Using Hess law, calculate ΔH for the equation: N2O3 ---> NO + NO2

Question

N2O3 ---> N2 + 3⁄2O2

ΔH = 83.7 kJ

1⁄2N2 + 1⁄2O2 ---> NO

ΔH = -90.2 kJ

NO2 --->1⁄2N2 + O2

ΔH = 33.2 kJ

Answer 1

To calculate the ΔH for the reaction N2O3 ---> NO + NO2, we need to manipulate the given equations and their corresponding enthalpy changes to cancel out the reactants and form the desired products:

N2O3 ---> N2 + 3⁄2O2 ΔH = 83.7 kJ (multiply by 2 to get 2NO)
2NO + O2 ---> 2NO2 ΔH = -33.2 kJ
N2 + O2 ---> 2NO ΔH = 90.2 kJ (multiply by -1 to reverse)

Adding these equations and their corresponding enthalpy changes gives:

N2O3 ---> NO + NO2 ΔH = -9.3 kJ

Therefore, the ΔH for the reaction N2O3 ---> NO + NO2 is -9.3 kJ.