Question

Calculate ∆H0 for the reaction
2 N2(g) + 5 O2(g) −→ 2 N2O5(g)
given the data
H2(g) + 1/2O2(g) −→ H2O(ℓ) ; ∆H0f = −282.1 kJ/mol
N2O5(g) + H2O(ℓ) −→ 2 HNO3(ℓ); ∆H0 = −79.7 kJ/mol
1/2N2(g) + 3/2O2(g) + 1/2H2(g) −→ HNO3(ℓ); ∆H0f = −170.9 kJ/mol

Answer 1

$\Delta H$ = 40 KJ/mol.

Please see the attachment

Explanation:

Answer 2

Thus ∆Ho for the reaction 2 N2(g) + 5 O2(g) −→ 2 N2O5(g) is 40 KJ / mol

Explanation:

Given data:

(1) H2(g) + 1/2O2(g) −→ H2O(ℓ) ; ∆H0f = −282.1 kJ/mol

(2) N2O5(g) + H2O(ℓ) −→ 2 HNO3(ℓ); ∆H0 = −79.7 kJ/mol

(3) 1/2 N2(g) + 3/2 O2(g) + 1/2 H2(g) → HNO3(ℓ); ∆H0f = −170.9 kJ/mol

Solution:

Given reaction:

(4) 2 N2(g) + 5 O2(g) −→ 2 N2O5(g)

• Multiply equation (1) by 2
• Multiply equation (2) by 2
• Multiply equation (3) by 4

Now

2H2(g)  → 2H2 + O2(ℓ) ; ∆H0f =  2 x  −282.1 kJ/mol  = 564.2 KJ/mol

4HNO3 → N2O5(g) + H2O(ℓ)  ; ∆H0 = 2 x 79.7 kJ/mol = 159.4 KJ / mol

2N2 + 6O2 + 2H2 → 4HNO3 ;  ∆H0f = −170.9 x 4 = - 683.6 KJ / mol

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2N2 + 5O2 → 2N2O5  

∆H4 = ∆H1 + ∆H2 + ∆H3

∆H4 = 564.2  + 159.4 - 683.6 = 40 KJ / mol

Thus ∆Ho for the reaction 2 N2(g) + 5 O2(g) −→ 2 N2O5(g) is 40 KJ / mol