Nitrogen dioxide, NO2(g), is an emission resulting from the burning of gasoline in the air in an automobile engine. nitrogen dioxide contributes to the formation of smog and acid rain. it can be converted to dinitrogen tetraoxide as shown below:

2NO2(g) --> N2O4(g)

a) use Hess's law and the following equations to determine the enthalpy change for this reaction.
1) N2(g) +2O2(g) --> 2NO2(g) ΔH^o = 66.4 k/J
2) N2(g) + 2O2(g) --> N2O4(g) ΔH^o = 11.1 k/j

b) write the thermochemical equation for the overall reaction.

a. -1 x (1) 2NO2(g) → N2(g) + 2O2(g) ΔH = -66.4 kJ

1 x (2) N2(g) + 2O2(g) → N4O4(g) ΔH = 11.1 kJ
Total 2NO2(g) + N2(g) + 2O2(g) → N2(g) + 2O2(g) + N4O4(g) ΔH = -55.3 kJ


b. 2NO2(g) → N2O4(g) ΔH = -55.3 kJ
or
2NO2(g) → N2O4(g) + 55.3 kJ

I looked at your earlier post and I was confused. However, here is what you do.

(1).......... N2(g) +2O2(g) --> 2NO2(g) ΔH^o = 66.4 kJ
(2)...........N2(g) + 2O2(g) --> N2O4(g) ΔH^o = 11.1 kJ
---------------------------------------------------------------------------
add...........2N2 + 4O2 ==> 2NO2 + N2O4 dH = 77.5 kJ
Now you have a 2NO2 on the right and you don't want that.
reverse(1)......2NO2 ==> N2 + 2O2 dH = -66.4 kJ
------------------------------------------------------------
..............N2 + 2O2 ==> N2O4 dH = 77.5 - 66.4 = ? kJ.
Check my thinking.

oops. I didn't get the equation you want. You want 2NO2 ==> N2O4 and I calculated N2 + O2 ==> N2O4 Sorry about that.

BY THE WAY, I LOOED UP AND FOUND +57.3 Kj (ENDOTHERMIC). i'LL PLAY AROUND WITH A LATER. IT'S BED TIME FOR ME.

I looked at some other sites and found the rxn actually is exothermic so a negative sign is correct for dH. Go with Tracy's work.

To determine the enthalpy change for the reaction and write the thermochemical equation, we can use Hess's law. Hess's law states that the change in enthalpy for a reaction is independent of the pathway taken and depends only on the initial and final states.

a) To use Hess's law, we need to manipulate the given equations to match the desired reaction equation.

1) We have the equation: N2(g) + 2O2(g) → 2NO2(g) ΔH° = 66.4 kJ

2) We have the equation: N2(g) + 2O2(g) → N2O4(g) ΔH° = 11.1 kJ

We want the desired reaction: 2NO2(g) → N2O4(g)

To achieve this, we can reverse equation 2 and multiply it by 2. The sign of enthalpy change also changes when we reverse an equation.

Reversed equation 2: N2O4(g) → N2(g) + 2O2(g) ΔH° = -11.1 kJ

Multiplying reversed equation 2 by 2: 2N2O4(g) → 2N2(g) + 4O2(g) ΔH° = -22.2 kJ

Next, we need to double equation 1 to match the coefficient of NO2:

2 × (N2(g) + 2O2(g) → 2NO2(g)) ΔH° = 2 × 66.4 kJ

This gives us: 2N2(g) + 4O2(g) → 4NO2(g) ΔH° = 132.8 kJ

Now, we can add equation 1 and the reversed and multiplied equation 2 to get the desired reaction equation:

(2N2(g) + 4O2(g) → 4NO2(g)) + (2N2O4(g) → 2N2(g) + 4O2(g))

Cancelling out common species, we obtain the overall reaction equation:

2NO2(g) → N2O4(g) ΔH° = 132.8 kJ - 22.2 kJ

b) The thermochemical equation for the overall reaction is:

2NO2(g) → N2O4(g) ΔH° = 110.6 kJ (132.8 kJ - 22.2 kJ)