For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.):

ΔH∘rxn 180.5kJ/mol
ΔS∘rxn 24.80J/(mol⋅K)
a. Calculate the temperature in kelvins above which this reaction is spontaneous

b. The thermodynamic values from part A will be useful as you work through part B:

ΔH∘rxn 180.5kJ/mol
ΔS∘rxn 24.80J/(mol⋅K)
Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C :

N2(g)+O2(g)→2NO(g)

a. 7278 K

b. 4.542 × 10⁻³¹

Explanation:

a.

Let´s consider the following reaction.

N₂(g) + O₂(g) ⇄ 2 NO(g)

The reaction is spontaneous when:

ΔG° < 0  [1]

Let's consider a second relation:

ΔG° = ΔH° - T × ΔS° [2]

Combining [1] and [2],

ΔH° - T × ΔS° < 0

ΔH° < T × ΔS°

T > ΔH°/ΔS°

T > (180.5 × 10³ J/mol)/(24.80 J/mol.K)

T >  7278 K

b.

First, we will calculate ΔG° at 25°C + 273.15 = 298 K

ΔG° = ΔH° - T × ΔS°

ΔG° = 180.5 kJ/mol - 298 K × 24.80 × 10⁻³ kJ/mol.K

ΔG° = 173.1 kJ/mol

We can calculate the equilibrium constant using the following expression.

ΔG° = - R × T × lnK

lnK = - ΔG° / R × T

lnK = - 173.1 × 10³ J/mol / (8.314 J/mol.K) × 298 K

K = 4.542 × 10⁻³¹

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.