Ca(HCO3)2(s) decomposes at elevated temperatures according to the stoichiometric equation Ca(HCO3)2(s)-> CaCO3(s) + H2O(g) + CO2(g)
a.) If pure Ca(HCO3)2(s) is put into a sealed vessel, the air is pumped out, and the vessel and its contents are heated, the total pressure is 0.235 bar. Determine Kp under these conditions.
b.) If the vessel initially also contains 0.105 bar H2O(g), what is the partial pressure of CO2(g) at equilibrium?
To answer these questions, we need to use the concept of equilibrium and the expression for equilibrium constant (Kp) for the given decomposition reaction.
a.) Determining Kp:
The equilibrium constant expression for the decomposition reaction is given by:
Kp = (P(CO2) * P(H2O)) / P(Ca(HCO3)2)
However, since the solid Ca(HCO3)2 does not contribute to the pressure, we can write the expression as:
Kp = (P(CO2) * P(H2O))
Given that the total pressure inside the sealed vessel is 0.235 bar, we substitute this value into the equilibrium constant expression:
0.235 bar = (P(CO2) * P(H2O))
b.) Determining the partial pressure of CO2 at equilibrium:
If the initial pressure of H2O(g) is 0.105 bar, then we can write:
0.235 bar = (P(CO2) * 0.105 bar)
Rearranging the equation to solve for P(CO2):
P(CO2) = 0.235 bar / 0.105 bar
P(CO2) ≈ 2.24 bar
Therefore, the partial pressure of CO2 at equilibrium is approximately 2.24 bar.
To solve these problems, we need to use the ideal gas law and the expression for equilibrium constant Kp.
a.) We know that the stoichiometric equation for the decomposition reaction is:
Ca(HCO3)2(s) -> CaCO3(s) + H2O(g) + CO2(g)
The equilibrium constant expression for this reaction is:
Kp = (P(CO2) * P(H2O)) / P(CaCO3)
Given the total pressure in the vessel is 0.235 bar, we can assume that the initial pressure of Ca(HCO3)2(s) is negligible compared to the other species. Therefore, the pressure of CaCO3(s) is considered to be constant and does not affect the equilibrium expression.
Let's set up the equation using the given information:
Kp = (P(CO2) * P(H2O)) / P(CaCO3)
Kp = (P(CO2) * P(H2O)) / (constant pressure)
Since the vessel is sealed, the total pressure inside is the sum of the partial pressures of CO2(g) and H2O(g):
P(Total) = P(CO2) + P(H2O)
Substituting this expression into the Kp equation, we get:
Kp = (P(CO2) * P(H2O)) / (P(Total) - P(CO2))
Substituting the given values:
Kp = (P(CO2) * 0.105 bar) / (0.235 bar - P(CO2))
b.) If the vessel initially contains 0.105 bar H2O(g), the equilibrium expression becomes:
Kp = (P(CO2) * 0.105 bar) / (0.235 bar - P(CO2))
To find the partial pressure of CO2(g) at equilibrium, we need to solve for it. Rearranging the equation, we get:
Kp * (0.235 bar - P(CO2)) = P(CO2) * 0.105 bar
0.235 Kp - Kp * P(CO2) = 0.105 P(CO2)
0.235 Kp = 0.105 P(CO2) + Kp * P(CO2)
0.235 Kp = P(CO2) * (0.105 + Kp)
P(CO2) = (0.235 Kp) / (0.105 + Kp)
So, the partial pressure of CO2(g) at equilibrium is (0.235 Kp) / (0.105 + Kp).
Ca(HCO3)2(s)-> CaCO3(s) + H2O(g) + CO2(g)
Total pressure is 0.235 bar. Since the mole ratio is 1:1, the partial pressure H2O and partial pressure of CO2 must be the same; therefore, 2x = 0.235 and x = 0.235/2.
Then Kp = P(CO2)*P(H2O) = ??
b).
.....CO2 + H2O
...0.105 bar..0
.... + x....+x
final is 0.105+x and x
Substitute into Kp expression and solve for x.