Using the enthalpies of formation given below, calculate ΔH°rxn in kJ, for the following reaction.
3NO2(g)+1H2O(l) -> 2HNO3(g)+1NO(g)
NO2 (g): 33.10 kJ/mol
H2O (l): -285.83 kJ/mol
HNO3 (l): -174.10 kJ/mol
NO (g): 90.29 kJ/mol
dHrxn = dHproducts - dHreactants
The Attempt at a Solution
Is this possible to do since the equation says HNO3 is a gas, but they give the enthalpy of formation for HNO3 as a liquid.