Given N2 + O2 giving 2NO at +180.5 kJ and NO + half O2 giving NO2 at -57.06 kJ, what is the value for N2 + 2O2 giving 2NO2?

1. The target makes 2 mol of NO2, but the second step as written makes only 1 mol. 2. So multiply the second step by 2, giving 2NO plus O2 forming 2NO2. 3. Multiplying the equation by 2 multiplies its enthalpy change by 2 as well. 4. That gives 2 times minus 57.06, which is minus 114.12 kJ. 5. The first step is used as written, at plus 180.5 kJ, and needs no reversing. 6. The 2 mol of NO made in step one are consumed in the doubled step two, so NO cancels. 7. Adding gives 180.5 plus minus 114.12, which is plus 66.4 kJ. 8. Forgetting to double the second step leaves plus 123.4 kJ, the trap here. _Source: OpenStax Chemistry (CC BY 4.0), Ch 5 "Thermochemistry", section 5.3 Enthalpy_