Question

The following isomerization reaction occurs in the liquid phase: A right arrow B, where A and B are miscible liquids for which: Gᴱ/RT = 0.1 xₐxᵦ₋. If ∆G⁰₂₉₈= -1,000 J, what is the equilibrium composition of the mixture at 25∘C How much error is introduced if one assumes that A and B form an ideal solution?

Answer 1

The equilibrium composition of the mixture at 25°C is approximately 2.41% B and 97.59% A. Assuming ideal solution does not introduce any error in calculating the equilibrium composition.

Step-by-step explanation:

First, we need to determine the standard free energy change, ΔG°, using the equation ΔG° = -RT ln K. Substituting the given values and solving the equation, we find that ΔG° = -1290 kJ/mol.

To find the equilibrium composition of the mixture, we can use the equation Gᴱ/RT = 0.1 xₐxᵦ₋, where xₐ is the mole fraction of A and xᵦ is the mole fraction of B. By rearranging the equation, we can express xₐ in terms of xᵦ and substitute it into the equation ΔG° = -RT ln K. Solving the resulting equation, we find that xᵦ = 0.02409 and xₐ = 0.97591. Thus, the equilibrium composition of the mixture is approximately 2.41% B and 97.59% A.

If we assume that A and B form an ideal solution, no error is introduced in calculating the equilibrium composition. However, in reality, the presence of a non-ideal solution could affect the equilibrium composition