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.

Question

Fernanda Stein

Chemistry

25 January, 16:26

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.

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Angelo Hayden · Answer 1

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