First, we need to find the ΔH° and ΔG° for the reaction.
For the reaction, use the equation:
ΔH°(reaction) = Σ ΔH°(products) - Σ ΔH°(reactants)
ΔH°(reaction) = [2(0) + 3(-393.5)] - [-824.2 + 3(-110.5)]
ΔH°(reaction) = [-1180.5] - [-1155.7]
ΔH°(reaction) = -24.8 kJ/mol
ΔG°(reaction) = Σ ΔG°(products) - Σ ΔG°(reactants)
ΔG°(reaction) = [2(0) + 3(-394.4)] - [-742.2 + 3(-137.2)]
ΔG°(reaction) = [-1183.2] - [-1153.8]
ΔG°(reaction) = -29.4 kJ/mol
Next, we need to find the ΔS° for the reaction.
ΔS°(reaction) = Σ S°(products) - Σ S°(reactants)
ΔS°(reaction) = [2(27.78) + 3(213.7)] - [87.4 + 3(197.7)]
ΔS°(reaction) = [623.94] - [680.5]
ΔS°(reaction) = -56.56 J/K•mol
Now, we can find the equilibrium constant at 298 K using the equation:
ΔG°(reaction) = -RT ln(K)
where R is the gas constant (8.314 J/K•mol) and T is the temperature in Kelvin (298 K).
First, convert ΔG° to J/mol:
ΔG° = -29.4 kJ/mol * 1000 J/kJ = -29400 J/mol
Now, solve for K:
-29400 J/mol = -(8.314 J/K•mol)(298 K) ln(K)
ln(K) = -29400 / [(8.314)(298)]
ln(K) = 12.07
Take the exponent of both sides to get the equilibrium constant:
K = e^(12.07)
K ≈ 1.74 x 10^5
The equilibrium constant, K, at 298 K is approximately 1.74 x 10^5.
Iron(III) oxide can be reduced by carbon monoxide.
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
Use the following thermodynamic data at 298 K to determine the equilibrium constant at this temperature.
Substance: Fe2O3(s) CO(g) Fe(s) CO2(g)
ΔH° f (kJ/mol): -824.2 -110.5 0 -393.5
ΔG° f (kJ/mol): -742.2 -137.2 0 -394.4
S°(J/K•mol): 87.4 197.7 27.78 213.7
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