Answer:
The temperature above which this reaction is spontaneous is 7,278.22 K.
Explanation:
The Gibbs free energy is given by:
ΔG°=ΔH° - TΔS°
ΔG° = Gibbs free energy at temperature T
ΔH° = Enthalpy of the reaction
ΔS° = Entropy of the reaction.
We have :
ΔH° = 180.5kJ/mol = 180500 J/mol, ΔS° = 24.80J/(mol⋅K)
for reaction to be spontaneous, ΔG° should be less than 0.
ΔG°<0
ΔG°=ΔH° - TΔS°
ΔH° - TΔS° < 0
ΔH° < 0 + TΔS°
[tex]\frac{\Delta H^o}{\Delta S^o}<T[/tex]
[tex]\frac{180500 J/mol}{24.80J/(mol K)}<T[/tex]
7,278.22 K < T
The temperature above which this reaction is spontaneous is 7,278.22 K.
The temperature is mathematically given as
K = 4.542 × 10^{-31} K
Question Parameters:
ΔH∘rxn 180.5kJ/mol
ΔS∘rxn 24.80J/(mol⋅K)
Generally, the equation for the Chemical Reaction is mathematically given as
N2(g) + O2(g) ----><---- 2 NO(g)
Therefore
dH - T * dS < 0
dH < T * dS T > dH/dS
T > (180.5 × 10^3l)/(24.80)
T > 7278 K
b)
Where,
dG = ΔH - T * dS
dG = 180.5 kJ/mol - 298 K * 24.80 × 10^{-3} kJ/mol.K
dG = 173.1 kJ/mol
In conclusion, the equation for the equilibrium constant is mathematically given as
dG = - R * T * lnK
lnK = - 173.1 × 10^3 / (8.314) * 298 K
K = 4.542 × 10^{-31}
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