Using Hess law, calculate ΔH for the equation: N2O3 ---> NO + NO2
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
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.
To calculate ΔH for the equation N2O3 ---> NO + NO2 using Hess's law, you need to manipulate the given equations in order to cancel out any species that appear on both the reactant and product side.
First, reverse the second equation and multiply it by 2 in order to balance the nitrogen atoms:
1/2N2 + 1/2O2 ---> NO
2NO ---> N2 + O2 (Multiply by 2)
Next, add the first and third equations together to cancel out the nitrogen and oxygen species:
N2O3 ---> N2 + 3/2O2
NO2 ---> 1/2N2 + O2
Adding these equations gives the desired equation:
N2O3 + NO2 ---> NO + NO2
Now, add the enthalpy changes of the individual reactions to calculate the overall enthalpy change:
ΔH = (ΔH1 + ΔH3) - ΔH2
= (83.7 kJ + 33.2 kJ) - (-90.2 kJ)
= 116.9 kJ - (-90.2 kJ)
= 207.1 kJ
Therefore, ΔH for the equation N2O3 ---> NO + NO2 is 207.1 kJ.