Given the standard reaction enthalpies below:
N₂(g) + O₂ $\rightarrow$ 2NO(g)
$\Delta$ H° = +180.5 kJ.mol^-1
2NO₂(g) $\rightarrow$ N₂(g) + 2O₂(g)
$\Delta$ H° = -66.4 kJ.mol^-1
Calculate the standard reaction enthalpy for the oxidation of nitric oxide to nitrogen dioxide,shown below:
2NO(g) + O₂(g) $\rightarrow$ 2NO₂(g)

Question

Given the standard reaction enthalpies below:
N₂(g) + O₂

\rightarrow

2NO(g)

\Delta

H° = +180.5 kJ.mol^-1
2NO₂(g)

\rightarrow

N₂(g) + 2O₂(g)

\Delta

H° = -66.4 kJ.mol^-1
Calculate the standard reaction enthalpy for the oxidation of nitric oxide to nitrogen dioxide,shown below:
2NO(g) + O₂(g)

\rightarrow

2NO₂(g)

Quizplus · Accepted Answer

The enthalpy change for the desired reaction can be found by reversing the second reaction and adding it to the first. This gives: 2NO(g) + O2(g) → 2NO2(g) with ΔH° = 180.5 kJ/mol - 66.4 kJ/mol = 114.1 kJ/mol. Since the reaction is exothermic, the sign is negative: -114.1 kJ/mol.

Given the Standard Reaction Enthalpies Below: N2(g)+ O2 →\rightarrow→ 2NO(g) Δ\DeltaΔ

Given the Standard Reaction Enthalpies Below:
N₂(g)+ O₂ $\rightarrow$ 2NO(g) $\Delta$