# Problem: Consider the following reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2(g)If BaCO3 is placed in an evacuated flask, what partial pressure of CO2 will be present when the reaction reaches equilibrium?

###### FREE Expert Solution

Recall:

*Solids and liquids are ignored

BaCO3(s) ⇌ BaO(s) + CO2(g)

Recall that ΔG˚rxn and K are related to each other:

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

We can use the following equation to solve for ΔG˚rxn:

The values for ΔG˚f can be looked up in textbooks or online:

ΔG˚f, BaCO3(s) = 1137.63 kJ/mol

ΔG˚f, BaO(s)= –520.41 kJ/mol

ΔG˚f, CO2(g) = –394.38 kJ/mol

Note that we need to multiply each ΔG˚f by the stoichiometric coefficient since ΔG˚f is in kJ/mol.

Solving for ΔG˚rxn:

ΔG˚rxn222.84 kJ/mol

We also need to convert ΔG˚rxn from kJ to J:

ΔG˚rxn222,840 J/mol

We can now solve for the equilibrium constant using these given values:

ΔG˚rxn = 222,840 J/mol

R = 8.314 J/mol • K

T = 298.15 K

Solving for K:

$\mathbf{\Delta G}{\mathbf{°}}_{\mathbf{rxn}}\mathbf{=}\mathbf{-}\mathbf{RTlnK}$

Ln has a base of e and when you divide a number by ln you will take e to the power of the number.

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###### Problem Details

Consider the following reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2(g)

If BaCO3 is placed in an evacuated flask, what partial pressure of CO2 will be present when the reaction reaches equilibrium?