For the reaction 2H2 + O2 ---> 2H2O, ΔH° is given at -571.6 kJ/mol. What would the energy change (ΔU°) of the reaction be at 25°C and 1.00 bar?
Here is my attempt:
ΔU° = q + w
At constant pressure: q = ΔH° = -571.6 kJ/mol
w = -PΔV = (-1.00 bar)(0) = 0
ΔU° = -571.6 kJ/mol + 0 = -571.6 kJ/mol
In this case, since there is no change in volume, ΔU° = ΔH°.
I am not sure about my approach and reasoning. Can anyone confirm? Thanks. All help is very much appreciated.