Calculate the ΔHo for the reaction, 6NO2(g) + 8NH3(g) → 7N2(g) + 12H2O(l).
N2(g) + 2O2(g) → 2NO2(g) ΔHo=+66.36 kJ
4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) ΔHo=-1530.36 kJ
To find the ΔHo for the given reaction, we need to manipulate the given equations and their respective enthalpy changes to get the desired reaction and find the enthalpy change for it.
First, we flip the first equation and halve the second equation to match the stoichiometric coefficients in the desired reaction:
½(2NO2(g) → N2(g) + O2(g)) ΔHo= -½(66.36 kJ) = -33.18 kJ
½(2N2(g) + 6H2O(l) → 4NH3(g) + 3O2(g)) ΔHo= ½(-1530.36 kJ) = -765.18 kJ
Now, we multiply the first equation by 3 and the second equation by 2 in order to match the stoichiometric coefficients of NO2 and NH3 with the desired reaction:
3(2NO2(g) → N2(g) + O2(g)) ΔHo= 3(-33.18 kJ) = -99.54 kJ
2(2N2(g) + 6H2O(l) → 4NH3(g) + 3O2(g)) ΔHo= 2(-765.18 kJ) = -1530.36 kJ
Finally, we sum up the modified equations to get the desired reaction:
-99.54 kJ + 12NH3(g) + 16NO2(g) → 14N2(g) + 24H2O(l) - 1530.36 kJ
The ΔHo for this reaction is the sum of the enthalpy changes for the modified equations:
ΔHo = -99.54 kJ + (-1530.36 kJ) = -1629.9 kJ