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.

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

i believe it is 40–40–20%

explanation: