Question 6:
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
(If that was jumbled up i Dont understand?
To calculate ΔH° for the reaction, we need to use the enthalpy of formation values for the compounds involved. The enthalpy of formation (ΔH°f) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states.
In this case, the reaction is:
O3(g) + NO(g) → O2(g) + NO2(g)
We can calculate ΔH° for the reaction by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products.
Reactants:
ΔH°f(O3) = 142.7 kJ/mol
ΔH°f(NO) = 90.3 kJ/mol
Products:
ΔH°f(O2) = 0 kJ/mol
ΔH°f(NO2) = 33.2 kJ/mol
Now we can calculate the enthalpy change (ΔH°) for the reaction using the formula:
ΔH° = ΣΔH°f(products) - ΣΔH°f(reactants)
ΔH° = [0 kJ/mol + 33.2 kJ/mol] - [142.7 kJ/mol + 90.3 kJ/mol]
ΔH° = 33.2 kJ/mol - 233 kJ/mol
ΔH° = -199.8 kJ/mol
Therefore, the ΔH° for the reaction O3(g) + NO(g) → O2(g) + NO2(g) is -199.8 kJ/mol.