Use Hess's law and the following equations to calculate ÄH for the reaction N2(g) + 2O2(g) N2O4(g).
2NO2(g)->N2O4(g) ÄH = –57.0 kJ/mol
N2(g)+O2(g)->2NO(g) ÄH = 180.6 kJ/mol
2NO2(g)->2NO(g)+O2(gÄH = 114.4 kJ/mol
To calculate the ΔH for the reaction N2(g) + 2O2(g) → N2O4(g) using Hess's law, we can utilize the given equations and their respective enthalpy changes (ΔH).
Step 1: Reverse the reaction and the corresponding equation
Since the reaction we want to calculate ΔH for is the reverse of the given equation N2O4(g) → 2NO2(g), we need to reverse the equation and its enthalpy change as well:
ΔH1 = +57.0 kJ/mol (reversed from -57.0 kJ/mol)
Step 2: Multiply the reversed equation and its ΔH by the appropriate factor
Since the reversed equation produces 2 mol of NO2, while we need only 1 mol in the reaction we want to calculate ΔH for, we need to adjust both the equation and its ΔH accordingly:
2NO2(g) → N2O4(g)
ΔH1 = (-57.0 kJ/mol) * 1/2 = -28.5 kJ/mol
Step 3: Use the third given equation and its ΔH to cancel out the intermediate species
The third given equation is:
2NO2(g) → 2NO(g) + O2(g)
ΔH2 = +114.4 kJ/mol
Step 4: Multiply the second equation and its ΔH by the appropriate factor
To cancel out the NO2 in the third equation, we need twice the amount. Therefore, we multiply the second equation and its ΔH accordingly:
2NO(g) + O2(g) → N2(g) + O2(g)
ΔH3 = 180.6 kJ/mol * 2 = +361.2 kJ/mol
Step 5: Add up all the ΔH values
By adding up the ΔH values from steps 1, 2, 3, and 4, we can find the overall ΔH for the reaction:
ΔH = ΔH1 + ΔH2 + ΔH3
ΔH = -28.5 kJ/mol + 114.4 kJ/mol + 361.2 kJ/mol = 447.1 kJ/mol
Therefore, the ΔH for the reaction N2(g) + 2O2(g) → N2O4(g) is 447.1 kJ/mol.
I didn't do this with a pencil and pad but try this
Add equn 1 to equan 2 to the reverse of equan 3 and see if that won't give you the equation you want. For the dH values, change the sign of those you reverse.