What is the process to calculate ΔH° for the reaction?

O2(g) + 2NO(g) --> N2O4(g)

O2(g) + 2NO(g) --> 2NO2(g) ΔH°1

N2O4(g) --> 2NO2(g) ΔH°2

a) ΔH°1 + ΔH°2
b) ΔH°1 - ΔH°2
c) ΔH°2 - ΔH°1
d) ΔH°1 - 2ΔH°2

I know according to Hess' Law that I need to take the ΔH° products - ΔH° reactants.

I am a guessing it would probably be c.

To get equation 1 you want to add equn 1 to rhe reverse of equn 2. reversing eqn 2 makes it a minus dH2. Look to me like c is not right

So if it is the reverse of equation 2. You would have to take ΔH°1 - ΔH°2?

yes. Write out eqn 1 and add it to the reverse of eqn 2 and see if it give you the reaction you want.

To calculate ΔH° for the overall reaction, you can apply Hess's Law, which states that the change in enthalpy of a reaction is independent of the pathway taken. In this case, you are given the enthalpy changes for two separate reactions:

O2(g) + 2NO(g) --> 2NO2(g) (ΔH°1)
N2O4(g) --> 2NO2(g) (ΔH°2)

Now, to determine the ΔH° for the overall reaction:

O2(g) + 2NO(g) --> N2O4(g)

You need to combine these two reactions in a way that allows for the cancellation of the common species (in this case, 2NO2). Here is the process:

1. Reverse the second reaction to match the stoichiometry of the NO2 in the first reaction:
2NO2(g) --> N2O4(g)

2. Multiply the first reaction by 1 to balance the stoichiometry:
O2(g) + 2NO(g) --> 2NO2(g)

3. Add the two reactions together:
O2(g) + 2NO(g) + 2NO2(g) --> 2NO2(g) + N2O4(g)

4. Cancel out the common species on both sides of the reaction equation:
O2(g) + 2NO(g) --> N2O4(g)

Now, by applying Hess's Law, the ΔH° for the overall reaction is the sum of the enthalpy changes for the individual reactions:

ΔH°overall = ΔH°1 + ΔH°2

In this case, answer choice a) ΔH°1 + ΔH°2 correctly represents the process to calculate ΔH° for the reaction.