Calculate ∆h0 for the reaction 2 n2(g) + 5 o2(g) −→ 2 n2o5(g) given the data h2(g) + 1 2 o2(g) −→ h2o(ℓ) ∆h 0 f = −283.7 kj/mol n2o5(g) + h2o(ℓ) −→ 2 hno3(ℓ) ∆h0 = −78.5 kj/mol 1 2 n2(g) + 3 2 o2(g) + 1 2 h2(g) −→ hno3(ℓ) ∆h0 f = −173 kj/mol answer in units of kj.

Based on Hess's Law: 

2 N2(g) + 6 O2(g) + 2 H2(g) −→ 4 HNO3(l) ∆Hf = (−171.9 kJ/mol)(4 mol) 
2 H2O(l) −→ 2 H2(g) + O2(g) ∆Hf = (-283.8 kJ/mol)(2 mol)(-1) →times -1, rxn is reversed 
4 HNO3(l)−→ 2 N2O5(g) + 2 H2O(l) ∆Hf = (-76.4 kJ/mol)(2 mol)(-1) →times -1, rxn is reversed 

2 N2(g) + 5 O2(g) −→ 2 N2O5(g) ∆H0 = 32.8 kJ

Yeah irbyoundo this you will get the a swer wuick