Question

At high temperatures, boron carbide vaporizes according to

B₄​C(s)→4B(g)+C(s). At 2500 K, the equilibrium pressure of B(g) is 0.0342 mm Hg over a mixture of 0.300 mol B₄​C(s) and 0.500 mol C(s). Calculate ΔG° for this process.
[Hint: Determine the equilibrium constant first]

Answer 1

The question involves calculating the standard free energy change (ΔG°) for the vaporization of boron carbide at 2500 K using the equilibrium pressure of B(g).

Step-by-step explanation:

The question asks to calculate the standard free energy change (ΔG°) for the vaporization of boron carbide (B₄C) into boron gas (B(g)) and solid carbon (C(s)) at 2500 K, where the equilibrium pressure of B(g) is given. To determine ΔG°, we first need to calculate the equilibrium constant (Kp) for the reaction.

We can use the equilibrium pressure of B(g), 0.0342 mm Hg, in the expression for Kp because the carbon remains as a solid and does not contribute to the pressure.

The reaction is: B₄C(s) → 4B(g) + C(s). At equilibrium, the pressure of B(g) is 0.0342 mm Hg. Since the reaction produces 4 moles of B(g) for each mole of B₄C(s) that vaporizes, and B₄C(s) and C(s) are solids with activities taken as 1, the Kp expression is Kp = (PB)^4 = (0.0342)^4 (when PB is in atm, we must convert 0.0342 mm Hg to atm). Once Kp is determined, ΔG° can be calculated using the equation ΔG° = -RTln(Kp).