What will be the enthalpy of formation of NO2 from the given bond dissociation enthalpy values ? The bond dissociation enthalpy values for O2 , NO and NO2 are as follows, O2(g) : 0 kj/mol , NO(g) : 90.25 kj/mol and NO2(g) : 33.18 kj/mol respectively.

Question

What will be the enthalpy of formation of NO2 from the given bond dissociation enthalpy values ? The bond dissociation enthalpy values for O2 , NO and NO2 are as follows, O2(g) : 0 kj/mol , NO(g) : 90.25 kj/mol and NO2(g) : 33.18 kj/mol respectively. (A)  +114.1 kj  (B)  +52.7 kj  (C)  -52.7 kj  (D)  -114.1 kj

Tardigrade Admin · Accepted Answer

2 N O(g)+O2(g) leftharpoons 2 N O2(g) Δ H° textreaction = ∑ Δ Hf text(products)-Δ Hf text(reactants) =(2×33.18)-(2×90.25+0) Δ H° textreaction=-114.1 KJ The values given are enthalpy of formation (Δ Hf°)

$+ 114.1 kj$

$+ 52.7 kj$

$- 52.7 kj$

$- 114.1 kj$

$2 N O_{(g)} + O_{2 (g)} ⇌ 2 N O_{2 (g)}$
$Δ H_{reaction}^{\circ} = \sum Δ H_{f (p ro d u c t s)} - Δ H_{f (re a c t an t s)}$
$= (2 \times 33.18) - (2 \times 90.25 + 0)$
$Δ H_{reaction}^{\circ} = - 114.1 K J$
The values given are enthalpy of formation $(Δ H_{f}^{\circ})$

Q. What will be the enthalpy of formation of NO2​ from the given bond dissociation enthalpy values ? The bond dissociation enthalpy values for O2​ , NO and NO2​ are as follows, O2(g)​ : 0kj/mol , NO(g)​ : 90.25kj/mol and NO2(g)​ : 33.18kj/mol respectively.

+114.1kj

+52.7kj

−52.7kj

−114.1kj

2NO(g)​+O2(g)​⇌2NO2(g)​ ΔHreaction∘​=∑ΔHf(products)​−ΔHf(reactants)​ =(2×33.18)−(2×90.25+0) ΔHreaction∘​=−114.1KJ The values given are enthalpy of formation (ΔHf∘​)

$+ 114.1 kj$

$+ 52.7 kj$

$- 52.7 kj$

$- 114.1 kj$

$2 N O_{(g)} + O_{2 (g)} ⇌ 2 N O_{2 (g)}$
$Δ H_{reaction}^{\circ} = \sum Δ H_{f (p ro d u c t s)} - Δ H_{f (re a c t an t s)}$
$= (2 \times 33.18) - (2 \times 90.25 + 0)$
$Δ H_{reaction}^{\circ} = - 114.1 K J$
The values given are enthalpy of formation $(Δ H_{f}^{\circ})$