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?

🤓 Based on our data, we think this question is relevant for Professor Stone's class at UMD.

FREE Expert Solution

Recall: 

K = productsreactantsK = PCO2

*Solids and liquids are ignored

BaCO3(s) ⇌ BaO(s) + CO2(g)

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


ΔG°rxn=-RTlnK


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


ΔG°rxn=ΔG°f, prod-ΔG°f, react



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°rxn=ΔG°f, prod-ΔG°f, react


ΔG°rxn=(1 mol BaO)520.41 kJ1 mol BaO+(1 mol  CO2)(394.38 kJ1 mol CO2)           -(1 mol BaCO3) 1137.63 kJ1 mol BaCO3

ΔG˚rxn222.84 kJ/mol


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


ΔG°rxn=222.84 kJmol×103 J1 kJ

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:


ΔG°rxn=-RTlnK


222840 Jmol=-8.314Jmol·K(298.15 K)(ln K)

lnK=222840  Jmol-2478.82 Jmol=-89.90


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?

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