Calculate the enthalpy of formation of carbon dioxide in the following reaction:
C(s) + O2(g) -> CO2(g)
Use the following equations:
a) H2O(l) --> H2(g) + 1/2 O2(g): ΔH° = +285.8 kJ/mol
b) C2H6(g) --> 2C(s) + 3H2(g): ΔH° = +84.7 kJ/mol
c) 2CO2(g) + 3O2(g) --> C2H6(g) + (7/2)O2(g): Δ° = +1560.7 kJ/mol
Won't work. Check your post. c is not balanced; i.e., no H on the left. Check the others, too. I didn't check them.
To calculate the enthalpy of formation of carbon dioxide (CO2), we need to use the given equations and apply the concept of Hess's Law, which states that the overall enthalpy change of a reaction is independent of the pathway taken.
The given reaction shows the formation of carbon dioxide (CO2) from its elements carbon (C) and oxygen (O2). However, to use the given equations, we need to rearrange the given reaction in terms of the given equations.
Rearranging the given reaction:
C(s) + O2(g) -> CO2(g)
First, we can break down CO2 into C and O2:
C(s) + O2(g) -> C(g) + O2(g)
Next, we can use equation (b) "C2H6(g) --> 2C(s) + 3H2(g)" to cancel out the C(g):
C(s) + O2(g) + C2H6(g) -> C2H6(g) + O2(g)
Finally, we can use equation (c) "2CO2(g) + 3O2(g) --> C2H6(g) + (7/2)O2(g)" to cancel out the C2H6(g):
C(s) + O2(g) + 2CO2(g) + 3O2(g) -> 2CO2(g) + (7/2)O2(g)
Now, we have a balanced equation we can use.
Recall that enthalpy change is additive, so we can calculate the enthalpy change of the overall reaction by summing the enthalpy changes of the individual reactions.
ΔH° = (+285.8 kJ/mol) + (+84.7 kJ/mol) + (+1560.7 kJ/mol)
ΔH° = 1931.2 kJ/mol
Therefore, the enthalpy of formation of carbon dioxide (CO2) in the given reaction is 1931.2 kJ/mol.