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?
To find the partial pressure of CO2 when the reaction reaches equilibrium, we need to use the equilibrium constant, Kp, which can be determined using the relationship between the Gibbs free energy change (ΔG°) and the equilibrium constant:
ΔG° = -RT * ln(Kp)
Where ΔG° is the Gibbs free energy change under standard conditions, R is the gas constant (8.314 J/mol*K), T is the temperature (298 K), and Kp is the equilibrium constant in terms of partial pressures. First, we need to solve for Kp:
-219,700 J/mol = -(8.314 J/mol*K) * (298 K) * ln(Kp)
Now solve for Kp:
ln(Kp) = -219,700 J/mol / (8.314 J/mol*K * 298 K)
ln(Kp) ≈ 8.96
Kp = e^(8.96)
Kp ≈ 7800
Now that we have the value for Kp, we can write the expression for the equilibrium constant in terms of partial pressures:
Kp = (PCO2)/(1) = PCO2
Since BaCO3 is a solid and BaO is a solid, their activities are equal to one, and they do not appear in the expression for Kp. The partial pressure of CO2 (PCO2) at equilibrium is equal to the Kp value:
PCO2 = Kp = 7800 atm
So, when the reaction reaches equilibrium, the partial pressure of CO2 in the flask will be around 7800 atm.
To determine if a reaction is spontaneous under standard conditions, we examine the sign of the Gibbs free energy change (ΔG). If ΔG is negative, the reaction is spontaneous, whereas if ΔG is positive, the reaction is non-spontaneous.
You have already calculated ΔG of the reaction to be 219.7 kJ. However, since you mentioned that the reaction is not spontaneous under standard conditions, it means that ΔG is positive (ΔG > 0).
Now, you want to determine the partial pressure of CO2 at equilibrium when BaCO3 is placed in an evacuated flask. To do this, we can use the relationship between ΔG and equilibrium constant (K).
The equation connecting ΔG and K is:
ΔG = -RTln(K)
Where:
ΔG = Gibbs free energy change
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
K = equilibrium constant
Since we are given the temperature (298K) and ΔG (219.7 kJ), we can rearrange the equation to solve for ln(K):
ln(K) = -ΔG / (RT)
Now, convert ΔG from kJ to J:
ΔG = 219.7 kJ × 1000 J/kJ = 219700 J
Substituting the values into the equation:
ln(K) = - (219700 J) / (8.314 J/(mol·K) × 298K)
ln(K) ≈ -30.02
To find K, we take the exponent of both sides:
K = e^ln(K) ≈ e^(-30.02)
K ≈ 1.016 × 10^(-13)
For the reaction at equilibrium, the expression for the equilibrium constant (Kc) can be written as:
Kc = [CO2] / [BaO]
Since the reaction starts with only BaCO3, we assume that the initial concentration of CO2 is 0. Therefore, at equilibrium, the concentration of CO2 will be equal to the partial pressure of CO2 (P(CO2)).
Now, let's assume the partial pressure of CO2 at equilibrium is P.
Therefore, Kc = P / [BaO]
Substituting the value of Kc:
1.016 × 10^(-13) = P / [BaO]
From the stoichiometry of the reaction, the mole ratio between CO2 and BaO is 1:1. Therefore, the partial pressure of CO2 at equilibrium will be equal to the partial pressure of BaO.
So, the partial pressure of CO2 present when the reaction reaches equilibrium in an evacuated flask will be approximately 1.016 × 10^(-13) atmospheres.
To determine if the reaction is spontaneous under standard conditions, we need to evaluate the ΔG° (standard Gibbs free energy change) of the reaction. The ΔG° is calculated using the equation:
ΔG° = ΔH° - TΔS°
Where:
ΔH° = standard enthalpy change
T = temperature in Kelvin (298 K in this case)
ΔS° = standard entropy change
Given that the ΔG° is 219.7 kJ, we can use this information to determine the spontaneity of the reaction. If ΔG° is positive, the reaction is not spontaneous under standard conditions.
Now, to calculate the partial pressure of CO2 at equilibrium, we can use the reaction's equilibrium constant (Kp). The equilibrium constant expression for the reaction is:
Kp = (PCO2 ^ 1)
Where:
PCO2 = partial pressure of CO2 in the flask
Since the reaction is given as:
BaCO3(s) → BaO(s) + CO2(g)
We know that the stoichiometry of the reaction implies a 1:1 ratio between BaCO3 and CO2. This means that the partial pressure of CO2 at equilibrium is equal to the initial pressure of BaCO3.
Therefore, the partial pressure of CO2 at equilibrium, when BaCO3 is placed in an evacuated flask, will be equal to the initial pressure of BaCO3.