how to calculate the enthalpy of combustion? ?
Use the balanced combustion reaction above to calculate the enthalpy of combustion of (delta H comb) for C6H6. can someone explain the steps in how to solve this? if you know how to do the other 2 questions that’d be helpful too.
- Roger the MoleLv 72 months ago
products - reactants (all units in kJ/mol)
(6 x −393.509) + (3 x −241.818) − (−79.9) − (15/2 x 0) = −3006.608 kJ/mol C6H6(l)
(2.905 × 10^15 kJ) / (3006.608 kJ/mol) × (78.1118 g C6H6/mol) = 7.547 × 10^13 g =
7.547 × 10^10 kg C6H6
(2.905 × 10^15 kJ) / (3006.608 kJ/mol) × (6 mol CO2 / 1 mol C6H6) ×
(44.00964 g CO2/mol) = 2.551 × 10^14 g = 2.551 × 10^11 kg CO2
- ChemTeamLv 72 months ago
You will use standard enthalpies of formation to solve this problem. You will need four of them.
You are given the value for C6H6 and the element oxygen's value is zero. All elements, in their standard state, have a formation enthalpy of zero.
The other two can be looked up in your text as well as online. In the link below, I use the values for CO2 and H2O in several problems.
From Hess' Law, here is the basic idea:
[sum of enthalpies of formation of products] − [sum of enthalpies of formation of reactants]
Here's the link:
Remember to multiply each formation value by the coefficient in the balanced equation.