The heat of combustion of liquid hexane (C6H14) to carbon dioxide and liquid water at 298 K is -4215 kJmol−1 .
Find ΔrU of the reaction.
dU = q + work
q = -4215 kJ/mol x 2 mol = ?
2C6H14 + 19O2 ==> 12CO2 + 14H2O
work is pdV. I assume p = 1 atm. dV is -7 mol x 22.4 L/mol = ?
Since the volume is contracting then the surroundings are doing work on the system; therefore, work will be negative. Substitute and solve.
Oh, hexane, you fiery little molecule! ΔrU, huh? Well, let me crunch some numbers and take a wild guess here. ΔrU, the change in internal energy, is -4215 kJ/mol. So, if we were to put hexane on stage with carbon dioxide and water, it's like watching a transformation, from a combustible liquid to some CO2 and H2O. And this transformation releases an impressive amount of heat. Now, to find ΔrU, we simply multiply -4215 kJ/mol by the number of moles of hexane involved in the reaction. Voilà! A numerical representation of the energetic shenanigans going on inside that reaction.
To find the change in internal energy (ΔrU) of the reaction, we can use the equation:
ΔrU = Σ(nU(products)) - Σ(nU(reactants))
First, we need to determine the number of moles of hexane (C6H14) in the balanced chemical equation for its combustion to carbon dioxide (CO2) and water (H2O).
The balanced equation for the combustion of hexane is:
C6H14 + 19/2 O2 -> 6 CO2 + 7 H2O
From the balanced equation, we can see that 1 mole of hexane (C6H14) produces 6 moles of CO2 and 7 moles of H2O.
Next, let's calculate the energy change associated with the products:
nU(products) = (6 mol CO2) * U(CO2) + (7 mol H2O) * U(H2O)
Since the given value, the heat of combustion of hexane, is in kJ/mol, ΔrU will also be in kJ.
From tables, the standard molar enthalpies of formation for carbon dioxide (U(CO2)) and water (U(H2O)) are:
U(CO2) = -393.5 kJ/mol
U(H2O) = -285.8 kJ/mol
Substituting these values into the equation:
nU(products) = (6 mol CO2) * (-393.5 kJ/mol) + (7 mol H2O) * (-285.8 kJ/mol)
Next, let's calculate the energy change associated with the reactants:
nU(reactants) = 1 mol C6H14 * U(C6H14)
From tables, the standard molar enthalpy of formation for hexane (U(C6H14)) is:
U(C6H14) = 0 kJ/mol
Substituting the values into the equation:
nU(reactants) = (1 mol C6H14) * (0 kJ/mol)
Now, substitute the values of nU(products) and nU(reactants) into the equation for ΔrU:
ΔrU = [(6 mol CO2) * (-393.5 kJ/mol) + (7 mol H2O) * (-285.8 kJ/mol)] - [(1 mol C6H14) * (0 kJ/mol)]
Calculating ΔrU:
ΔrU = (-2361 kJ/mol) + (-2000.6 kJ/mol)
ΔrU = -4361.6 kJ/mol
Therefore, the change in internal energy (ΔrU) of the reaction is -4361.6 kJ/mol.
To find ΔrU (change in internal energy) of the reaction, we need to use the equation:
ΔU = Σ(ni * ΔUi)
where ΔU is the change in internal energy, ni is the stoichiometric coefficient (number of moles) of the reactant or product, and ΔUi is the molar internal energy change for each species.
Let's first determine the stoichiometric coefficients for the reaction:
C6H14 + 19/2 O2 -> 6 CO2 + 7 H2O
From the balanced equation, we can see that the stoichiometric coefficient of hexane (C6H14) is 1, while the stoichiometric coefficients for carbon dioxide (CO2) and water (H2O) are 6 and 7, respectively.
Now we can calculate the ΔrU using the given ΔU values:
ΔrU = (nCO2 * ΔUCO2) + (nH2O * ΔUH2O) - (nC6H14 * ΔUC6H14)
ΔrU = (6 * ΔUCO2) + (7 * ΔUH2O) - (1 * ΔUC6H14)
To find the molar internal energy change for each species, we multiply the given ΔU value (in kJ/mol) by the stoichiometric coefficient:
ΔUCO2 = (-4215 kJ/mol) * 6
ΔUH2O = (-4215 kJ/mol) * 7
ΔUC6H14 = (-4215 kJ/mol) * 1
Now we can substitute the values into the equation:
ΔrU = (6 * (-4215 kJ/mol)) + (7 * (-4215 kJ/mol)) - (1 * (-4215 kJ/mol))
Calculating the expression:
ΔrU = -25380 kJ/mol + (-29505 kJ/mol) + (4215 kJ/mol)
ΔrU = -50670 kJ/mol + 4215 kJ/mol
ΔrU = -46455 kJ/mol
Therefore, the ΔrU of the reaction is -46455 kJ/mol.