Final answer:
To determine the standard molar enthalpy of formation of NO(g), the given reactions were manipulated according to Hess's Law, resulting in an enthalpy change of -47.7 kJ/mol for forming 1 mole of NO(g) from its elements in their standard states.
Step-by-step explanation:
The calculation of the standard molar enthalpy of formation of NO(g) requires using Hess's Law, which states that the change in enthalpy for a given reaction is the same whether it occurs in one step or a series of steps. We can use the provided reactions and their enthalpy changes to find the enthalpy change for the formation of NO(g).
First, based on the provided information:
- N₂(g) + 2O₂(g) → 2NO₂(g), ΔH° = +66.4 kJ
- 2NO(g) + O₂(g) → 2NO₂(g), ΔH° = -114.1 kJ
The goal is to derive the reaction N₂(g) + 1/2O₂(g) → NO(g), which represents the formation of 1 mole of NO(g). To achieve this, we invert the second equation and divide it by 2 to align with the required stoichiometry for the formation equation. This will reverse the sign of the enthalpy change as well. The revised equation and associated enthalpy change is:
- 1/2O₂(g) + NO(g) → NO₂(g), ΔH° = +57.05 kJ
Now, subtract this reaction from the first reaction to cancel out the NO₂(g) and isolate NO(g):
- N₂(g) + 2O₂(g) → 2NO₂(g), ΔH° = +66.4 kJ
- 1/2O₂(g) + NO(g) → NO₂(g), ΔH° = +57.05 kJ (times 2)
After canceling the NO₂(g) species and combining the enthalpy changes, the resulting enthalpy for the formation of NO(g) is:
N₂(g) + 1/2O₂(g) → NO(g), ΔH°f[NO(g)] = 66.4 kJ - 2(57.05 kJ) = 66.4 kJ - 114.1 kJ = -47.7 kJ/mol
Therefore, the standard molar enthalpy of formation of NO(g) is -47.7 kJ/mol.