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

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j