Question

If baco3 is placed in an evacuated flask, what partial pressure of co2 will be present when the reaction reaches equilibrium? express your answer using two significant figures.

Answer 1

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 = $e^{-88.8}$ = 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 
$P_{CO_{2}}$ = 2.7 x 10⁻³⁹ atm   

Answer 2

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