Description
Calculate ∆H(rxn)for the reaction:
5N2O4(l) +4N2H3CH3(l) = 12H2O(g) + 9N2(g) + 4CO2(g)Given:
N2H3CH3(l) ∆Hf(rxn) = 54.0 kJ/mole
N2O4(l) ∆Hf(rxn) = -20.0kJ/mole
Explanation & Answer
dH rxn = sum(dHf product*coeff) - sum(dHf reactant*coeff)
dHrxn = [12*(-241.95) + 9*0 + 4*(-393.51)] - [5*(-29)+ 4*(54)] = -4548.44