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pashok25 [27]
4 years ago
11

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.
Chemistry
1 answer:
8_murik_8 [283]4 years ago
7 0
Based on Hess's Law: 

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

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