We are considering the decomposition of barium carbonate to calculate the equilibrium pressure of CO_{2} at 298K.

The equation for the decomposition of BaCO_{3} is the following:

BaCO_{3} (s) → BaO (s) + CO_{2}(g)

The equilibrium expression for the reaction is:

$\overline{){\mathbf{Kp}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathbf{P}}}_{{\mathbf{CO}}_{\mathbf{2}}}}$Solids are not included in the equilibrium expression

The **Standard Gibbs Free Energy of formation** of a compound is a change of Gibbs Free energy that accompanies the formation of 1 mole of that substance from its component elements, at their standard states (the most stable form of the element at 25C and 100 kpa).

(Data from books and the internet)

**∆G ^{o}_{f (BaO)} = -525.1 kJ/mol**

**∆G ^{o}_{f} _{(CO2)} = -394.4 kJ/mol**

**∆G ^{o}_{i(BaCO3)} = -1134.4 kJ/mol**

**The formula for the Gibbs Standard Free energy is given as:**

_{$\overline{){\mathbf{\u2206}}{{\mathbf{G}}}^{{\mathbf{o}}}{\mathbf{=}}{\mathbf{}}{\mathbf{\sum}}{\mathbf{n}}{\mathbf{\u2206}}{{\mathbf{G}}}^{{\mathbf{o}}}{\mathbf{f}}{\mathbf{}}{\mathbf{-}}{\mathbf{}}{\mathbf{}}{\mathbf{\sum}}{\mathbf{n}}{\mathbf{\u2206}}{{\mathbf{G}}}^{{\mathbf{o}}}{\mathbf{i}}}$}

Consider the decomposition of barium carbonate: BaCO_{3} (s) ⇄ BaO(s) + CO_{2} (g)

Using data from Appendix C in the textbook, calculate the equilibrium pressure of CO_{2} at 298 K.

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