Octane (C8H18) undergoes combustion according to the following thermochemical equation:
2 C8H18(l) + 25 O2(g) ---> 16 CO2(g) + 18 H2O(l) DHo = -11,200 kJ
Given : DHfo [CO2(g)] = -393.5 kJ/mol and DHfo[H2O(l)] = -285.8 kJ/mol.
Calculate the enthalpy of formation of one mole of liquid octane.
1) -6296 kJ
2. ) -120 kJ
3. ) -11,440 kJ
4. ) -5144 kJ
5. ) -240 kJ
I calculated 2 to be correct is it?
Since the heat of formation for 2 moles of octane is 11,200, you divide by 2 to get 5600kJ for 1 mole. According to the equation above, 1 mole of octane will give you 8 moles of CO2 and 9 moles of H2O. The heat of formation for O2 is zero because it naturally exists and has a bond energy of 0. Recall that heat of formation is=
heat of formation of products - heat of formation of reactants
So you get:
[8(393.5) + 9(285.8)] - [enthalpy of reaction]=-240kJ / 2 =>
The answer is b) -120kJ
Octane (C8H18) undergoes combustion according to the following equation. How many grams of octane are needed to produce 9620.5 kJ of heat?(Given M(C8H18) = 114 g/mol) %3D
To find the enthalpy of formation of one mole of liquid octane, we need to use the given thermochemical equation and the enthalpy of formation values for carbon dioxide and water.
First, let's determine the total change in enthalpy (ΔH) for the combustion reaction:
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(l)
ΔH = -11,200 kJ
Now, let's calculate the enthalpy change for the formation of one mole of liquid octane (C8H18):
8 C8H18(l) + 25 O2(g) → ΔH1 (formation of octane)
16 CO2(g) + 18 H2O(l) → ΔH2 (formation of CO2 and H2O)
Since we're interested in the formation of one mole of octane, we'll set the coefficients of all other substances to 1:
C8H18(l) + 25/8 O2(g) → ΔH1/8 (formation of octane)
CO2(g) + H2O(l) → ΔH2 (formation of CO2 and H2O)
Now, let's rearrange the equation and solve for ΔH1/8:
ΔH = -11,200 kJ = ΔH1/8 - 25/8 ΔH2
ΔH1/8 = -11,200 kJ + 25/8 ΔH2
Next, let's substitute the enthalpy of formation values:
ΔH1/8 = -11,200 kJ + 25/8 (-393.5 kJ/mol + 18 (-285.8 kJ/mol))
ΔH1/8 = -11,200 kJ + 25/8 (-393.5 kJ/mol + (-5144.4 kJ/mol))
ΔH1/8 = -11,200 kJ + 25/8 (-5651.9 kJ/mol)
ΔH1/8 = -11,200 kJ + 25/8 (-5651.9 kJ/mol)
ΔH1/8 = -11,200 kJ - 25/8 (5651.9 kJ/mol)
ΔH1/8 = -11,200 kJ - 17,651.87 kJ/mol
ΔH1/8 = -28,851.87 kJ/mol
Therefore, the enthalpy of formation of one mole of liquid octane is approximately -28,852 kJ/mol.
None of the answer choices given match this calculation, so it seems there might be a mistake in the provided answer choices.
To calculate the enthalpy of formation of one mole of liquid octane (C8H18), you will need to use the enthalpy change of the combustion reaction and the enthalpies of formation of the products and reactants.
First, let's identify the stoichiometric coefficients of octane in the balanced combustion equation. In the equation:
2 C8H18(l) + 25 O2(g) ---> 16 CO2(g) + 18 H2O(l)
The coefficient of octane (C8H18) is 2.
The enthalpy change of the combustion reaction (ΔH°) is given as -11,200 kJ. This value represents the enthalpy change for the combustion of 2 moles of octane. Therefore, we need to divide this value by 2 to find the enthalpy change for the combustion of one mole of octane.
ΔH° for 1 mole of octane = (-11,200 kJ) / 2 = -5,600 kJ/mol.
To calculate the enthalpy of formation of one mole of liquid octane (ΔH°f[C8H18]), we need to consider the enthalpies of formation of the products and reactants involved in the combustion reaction.
The enthalpies of formation for CO2(g) and H2O(l) are given as follows:
ΔH°f[CO2(g)] = -393.5 kJ/mol
ΔH°f[H2O(l)] = -285.8 kJ/mol
Using these values, we can calculate the enthalpy of formation of octane:
ΔH°f[C8H18] = (16 × ΔH°f[CO2(g)]) + (18 × ΔH°f[H2O(l)]) - ΔH° for 1 mole of octane
Plugging in the values:
ΔH°f[C8H18] = (16 × -393.5 kJ/mol) + (18 × -285.8 kJ/mol) - (-5,600 kJ/mol)
ΔH°f[C8H18] = -6296 kJ/mol
Therefore, the enthalpy of formation for one mole of liquid octane is -6296 kJ/mol.
So, the correct answer is 1) -6296 kJ.