Use the following thermodynamic data at 298 K to determine the equilibrium constant at this temperature for the reaction of iron in steam to form hydrogen and iron(III) oxide.
2Fe(s) + 3H2O(g) 3H2(g) + Fe2O3(s)
Substance Fe(s) H2O(g) H2(g) Fe2O3(s)
ÄH°f (kJ/mol) 0 -242 -0 -824
ÄG°f (kJ/mol) 0 -229 -0 -742
S°(J/K·mol) 27 189 131 87
1.8 x 10-3
1.58 x 1022
4.36 x 10-9
4.36 x 109
i keep getting the wrong answer, well none of these answers, pls show working so i can figure it out myself.
To determine the equilibrium constant (K) for the reaction, we can use the standard free energy change (∆G°) at 298 K and the equation:
∆G° = -RT ln(K)
Where:
∆G° is the standard free energy change at 298 K
R is the gas constant (8.314 J/K·mol)
T is the temperature in Kelvin (298 K)
ln represents the natural logarithm
First, we need to calculate the ∆G° for the reaction. The ∆G° of the reaction can be calculated by summing the ∆G°f values of the products and subtracting the sum of the ∆G°f values of the reactants:
∆G° = (3 mol of H2(g) x ∆G°f of H2(g)) + (1 mol of Fe2O3(s) x ∆G°f of Fe2O3(s))
- (2 mol of Fe(s) x ∆G°f of Fe(s)) - (3 mol of H2O(g) x ∆G°f of H2O(g))
∆G° = (3 mol x 0 kJ/mol) + (1 mol x -742 kJ/mol) - (2 mol x 0 kJ/mol) - (3 mol x -229 kJ/mol)
∆G° = -742 kJ/mol + 687 kJ/mol
∆G° = -55 kJ/mol
Now, we can use the formula ∆G° = -RT ln(K) to solve for K:
-55 kJ/mol = -(8.314 J/K·mol) x (298 K) ln(K)
-55 kJ/mol = -2474.472 J ln(K)
Let's convert the units and solve for ln(K):
-55 kJ/mol = -2474472 J ln(K)
ln(K) = (-55 kJ/mol) / (-2474472 J)
ln(K) ≈ 2.2195
To find K, we can take the exponential of both sides:
K = e^(ln(K))
K = e^(2.2195)
K ≈ 9.204
Therefore, the equilibrium constant at 298 K for the reaction of iron in steam to form hydrogen and iron(III) oxide is approximately 9.204. None of the provided answer choices match this result. It is possible that there was an error while calculating the ∆G° or in the conversion of units.