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?
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
To determine the equilibrium partial pressure of CO2 for the given reaction, we can use the equilibrium constant (Kp) and the ideal gas law.
First, let's find the expression for the equilibrium constant (Kp) for the reaction:
BaCO3(s) -> BaO(s) + CO2(g)
The stoichiometric coefficients of the balanced equation imply that the molar ratio of the products to the reactant is 1:1. Therefore, Kp can be expressed as:
Kp = (P(CO2)) / (P^0)
where P(CO2) is the equilibrium partial pressure of CO2, and P^0 is the standard pressure (which is typically 1 atm).
Since we know that the reaction is not spontaneous under standard conditions, ΔG must be positive. The relationship between ΔG and the equilibrium constant (K) is given by:
ΔG = -RT ln(K)
where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin. Let's convert ΔG from kJ to J by multiplying by 1000:
ΔG = 219.7 kJ × 1000 J/kJ = 219700 J
Now we can rearrange the equation to solve for ln(K):
ln(K) = -(ΔG / RT)
Substituting the values:
ln(K) = -(219700 J) / (8.314 J/mol·K × 298K)
= -90.49
Taking the exponential of both sides:
K = e^(ln(K)) = e^(-90.49)
= 1.26 × 10^(-39)
Finally, we can use the equilibrium constant (Kp) to find the equilibrium partial pressure of CO2 (P(CO2)):
Kp = (P(CO2)) / (P^0)
(1.26 × 10^(-39)) = (P(CO2)) / (1 atm)
P(CO2) = (1.26 × 10^(-39)) × (1 atm)
≈ 0 atm (negligible)
Therefore, when the reaction reaches equilibrium, the partial pressure of CO2 in the evacuated flask is practically zero.
To determine if a reaction is spontaneous under standard conditions, we need to calculate the standard Gibbs free energy change (ΔG°) of the reaction. You mentioned that you have already calculated the ΔG° value as 219.7 kJ.
Now, to determine the partial pressure of CO2 when the reaction reaches equilibrium, we need to use the equilibrium constant (Kp) of the reaction.
The general equation for the equilibrium constant relating the partial pressures of the gases involved in a reaction is:
Kp = (P_CO2) / (P_BaO * P_CO2),
where P_CO2 is the partial pressure of CO2, and P_BaO and P_CO2 are the partial pressures of BaO and CO2, respectively.
In this case, since the reaction is the decomposition of BaCO3, we can assume that the amount of BaO is negligible. Therefore, we can simplify the equation:
Kp = (P_CO2) / (P_CO2).
Simplifying further, Kp = 1.
At equilibrium, the reaction quotient (Q) is equal to the equilibrium constant (Kp). Therefore, we can write:
Q = (P_CO2) / (P_CO2) = Kp.
Substituting Kp = 1 into the equation, we get:
(P_CO2) / (P_CO2) = 1.
This equation shows that the partial pressure of CO2 will be the same as the partial pressure of CO2 at equilibrium. In other words, the partial pressure of CO2 when the reaction reaches equilibrium will be equal to the partial pressure of CO2 at standard conditions.
However, since the reaction is not spontaneous under standard conditions (ΔG° = 219.7 kJ), the reaction will not proceed to reach equilibrium. Therefore, no CO2 will be present when the reaction is in an evacuated flask.
In summary, the partial pressure of CO2 will be zero (0) when the reaction reaches equilibrium in an evacuated flask, as no CO2 will be present due to the non-spontaneous nature of the reaction under standard conditions.