Answer :
The equation representing the decomposition of BaCO₃ is as follow:
BaCO₃(s) ⇄ BaO(s) + CO₂(g)
Standard Gibbs-free energy (ΔG⁰) = 220 kJ/mol
Gas constant (R) = 8.314 x 10⁻³ kJ/K.mol
Temperature (T) =25 + 273 = 298 K
ΔG⁰ = - R T lnK
ln K = - ΔG⁰ / RT = - 88.8
K = [tex] e^{-88.8} [/tex] = 2.7 x 10⁻³⁹
So the equilibrium constant of the reaction at 298 K is 2.7 x 10⁻³⁹
From the equation:
Equilibrium constant Kp = Pco₂
[tex] P_{CO_{2}} [/tex] = 2.7 x 10⁻³⁹ atm
BaCO₃(s) ⇄ BaO(s) + CO₂(g)
Standard Gibbs-free energy (ΔG⁰) = 220 kJ/mol
Gas constant (R) = 8.314 x 10⁻³ kJ/K.mol
Temperature (T) =25 + 273 = 298 K
ΔG⁰ = - R T lnK
ln K = - ΔG⁰ / RT = - 88.8
K = [tex] e^{-88.8} [/tex] = 2.7 x 10⁻³⁹
So the equilibrium constant of the reaction at 298 K is 2.7 x 10⁻³⁹
From the equation:
Equilibrium constant Kp = Pco₂
[tex] P_{CO_{2}} [/tex] = 2.7 x 10⁻³⁹ atm
The correct answer is 2.7 x 10^-39 atm
The expression:
when BaCO3 is placed in an evacuated flask, So the reaction equation of the decomposition of the BaCO3 is as the following:
BaCO3(s) ⇄ BaO(s) + CO2(g)
so the Kp expression for this reaction is:
Kp = P(CO2)
because Kp expression is including only (g) gas phase.
-now we need to get the value of K by using the following formula:
ΔG⁰ = - R T lnK
when Standard Gibbs-free energy (ΔG⁰) = 220 kJ/mol
and Gas constant (R) = 8.314 x 10⁻³ kJ/K.mol
and Temperature (T) is in kelvin =25 + 273 = 298 K
ln K = - ΔG⁰ / RT = - 88.8
∴ K = e^{-88.8} = 2.7 x 10⁻³⁹
so, from the Kp expression ∴ P(CO2) = Kp
∴ P (CO2) = 2.7 x 10^-39 atm