Calculate ΔH° for the reaction below: 4NO2(g) 4NO(g) + 2O2(g) ΔH° = ?

Question

Consider the following thermochemical equation:

2NO(g) + O₂(g)  2NO₂(g) ΔH° = -113.2 kJ

Calculate ΔH° for the reaction below:

4NO₂(g)  4NO(g) + 2O₂(g) ΔH° = ?

Answer key: +226.4 kJ

This was a complete guess since I noticed the moles for the second reactions was doubled from the first one. How did 226.4 kJ came to be?

Answer 1

Hi Kent,

Step #1 - Gather Necessary Data

To begin, lets gather the necessary enthalpic data from a chemical properties table. This data is taken from STANDARD THERMODYNAMIC PROPERTIES OF CHEMICAL SUBSTANCES, A PDF file that can easily found on the net. You will notice my answers are slightly different from yours is due to the subtle changes in the enthalpy data.

^ΔH (kJ/mol):

NO₂ (g) = 33.2

NO (g) = 91.3

O₂ (g) = 0

Step #2 - 1st Chemical Equation

2NO(g) + O₂(g) ↔ 2NO₂(g)

^ΔH = ΣH_p - ΣH_r = 2*33.2-2*91.3-1*0 = -116.2 kJ

Step #3 - 2nd Chemical Equation

4NO₂(g) ↔ 4NO(g) + 2O₂(g)

^ΔH = ΣH_p - ΣH_r = 4*91.3+2*0-4-4*33.2 = 232.4 kJ

As you can see, the 232.4 kJ is derived from the reverse reaction. One can easily double the final answer of the first chemical equation and change the signs to reflect the direction in the reaction.