Use the following data
2C(s) + 4H2(g) --> C3H8(g)
DH= -103.85 kJ/molrxn
C(s) + O2(g) --> CO2(g)
DH= -393.51 kJ/molrxn
H2(g) + 1/2O2(g) --> H2O(g)
DH= -241.82 kJ/molrxn
to calculate the heat of combustion of propane, C3H8
C3H8(g) + 5O2(g) --3CO2(g) + 4H20(g)
You have a typo in #1, that should be 3C(s).
Reverse #1, add to 3x #2, and to 4x #3.
To calculate the heat of combustion of propane (C3H8), you can use the given data for the standard enthalpies of formation of the reactants and products in the balanced combustion equation.
The balanced equation for the combustion of propane is:
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)
Given Data:
DH of C(s) + O2(g) --> CO2(g) = -393.51 kJ/molrxn
DH of H2(g) + 1/2O2(g) --> H2O(g) = -241.82 kJ/molrxn
DH of 2C(s) + 4H2(g) --> C3H8(g) = -103.85 kJ/molrxn
Now, we need to use Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the steps into which the reaction can be divided.
Let's break down the combustion of propane into its component reactions:
1. C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g) (Given)
2. C3H8(g) --> 3C(s) + 4H2(g) (Reverse of C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g))
3. 3C(s) + 4H2(g) --> 3CO2(g) + 4H2O(g) (Reverse of 2C(s) + 4H2(g) --> C3H8(g))
4. 3CO2(g) + 4H2O(g) --> C3H8(g) + 5O2(g) (Reverse of 1. C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g))
Now, let's reverse or manipulate the given reactions to align with the desired reaction.
Reactions:
1. C(s) + O2(g) --> CO2(g) (Given)
2. H2(g) + 1/2O2(g) --> H2O(g) (Given)
3. 2C(s) + 4H2(g) --> C3H8(g) (Given)
Now, let's multiply and manipulate these reactions to get the desired reaction of C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g).
Multiply Reaction 1 by 3:
3C(s) + 3O2(g) --> 3CO2(g)
Multiply Reaction 2 by 4:
4H2(g) + 2O2(g) --> 4H2O(g)
Double Reaction 3:
4C(s) + 8H2(g) --> 2C3H8(g)
Now, add all the manipulated reactions together:
3C(s) + 3O2(g) + 4H2(g) + 2O2(g) + 4C(s) + 8H2(g) --> 3CO2(g) + 4H2O(g) + 2C3H8(g)
Simplify:
7C(s) + 5O2(g) + 12H2(g) --> 3CO2(g) + 4H2O(g) + 2C3H8(g)
Now, let's take the enthalpy changes of the individual reactions and add them up:
DH(Reactions) = DH(1) + DH(2) + DH(3)
DH(Reactions) = -393.51 kJ/molrxn + (-241.82 kJ/molrxn) + (-103.85 kJ/molrxn)
DH(Reactions) = -739.18 kJ/molrxn
Therefore, the heat of combustion of propane (C3H8) is -739.18 kJ/molrxn.