ΔG = -RTlnP
ΔG = 219.7 kJ
R = 8.314 J/mol*K = 0.008314 KJ/mol*K
lnP = [Pn(BaO)][Pn(CO2)]/ [Pn(BaCO3)]
= (1*1)/1 = 1
lnP = 219.7/ -(0.008314*298) = -88.68
P = e^(-88.68) = 3.07*10^(-39)
CO2 partial pressure = 3.07*10^(-39) 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?
1 answer