From the following data, calculate ΔH° for the reaction
O3(g) +NO(g)----->O2(g)+NO2 (g)
Compound ΔH°f (kJ/mol)
O3 142.7
NO 90.3
O2 0
NO2 33.2
dHorxn = (n*dHo products) - (n*dHo reactants)
Post your work if you get stuck.
To calculate ΔH° for the reaction, we need to use the enthalpy of formation values for each compound involved.
The reaction is:
O3(g) + NO(g) → O2(g) + NO2(g)
First, we need to determine the ΔH° for the reactants:
ΔH°(reactants) = ΣΔH°f(products) - ΣΔH°f(reactants)
ΔH°(reactants) = (ΔH°f(O2) + ΔH°f(NO2)) - (ΔH°f(O3) + ΔH°f(NO))
Substituting the given values:
ΔH°(reactants) = (0 kJ/mol + 33.2 kJ/mol) - (142.7 kJ/mol + 90.3 kJ/mol)
ΔH°(reactants) = 33.2 kJ/mol - 233 kJ/mol
ΔH°(reactants) = -199.8 kJ/mol
Therefore, ΔH° for the reaction O3(g) + NO(g) → O2(g) + NO2(g) is -199.8 kJ/mol.
To calculate ΔH° for the reaction, you need to use the formula:
ΔH° = ΣΔH°products - ΣΔH°reactants
First, let's determine the ΔH° for the products:
For O2(g), the value is 0 kJ/mol.
For NO2(g), the value is 33.2 kJ/mol.
Next, let's determine the ΔH° for the reactants:
For O3(g), the value is 142.7 kJ/mol.
For NO(g), the value is 90.3 kJ/mol.
Now, we can substitute these values into the formula:
ΔH° = (0 + 33.2) - (142.7 + 90.3)
ΔH° = 33.2 - 233
ΔH° = -199.8 kJ/mol
Therefore, the ΔH° for the reaction O3(g) + NO(g) --> O2(g) + NO2(g) is -199.8 kJ/mol.