Final answer:
Using enthalpy data provided and Hess's Law, the enthalpy change for the reaction NO(g) + O₂(g) → NO₂(g) is calculated to be -114.1 kJ.
Step-by-step explanation:
The question is asking to calculate the enthalpy change (ΔH) for the reaction that forms nitrate (NO2) from nitrogen monoxide (NO) and oxygen (O2). According to the information given, we have a few related reactions and their enthalpy changes.
First, we need to establish the enthalpy of formation for NO(g) which is derived from a series of reactions. From the provided data, the formation of NO2(g) from N2(g) and O2(g) has an enthalpy change (ΔH°) of 66.4 kJ and the reaction of NO(g) with O2(g) to form NO2(g) has ΔH° of -114.1 kJ. To find the ΔH of the formation of NO(g), we can reverse the second reaction and then use Hess's Law which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of its individual steps, regardless of the pathway taken.
By reversing the reaction 2NO(g) + O2 → 2NO2(g), we get 2NO2(g) → 2NO(g) + O2 with a ΔH° of +114.1 kJ. Then, N2(g) + 2O2(g) → 2NO2(g) has ΔH° of 66.4 kJ. When we add these two reactions, the NO2 cancels out, leaving us with the formation reaction for NO(g):
N2(g) + O2(g) → 2NO(g)
ΔH° = 66.4 kJ + 114.1 kJ = 180.5 kJ
Finally, since the desired reaction NO(g) + O2 → NO2(g) is the reverse of the second step we originally reversed, it will have the opposite sign for ΔH°, yielding:
NO(g) + O2 → NO2(g)
ΔH° = -114.1 kJ