Answer:
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
To determine the Keq, we need the chemical reaction in the system. In this case it would be:
CO + 2H2 = CH3OH
The Keq is the ration of the amount of the product and the reactant. We use the ICE table for this. We do as follows:
CO H2 CH3OH
I .42 .42 0
C -0.13 -2(0.13) 0.13
-----------------------------------------------
E = .29 0.16 0.13
Therefore,
Keq = [CH3OH] / [CO2] [H2]^2 = 0.13 / 0.29 (0.16^2)
Keq = 17.51
