To calculate ΔH for the reaction C2H6 → C2H2 + 2H2 using Hess's Law, we need to use the given equations and manipulate them to obtain the desired reaction.
First, we need to reverse the direction of the second equation:
2CO2 + H2O → C2H2 + 5/2O2
Then, we multiply the first equation by 2 and the third equation by 2:
2(C2H2 + 5/2O2 → 2CO2 + H2O)
2(C2H6 → C2H2 + 2H2)
Now, we can add the equations together, canceling out any species that appear on both sides of the reaction:
2C2H6 + 7O2 → 4CO2 + 6H2O
2C2H2 + 5O2 → 4CO2 + 2H2O
Next, we want to eliminate the extra O2 and H2O in the second equation. We can do this by doubling the first equation and canceling out the appropriate species:
4C2H2 + 10O2 → 8CO2 + 4H2O
2C2H2 + 5O2 → 4CO2 + 2H2O
Finally, we subtract the second equation from the first equation:
2C2H6 + 7O2 - (2C2H2 + 5O2) → 4CO2 + 6H2O - (4CO2 + 2H2O)
Simplifying the equation, we get:
2C2H6 + 7O2 - 2C2H2 - 5O2 → 4CO2 + 6H2O - 4CO2 - 2H2O
Cancelling out the like terms:
2C2H6 - 2C2H2 + 7O2 - 5O2 → 6H2O - 2H2O
Simplifying further:
2C2H6 - 2C2H2 + 2O2 → 4H2O
The ΔH for the reaction C2H6 → C2H2 + 2H2 can be calculated by summing the enthalpies from the reactions we used to manipulate the equations:
ΔH = -(3120 kJ) - (-1300 kJ) - (-572 kJ)
ΔH = -3120 kJ + 1300 kJ - 572 kJ
ΔH = -2392 kJ
Therefore, the ΔH for the reaction C2H6 → C2H2 + 2H2 is -2392 kJ.