C6H12O6 + 6O2 ==> 6CO2 + 6H2O
dHrxn = (n*dHf products) - (n*dHf reactants) and dHrxn will be in kJ/mol.
Divide by 1 and divide by molar mass C6H12O6 to get kJ/gram. To convert to fuel value change the negative sign to a + sign.
Do the same for tristearin. The equation is
2C57H110O6 + 163O2 ==> 114CO2 + 110H2O
Here the answer is in kJ/mol so you divide by 2 (to convert from 2 mols to 1 mol) the divide by molar mass of trisearin. Then change the - number to + number.
Compare the two. The larger + number wins.
Fuel values, reported in units of KJ/g, are useful in comparing the efficiency of various energy sources on a mass basis. Determine the fuel value of the glucose and the tristearin. Which one, carbohydrates or fats, have the highest fuel value (and hence is the most efficient energy source on a mass basis)?
Glucose C6H12O6, DHf -1260.04 KJ/mol
Tristearin C57H110O6, DHf -390.70 KJ/mol
1 answer
1 answer