Question

use the following data to calculate the enthalpy change (KJ) for reaction 3:

N2O4 (g) --> 2 NO2 (g) (1) delta H= 24 KJ
NO (g) + 1/2 O2 (g) --> NO2 (g) delta H= -68 Kj
2 NO (g) + O2 (g) --> N2O4 (g) delta H= ???

Answer 1

enthalpy (3) = 2x(2) + [the reverse enthalpy of (1)]
enthalpy (3) = 2x(-68) - 24
enthalpy (3) = -160kJ