The standard enthalpy of formation of n-octane is -249.95 kJ/mol. Compute the amount of heat liberated when 5.24 g of n-octane is burned completely with excess oxygen to form carbon dioxide and liquid water.
Hint:
Write the balanced combustion reaction. You can find the standard enthalpies of formation for O2, CO2, and H2O from many sources, including the CRC Handbook of Chemistry and Physics.
Solution:
Balanced combustion reaction: C8H18 + 12.5 O2 --> 8 CO2 + 9 H2O
From the CRC Handbook of Chemistry & PhysicsStandard enthalpies of formation (25C):
n-octane -249.95, oxygen 0.00, CO2 -393.5, H2O(l) -285.8 kJ/mol
ÄHrxn = -5470.25 kJ/mol
moles of octane burned = (5.24 g)/(114.231 g/mol) = X mol
Total heat released = 5470.25 kJ/mol * X mol = Answer kJ
alchol releases 29.7kJ/g when it burns. Convert this value to the number of calories per gram
To compute the amount of heat liberated when n-octane is burned completely, we need to use the enthalpy of formation and the balanced chemical equation for the combustion reaction.
First, let's write the balanced chemical equation for the combustion of n-octane (C8H18):
C8H18 + 12.5O2 → 8CO2 + 9H2O
From the balanced equation, we can see that every mole of n-octane reacts with 12.5 moles of oxygen to produce 8 moles of carbon dioxide and 9 moles of water.
Next, we need to calculate the number of moles of n-octane in 5.24 g. To do this, divide the mass by the molar mass of n-octane (114.22 g/mol):
moles of n-octane = 5.24 g / 114.22 g/mol
Once we have the moles of n-octane, we can use the enthalpy of formation to calculate the amount of heat liberated:
heat liberated = moles of n-octane × enthalpy of formation
Let's plug in the values:
moles of n-octane = 5.24 g / 114.22 g/mol ≈ 0.0459 mol
enthalpy of formation of n-octane = -249.95 kJ/mol
heat liberated = 0.0459 mol × -249.95 kJ/mol
Finally, we can calculate the amount of heat liberated:
heat liberated ≈ -11.46 kJ
Therefore, when 5.24 g of n-octane is burned completely, approximately -11.46 kJ of heat is liberated.