Calculate the standard reaction enthalpy for
the reaction
NO2(g) ! NO(g) + O(g)
Given:
O2(g) ! 2O(g) �H� = +498.4 kJ/mol
NO(g) + O3(g) ! NO2(g) + O2(g)
�H� = −200 kJ/mol
3O2(g) ! 2O3(g) �H� = +285.4 kJ/mol
The trick here is to add equations, as appropriate, (multiply and or reverse) to obtain the desired equation. Then follow suit with the delta H values.
To calculate the standard reaction enthalpy for the given reaction, we need to use Hess's Law. Hess's Law states that the total enthalpy change in a chemical reaction is the same regardless of the route taken.
First, let's analyze the given reactions:
1. O2(g) → 2O(g)
ΔH1 = +498.4 kJ/mol
2. NO(g) + O3(g) → NO2(g) + O2(g)
ΔH2 = -200 kJ/mol
3. 3O2(g) → 2O3(g)
ΔH3 = +285.4 kJ/mol
By manipulating these equations, we can obtain the target reaction:
NO2(g) → NO(g) + O(g)
To get the target reaction, we'll cancel out the common species on both sides of the equations. We'll multiply the equations as needed to achieve this:
2 x (1) + (2) + 3 x (3):
2 x O2(g) → 4 x O(g)
NO(g) + O3(g) → NO2(g) + O2(g)
3 x 3O2(g) → 2 x 2O3(g)
Now add up the equations:
2O2(g) + NO(g) + O3(g) + 3O2(g) → 4O(g) + NO2(g) + O2(g) + 2O3(g)
Or simplified:
5O2(g) + NO(g) + O3(g) → 4O(g) + NO2(g) + 2O3(g)
Now, let's add up the enthalpy changes for these equations to find the enthalpy change for the target reaction.
ΔH_total = ΔH1 + ΔH2 + ΔH3
ΔH_total = (+498.4 kJ/mol) + (-200 kJ/mol) + (+285.4 kJ/mol)
ΔH_total = 583.8 kJ/mol
Therefore, the standard reaction enthalpy for the given reaction is 583.8 kJ/mol.