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

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.

To determine the equilibrium constant at 298 K, you can use the equation:

ΔG° = ΔH° - TΔS°

where:
ΔG° is the standard Gibbs free energy change,
ΔH° is the standard enthalpy change,
ΔS° is the standard entropy change, and
T is the temperature in Kelvin.

For the given reaction:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

The ΔG° for the reaction can be calculated as follows:

ΔG° = ΣΔG°f(products) - ΣΔG°f(reactants)

ΔG° = [2ΔG°f(Fe(s)) + 3ΔG°f(CO2(g))] - [ΔG°f(Fe2O3(s)) + 3ΔG°f(CO(g))]

Substituting the given values:

ΔG° = [2(0) + 3(-394.4 kJ/mol)] - [-742.2 kJ/mol + 3(-137.2 kJ/mol)]

ΔG° = -1183.2 kJ/mol - (-1163.8 kJ/mol)

ΔG° = -19.4 kJ/mol

Now, we can substitute the values into the equation:

ΔG° = ΔH° - TΔS°

-19.4 kJ/mol = -824.2 kJ/mol - T(87.4 J/K·mol + 3(197.7 J/K·mol))

Converting the units to match:
-19.4 kJ/mol = -824.2 kJ/mol - T(87,400 J/K·mol + 3(197,700 J/K·mol))

Simplifying:
-19.4 kJ/mol = -824.2 kJ/mol - T(682,900 J/K·mol)

Rearranging the equation:
-19.4 kJ/mol + 824.2 kJ/mol = T(682,900 J/K·mol)

804.8 kJ/mol = T(682,900 J/K·mol)

Now, we can solve for T:

T = (804.8 kJ/mol) / (682,900 J/K·mol)
T ≈ 1.18 K

Therefore, the equilibrium constant (K) at 298 K for the reaction Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) is approximately 1.18.

To determine the equilibrium constant at 298 K for the reaction between iron(III) oxide and carbon monoxide, we need to use the thermodynamic data provided.

The equilibrium constant (K) is defined by the relationship:

K = e^(-ΔG°/RT)

where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol⋅K)), and T is the temperature in Kelvin.

We can calculate ΔG° for the reaction using the data given:

ΔG° = ∑(n × ΔG°f,products) - ∑(n × ΔG°f,reactants)

where n is the stoichiometric coefficient of each substance in the balanced equation, and ΔG°f is the standard Gibbs free energy of formation.

For the given reaction:

ΔG° = [2 × ΔG°f(Fe)] + [3 × ΔG°f(CO2)] - [ΔG°f(Fe2O3)] - [3 × ΔG°f(CO)]

Substituting the given values:

ΔG° = [2 × 0] + [3 × (-394.4)] - [-742.2] - [3 × (-137.2)]
ΔG° = 0 + (-1183.2) + 742.2 + 411.6
ΔG° = -29.4 kJ/mol

Now, we can substitute the value of ΔG° into the equation for K:

K = e^(-ΔG°/RT)

Substituting the given temperature (298 K) and the value of ΔG°:

K = e^(-(-29.4 × 10^3 J/mol) / (8.314 J/(mol⋅K) × 298 K))
K = e^(88.6 / (8.314 × 298))
K = e^(0.411)

Using the scientific calculator, we find:

K = 1.507

Therefore, the equilibrium constant (K) at 298 K for the reaction between iron(III) oxide and carbon monoxide is approximately 1.507.