We’re being asked to determine the **equilibrium pressure CO _{2 }**at

**BaCO _{3} (s) **

The K expression for this reaction is:

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}{{\mathbf{P}}}_{{\mathbf{CO}}_{\mathbf{2}}}}$

**Note: Solids and liquids are not included in the equilibrium constant expression.*

Recall that * ΔG˚_{rxn} and K* are related to each other:

$\overline{){\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{RTlnK}}}$

We can use the following equation to solve for * ΔG˚_{rxn}*:

$\overline{){\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{\Delta H}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{-}}{\mathbf{T\Delta S}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}}$

We’re given the **ΔH˚ _{rxn} and ΔS˚_{rxn}** of the reaction from the appendix:

ΔH˚** _{rxn}** =

ΔS˚** _{rxn}** =

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 1300 K .

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