHrxn = [(n*dHfo CO2 + (n*dHfo H2O]-[n*dHfo C4H10]
You know all of the numbers except for dHo C4H10. Substitute and solve for that.
To calculate the enthalpy of formation of butane (ΔH°f[C4H10]), we can use the equation:
ΔH°rxn = ΣΔH°f[products] - ΣΔH°f[reactants]
First, let's calculate the enthalpy change for the products:
4CO2(g): (4 mol) x (-393.5 kJ/mol) = -1574 kJ/mol
5H2O(g): (5 mol) x (-241.8 kJ/mol) = -1209 kJ/mol
Now, let's calculate the enthalpy change for the reactant:
C4H10(g): (1 mol) x ΔH°f[C4H10]
Substituting the values into the equation:
-2613 kJ/mol = (-1574 kJ/mol) + (-1209 kJ/mol) + (ΔH°f[C4H10])
Rearranging the equation to solve for ΔH°f[C4H10]:
ΔH°f[C4H10] = -2613 kJ/mol + 1574 kJ/mol + 1209 kJ/mol
= -830 kJ/mol
Therefore, the value of the enthalpy of formation of butane (ΔH°f[C4H10]) is approximately -830 kJ/mol.
To calculate the enthalpy of formation of butane (C4H10), we need to use the equation for the reaction and the enthalpies of formation of carbon dioxide (CO2) and water (H2O) provided.
The reaction equation is:
C4H10(g) + (13/2) O2(g) -> 4CO2(g) + 5H2O(g)
From this equation, we can see that 4 moles of CO2 and 5 moles of H2O are produced for every mole of butane burned.
The enthalpy change of the reaction (ΔH°rxn) is given as -2613 kJ/mol.
We can use the enthalpies of formation of CO2 and H2O to calculate the amount of energy released or absorbed by the reaction.
The enthalpy change of the reaction (∆H°rxn) is equal to the sum of the enthalpies of formation of the products (CO2 and H2O) minus the sum of the enthalpies of formation of the reactant (butane):
ΔH°rxn = Σ(∆H°f, products) - Σ(∆H°f, reactants)
Using the given values:
ΔH°f[CO2(g)] = -393.5 kJ/mol
ΔH°f[H2O(g)] = -241.8 kJ/mol
We can substitute these values into the equation:
-2613 kJ/mol = (4 * ΔH°f[CO2(g)]) + (5 * ΔH°f[H2O(g)]) - ∆H°f[C4H10(g)]
Now, let's rearrange the equation to solve for the enthalpy of formation of butane (∆H°f[C4H10(g)]):
∆H°f[C4H10(g)] = (4 * ΔH°f[CO2(g)]) + (5 * ΔH°f[H2O(g)]) - 2613 kJ/mol
Plugging in the values:
∆H°f[C4H10(g)] = (4 * -393.5 kJ/mol) + (5 * -241.8 kJ/mol) - 2613 kJ/mol
Calculating the value:
∆H°f[C4H10(g)] = -1574 kJ/mol + (-1209 kJ/mol) - 2613 kJ/mol
∆H°f[C4H10(g)] = -1396 kJ/mol
Therefore, the value of the enthalpy of formation of butane (C4H10) is -1396 kJ/mol.