Give delta H for the combustion of propane: C3H8+5O2+3CO2+4H20 +530KCAL
delta h= -530 KCAL But how do i figure out how to get this answer? Thanks!
C3H8 + 5O2 ==> 3CO2 + 4H2O
dHrxn = (n*dHf products) - (n*dHf reactants)
Look up the delta Hf in tables, usually in your text, multiply by moles in the balanced equation and follow the step above. Most tables I've seen give dHf in joules so you will need to convert at the end but your tables may be in cal or kcal.
To calculate the enthalpy change (ΔH) for the combustion of propane (C3H8), you need to use the balanced chemical equation for the reaction and the enthalpy of formation values for the reactants and products involved.
The balanced chemical equation for the combustion of propane is as follows:
C3H8 + 5O2 → 3CO2 + 4H2O
The enthalpy change (ΔH) for the reaction can be calculated using the enthalpy of formation values (∆Hf) for the reactants and products. The ΔHf values represent the enthalpy change when one mole of a substance is formed from its constituent elements, with all species in their standard states.
ΔHf values (in kcal/mol):
C3H8: -103.851 kcal/mol
O2: 0 kcal/mol
CO2: -94.053 kcal/mol
H2O: -68.317 kcal/mol
Using these values, you can calculate the ΔH for the reaction as follows:
Reactants:
3 mol of C3H8 → 3 × (-103.851) kcal/mol = -311.553 kcal/mol
5 mol of O2 → 5 × 0 kcal/mol = 0 kcal/mol
Products:
3 mol of CO2 → 3 × (-94.053) kcal/mol = -282.159 kcal/mol
4 mol of H2O → 4 × (-68.317) kcal/mol = -273.268 kcal/mol
ΔH = (ΣΔHf of products) - (ΣΔHf of reactants)
ΔH = (-282.159 kcal/mol + (-273.268 kcal/mol)) - (-311.553 kcal/mol + 0 kcal/mol)
ΔH = -555.427 kcal/mol + 311.553 kcal/mol
ΔH = -243.874 kcal/mol
So, the enthalpy change (∆H) for the combustion of propane is -243.874 kcal/mol.
Note: The provided value of -530 kcal is likely the heat released per mole (heat of combustion) rather than the enthalpy change (∆H). These values can be converted using the heat of formation values.
To determine the enthalpy change (ΔH) for the combustion of propane, you need to account for the enthalpy changes associated with breaking and forming chemical bonds.
Step 1: Write out the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 → 3CO2 + 4H2O
Step 2: Calculate the enthalpy change for breaking all the bonds in the reactants.
Each C-H bond requires around 100 kcal/mol to break.
For propane (C3H8), there are 8 C-H bonds, so 8 x 100 kcal = 800 kcal.
Each O=O (oxygen-oxygen double bond) requires around 120 kcal/mol to break.
There are 5 O=O bonds in the reactants, so 5 x 120 kcal = 600 kcal.
Step 3: Calculate the enthalpy change for forming all the bonds in the products.
Each C=O bond (carbon-oxygen) releases around 100 kcal/mol when formed.
There are 3 C=O bonds formed in CO2, so 3 x 100 kcal = 300 kcal.
Each O-H bond (oxygen-hydrogen) releases around 100 kcal/mol when formed.
There are 4 O-H bonds formed in H2O, so 4 x 100 kcal = 400 kcal.
Step 4: Calculate the net enthalpy change (ΔH) by subtracting the energy released from the energy consumed:
ΔH = (Energy released in the formation of new bonds) - (Energy consumed in breaking old bonds)
ΔH = (300 kcal + 400 kcal) - (800 kcal + 600 kcal)
ΔH = 700 kcal - 1400 kcal
ΔH = -700 kcal
Therefore, the enthalpy change (ΔH) for the combustion of propane is -700 kcal. In the given question, it is rounded to -530 kcal, which may be due to the use of approximate values and rounding during calculations.