Consider the standard enthalpy of formation of nitrogen dioxide: % N2(g) + O2(g) → 2 NO2(g) (ΔHf = 33.9 kJ/mol) Use the properties of thermochemical equations to find ΔH for each related reaction (3 pts): N2(g) + 2 O2(g) → 2 NO2(g) ΔH = ? 2 NO2(g) → N2(g) + 2 O2(g) ΔH = ? 2 N2(g) + 4 O2(g) → 4 NO(g) ΔH
regardless of the pathway taken, provided the initial and final states are the same.
The correct Answer and Explanation is:
Given data:
- Standard enthalpy of formation of NO₂:
N2(g)+O2(g)→2 NO2(g)ΔHf=33.9 kJ/molN2(g)+O2(g)→2NO2(g)ΔHf=33.9kJ/mol
1. Find ΔHΔH for:
N2(g)+2 O2(g)→2 NO2(g)N2(g)+2O2(g)→2NO2(g)
Solution:
The given formation reaction involves 1 mol of N₂ and 1 mol of O₂ producing 2 mols of NO₂.
The target reaction involves 1 mol of N₂ and 2 mols of O₂ producing 2 mols of NO₂.
Since the formation reaction produces 2 mols of NO₂ from 1 mol N₂ and 1 mol O₂, to produce 2 mols of NO₂ with 1 mol N₂ and 2 mols O₂, we can think of the second O₂ as an additional reactant that does not participate in the formation directly, but we can write the reaction as:
N2(g)+2 O2(g)→2 NO2(g)N2(g)+2O2(g)→2NO2(g)
Applying Hess’s Law:
ΔH=ΔHformation of 2 mols NO2+additional reaction involving O2ΔH=ΔHformation of 2 mols NO2+additional reaction involving O2
But since the original formation reaction accounts for 2 mols NO₂ from 1 mol N₂ and 1 mol O₂, and the second O₂ is just an extra reactant, the enthalpy change for adding an extra O₂ molecule (which is just the standard enthalpy of O₂, which is zero as an element in its standard state), the total enthalpy change for this reaction is:
ΔH=33.9 kJΔH=33.9kJ
Answer:
ΔH=33.9 kJΔH=33.9kJ
2. Find ΔHΔH for:
2 NO2(g)→N2(g)+2 O2(g)2NO2(g)→N2(g)+2O2(g)
Solution:
This is the reverse of the formation reaction:
N2(g)+2 O2(g)→2 NO2(g)N2(g)+2O2(g)→2NO2(g)
The enthalpy change for the reverse reaction is simply the negative of the forward reaction:
ΔH=−33.9 kJΔH=−33.9kJ
3. Find ΔHΔH for:
2 N2(g)+4 O2(g)→4 NO(g)2N2(g)+4O2(g)→4NO(g)
Solution:
This reaction involves formation of nitric oxide (NO), not NO₂. To find this, we need the standard enthalpy of formation of NO:
- N2(g)+O2(g)→2 NO(g)N2(g)+O2(g)→2NO(g)
ΔHf≈90.3 kJ/molΔHf≈90.3kJ/mol (from standard data)
Since the reaction involves 2 mol of N₂ and 4 mol of O₂ producing 4 mols of NO, the enthalpy change is:
ΔH=2×90.3 kJ=180.6 kJΔH=2×90.3kJ=180.6kJ
Answer:
ΔH=180.6 kJΔH=180.6kJ
Summary:
| Reaction | ΔHΔH (kJ) |
|---|---|
| N2(g)+2 O2(g)→2 NO2(g)N2(g)+2O2(g)→2NO2(g) | 33.9 |
| 2 NO2(g)→N2(g)+2 O2(g)2NO2(g)→N2(g)+2O2(g) | -33.9 |
| 2 N2(g)+4 O2(g)→4 NO(g)2N2(g)+4O2(g)→4NO(g) | 180.6 |
Explanation:
This problem demonstrates the application of Hess’s Law and thermochemical properties. The key idea is that enthalpy changes are state functions, meaning the total enthalpy change depends only on initial and final states, not the pathway taken. The formation reaction for NO₂ provides a baseline, from which other reactions can be derived by reversing or scaling the equations. For reactions involving the formation or decomposition of compounds, the enthalpy of formation is central. Reversing a reaction changes the sign of ΔHΔH, and scaling the coefficients scales the enthalpy accordingly. For reactions involving different compounds, standard enthalpies of formation from tables are used.
