To find the partial pressure of CO2 when the reaction reaches equilibrium, we need to use the equilibrium constant, Kp, which can be determined using the relationship between the Gibbs free energy change (ΔG°) and the equilibrium constant:
ΔG° = -RT * ln(Kp)
Where ΔG° is the Gibbs free energy change under standard conditions, R is the gas constant (8.314 J/mol*K), T is the temperature (298 K), and Kp is the equilibrium constant in terms of partial pressures. First, we need to solve for Kp:
-219,700 J/mol = -(8.314 J/mol*K) * (298 K) * ln(Kp)
Now solve for Kp:
ln(Kp) = -219,700 J/mol / (8.314 J/mol*K * 298 K)
ln(Kp) ≈ 8.96
Kp = e^(8.96)
Kp ≈ 7800
Now that we have the value for Kp, we can write the expression for the equilibrium constant in terms of partial pressures:
Kp = (PCO2)/(1) = PCO2
Since BaCO3 is a solid and BaO is a solid, their activities are equal to one, and they do not appear in the expression for Kp. The partial pressure of CO2 (PCO2) at equilibrium is equal to the Kp value:
PCO2 = Kp = 7800 atm
So, when the reaction reaches equilibrium, the partial pressure of CO2 in the flask will be around 7800 atm.
Consider the following reaction occurring at 298K
BaCO3(s)-> BaO(s) + CO2 (g)
show that the reaction is not spontaneous under standard condition by calculating delta G rxn.
I calculated delta G and got 219.7 kJ, now the question asks 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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