Question

Given enthalpy of formation data:

Molecule:
NO^-(g): -33.18
N2O4(g): 9.16
2NO2(g) → N2O4(g):
a) -42.34
b) -24.02
c) 24.02
d) 42.34

Answer 1

To calculate the standard molar enthalpy of formation of NO(g), Hess's law is used to combine enthalpy changes from given reactions, arriving at a value of -90.25 kJ/mol for NO(g).

Step-by-step explanation:

To calculate the standard molar enthalpy of formation of NO(g), we need to use the provided enthalpy of formation data and apply Hess's law to find the enthalpy change for the formation of NO(g) from its elements in their standard states.

To start, we have the enthalpy of formation for various substances involved in the process:

• N₂(g) + 2O₂(g) → 2NO₂(g), ΔH° = +66.4 kJ
• 2NO(g) + O₂(g) → 2NO₂(g), ΔH° = -114.1 kJ

Using Hess's law, we can set up an equation where we calculate the enthalpy change of the reaction where N₂(g) and O₂(g) form NO(g) by rearranging and combining the given reactions to target the formation of 1 mole of NO(g). This involves halving the second reaction and subtracting the enthalpy of the first reaction divided by 2 from it:

ΔH°f[NO(g)] = (ΔH°[2NO(g) → 2NO₂(g)] / 2) - (ΔH°[N₂(g) + 2O₂(g) → 2NO₂(g)] / 2)

ΔH°f[NO(g)] = (-114.1 kJ / 2) - (66.4 kJ / 2) = -57.05 kJ - 33.2 kJ = -90.25 kJ/mol.

Therefore, the standard molar enthalpy of formation of NO(g) is -90.25 kJ/mol.