Equilibrium total pressure in vessel is (in bar)

Question

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Question Explanation: Finding equilibrium pressure using Gibbs free energy.

Concept: Gibbs Free Energy and Equilibrium.

Solution:

$T = 27 + 273 = 300 K$ .

First, calculate $Δ G^{\circ}$ using the relation $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ .

$Δ G^{\circ} = 57900 - (300 \times 200) = 57900 - 60000 = - 2100 c a l / m o l$ .

Use $Δ G^{\circ} = - RT ln K_{p}$ to find the equilibrium constant ( $K_{p}$ ).

$- 2100 = - 2 \times 300 \times ln K_{p} \Rightarrow ln K_{p} = 3.5$ .

Since $ln 2 = 0.7$ , then $3.5 = 5 \times 0.7 = 5 ln 2 = ln 2^{5}$ . Thus, $K_{p} = 32$ .

Relate $K_{p}$ to the partial pressures of the gaseous products to find the total pressure.

Reaction: $2 A (s) \to 2 B (g) + C (g)$ . If $P_{C} = p$ , then $P_{B} = 2 p$ .

$K_{p} = (P_{B})^{2} (P_{C}) = (2 p)^{2} (p) = 4 p^{3} = 32 \Rightarrow p^{3} = 8 \Rightarrow p = 2 ba r$ .

Total Pressure $P_{t o t a l} = 2 p + p = 3 p = 3 \times 2 = 6 ba r$ .

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