How to calculate the equilibrium pressure of CO2 due to the decomposition of barium carbonate (BaCO3) at 25°C

Answer 1

First, write the reaction of interest:

BaCO3(s) -> BaO(s) + CO2(g)

We can write the equilibrium constant for this reaction as:

K = (a_BaO*a_CO2)/a_BaCO3

where a_xxx means the thermodynamic activity of xxx.

But by convention, the activities of pure solids are taken to be equal to 1, so:

K = a_CO2

where a_CO2 is the activity of CO2 gas at equilibrium.

If we assume CO2 behaves as an ideal gas, we can write the activity in terms of the equilibrium partial pressure of CO2:

a_CO2 = p_CO2/1 atm

where we have chosen the standard state for CO2 as the pure gas at 1 atm pressure.

You should also know the fundamental relationship between the standard free energy change of a reaction and the equilibrium constant for that reaction:

delta-G = -R*T*ln(K)

where delta-G is the standard Gibbs free energy change for the reaction, T is the thermodynamic (absolute) temperature, and R is the universal gas constant (= 8.3145 J/(K*mol)

In this case, we have that:

ln(p_CO2/1 atm) = -delta-G/(R*T)
or
p_CO2 = exp(-delta-G/(R*T)) atm

To get the free energy change for the reaction, you need to look up the standard free energies of formation of BaCO3, BaO, and CO2, and calculate delta-G for the reaction. I happen to have a handy reference (see source) that lists the following values for the free energies of formation at 25 C of these compounds as:

BaCO3(s): -1132.21 kJ/mol
BaO(s): -548.1 kJ/mol
CO2(g): -393.51 kJ/mol

So the free energy change for the decomposition reaction is:

(-393.51 + -548.1 - -1132.21) kJ/mol = 190.6 kJ/mol

Plugging this value, along with the values for R and T, into the expression above:

p_CO2 = exp((-1.906*10^5 J/mol)/(8.3145 J/(mol*K) * 298.15 K) atm

p_CO2 = 4.06 * 10^-34 atm