The equilibrium 3Fe(s)+ C(s) Fe₃C(s)is established in a solid solution.For such a solution, one can write an equilibrium constant expression in the usual way except that the concentrations that refer to solids in the solid solution are included.Determine the equilibrium constant for the formation of cementite from iron and carbon at 680°C.(Given: for this reaction at 25°C, $\Delta$H° = 21 kJ/mol and $\Delta$S° = 20.4 J/mol·K)

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the formation of cementite from iron and carbon is:
K = ([Fe]/[C][Fe]^(2/3))^(1/2)
At 680°C, the standard enthalpy change and standard entropy change can be converted to their values at this temperature using the equations:
ΔH = ΔH° + (T - 298)ΔCp
ΔS = ΔS° + ΔCp ln(T/298)
where ΔCp is the difference in heat capacity between the products and reactants, which can be taken as approximately constant over the temperature range of interest. Using the given values of ΔH° and ΔS°, and assuming ΔCp = 0, we get:
ΔH = 21 kJ/mol + (680 - 298)ΔCp = 42 kJ/mol
ΔS = 20.4 J/mol·K + ΔCp ln(680/298) = 63.8 J/mol·K
The standard Gibbs energy change at this temperature is then:
ΔG° = ΔH - TΔS = -91.8 kJ/mol
Since the equilibrium constant depends only on the Gibbs energy change, we can use the expression:
K = exp(-ΔG°/RT)
where R is the gas constant and T is the temperature in Kelvin. Substituting the values, we get:
K = exp(-(-91.8×10³)/(8.314×10^-3×680)) = 0.82
Therefore, the equilibrium constant for the formation of cementite from iron and carbon at 680°C is 0.82, which corresponds to choice (D).

The Equilibrium 3Fe(s)+ C(s) Fe3C(s)is Established in a Solid

The Equilibrium 3Fe(s)+ C(s) Fe₃C(s)is Established in a Solid