If BaCO3 is placed in an evacuated flask, what partial pressure of CO2 will be present when the reaction reaches equilibrium?
298K BaCO3(s)-->BaO(s)+CO2(g)
2.7×10−39 atm
Can't you look up values of delta Go at 25C, find delta Go for the rxn;
delta Go rxn = (sum delta Go product)-(sum delta Go reactants).
Tben delta Go = -RTlnK
Use delta Go to calculate K and from there
Kp = PCO2
no the answer was wrong
Ally did you convert the kJ you would have most likely calculated for your change G0rxn to J to make the R constant work properly?
To determine the partial pressure of CO2 at equilibrium, we need to consider the principles of equilibrium and the stoichiometry of the reaction.
The first step is to write the balanced chemical equation:
BaCO3(s) → BaO(s) + CO2(g)
The equation tells us that one mole of BaCO3 produces one mole of BaO and one mole of CO2.
Since the reaction is taking place in an evacuated flask (meaning there is no initial pressure of CO2), the partial pressure of CO2 at equilibrium will depend on the number of moles of CO2 produced and the volume of the flask.
Given that the reaction has reached equilibrium at 298K, we can consider the ideal gas law: PV = nRT, where P is the partial pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
First, we need to calculate the number of moles of CO2 produced. Since one mole of BaCO3 produces one mole of CO2, we can assume that the number of moles of CO2 is equal to the number of moles of BaCO3.
Next, we need to know the volume of the flask. Without that information, it is not possible to calculate the partial pressure of CO2.
To summarize, we need to know the moles of BaCO3 and the volume of the flask to calculate the partial pressure of CO2 at equilibrium.