What is the enthalpy change when 100.0 g octane (C8H18, MW = 114.23 g/mol) is combusted at 25C?
2C8H18 + 50O2 --> 16CO2 + 18H2O
dHrxn = (n*dHo products) - (n*dHo reactants)
dHo for CO2 and H2O are available from tables in your text or online. dHo for O2 = 0. Post your work if you get stuck.
To calculate the enthalpy change when octane (C8H18) is combusted, we need to know the balanced chemical equation for the combustion reaction of octane.
The balanced chemical equation for the combustion of octane is:
2C8H18 + 25O2 -> 16CO2 + 18H2O
From the balanced equation, we can see that 2 moles of octane (C8H18) react with 25 moles of oxygen (O2) to produce 16 moles of carbon dioxide (CO2) and 18 moles of water (H2O).
First, let's calculate the number of moles of octane (C8H18) in 100.0 g:
Molar mass of octane (C8H18) = (12.01 g/mol × 8) + (1.01 g/mol × 18) = 114.23 g/mol
Number of moles = mass / molar mass = 100.0 g / 114.23 g/mol = 0.875 mol
Next, let's determine the enthalpy change (ΔH) for the combustion of 1 mole of octane.
From the balanced equation, we can see that the combustion of 2 moles of octane produces 16 moles of carbon dioxide and 18 moles of water. We can refer to a thermochemical table to find the enthalpy change for the combustion of 1 mole of octane.
Assuming the standard enthalpy change of combustion, ΔH°comb = -5470 kJ/mol, we can use stoichiometry to calculate the enthalpy change for 0.875 moles of octane:
ΔH = ΔH°comb × (0.875 mol / 2 mol) = -5470 kJ/mol × 0.875 mol / 2 mol = -2391.87 kJ
Therefore, the enthalpy change when 100.0 g of octane is combusted is approximately -2391.87 kJ.
To find the enthalpy change when octane is combusted, you need to use the balanced chemical equation for the combustion of octane and the standard enthalpy of combustion.
The balanced chemical equation for the combustion of octane is:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
The standard enthalpy of combustion (∆Hc°) is the enthalpy change that occurs when one mole of a substance is completely burned in excess oxygen under standard conditions. The standard enthalpy of combustion for octane is -5471 kJ/mol.
To calculate the enthalpy change when 100.0 g of octane is combusted, you need to convert the mass of octane to moles.
First, calculate the number of moles of octane using its molar mass (114.23 g/mol):
moles = mass / molar mass
moles = 100.0 g / 114.23 g/mol
Next, use the stoichiometry of the balanced chemical equation to determine the moles of octane consumed in the reaction. Since the coefficient of octane is 2 in the balanced equation, you multiply the moles of octane by 2:
moles of octane consumed = moles of octane × 2
Now that you have the moles of octane consumed, you can calculate the enthalpy change using the equation:
ΔH = moles of octane consumed × ΔHc°
Substitute the values you obtained into the equation to find the enthalpy change when 100.0 g of octane is combusted at 25°C.