When Solid NaHCO3 was placed in a rigid container witha volume of 2.5L and heated to 160 degrees Celsius, the equilibrium was reached.
2 NaHCO3(s) -> Na2CO3 (s) + CO2(g) + H2O(g)
At equilibrium, some of the starting material remains and the total pressure of the system is 7.68 bar.
A.) What are the partial pressures of each of the gases when equilibrium is reached?
B.) What is the value of Kp?
C.) What is Kc at 160 degrees Celsius
D.) The reaction is repeated, but this time, the container initially contains not only the solid reactant, but also CO2 with a partial pressure of 1 bar. When eq. is reached, what is the partial pressures of CO2 and H20?
THANK YOU FOR HELPING!!
My assumption is that you are to use bar as the standard pressure and not atm. If you want atm change bar to atm and recalculate accordingly.
2NaHCO3(s) ==> Na2CO3(s) + H2O(g) + CO2(g)
a. Ptotal = 7.68 bar = pH2O + pCO2
pH2O = pCO2 = 7.68/2 = ?bar
b. Substitute the pH2O and pCO2 into Kp expression and solve for Kp.
c. Kp = Kc(RT)^delta n.
d.
........2NaHCO3 ==> Na2CO3 + H2O + CO2
I.......solid.......solid.....0.....1
C.......solid-x.....solid+x..+x....+x
E.......solid-x.....solid+x..+x....1+x
You will note that solid +x and solid-x have no meaning with pressure but I used that anyway trying very hard not to confuse you.
Substitute the E line into the Kp expression and solve for x and 1+x
How do you know it goes to the right and not the left; that's because H2O is zero initially so the reaction MUST go to the right in order to increase H2O from zero.
Post your work if you get stuck.
Post your work if you get stuck.
To find the answers to these questions, we need to use the ideal gas law and equilibrium constants. Here's how we can solve each part:
A.) To find the partial pressures of each gas at equilibrium, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, let's calculate the number of moles of each gas using the balanced equation:
2 NaHCO3(s) -> Na2CO3 (s) + CO2(g) + H2O(g)
Since 2 moles of NaHCO3 produce 1 mole of CO2, 1 mole of CO2 is formed.
Therefore, the number of moles of Na2CO3 and H2O formed will also be 1 mole.
Assuming the volume of the rigid container is constant at 2.5L, we can substitute the values into the ideal gas law equation: PV = nRT.
For CO2:
P_CO2 x V = n_CO2 x R x T
P_CO2 x 2.5L = 1 mol x R x (160 + 273.15) K
Solving for P_CO2 gives us the partial pressure of CO2.
For H2O:
P_H2O x 2.5L = 1 mol x R x (160 + 273.15) K
Solving for P_H2O gives us the partial pressure of H2O.
B.) The value of Kp can be determined using the partial pressures of each gas at equilibrium. Kp is equal to the product of the partial pressures of the products (raised to the power of their stoichiometric coefficients) divided by the product of the partial pressures of the reactants (raised to the power of their stoichiometric coefficients).
In this case, Kp = (P_CO2 x P_H2O) / (P_NaHCO3)^2, since the stoichiometric coefficient of NaHCO3 is 2.
Substitute the partial pressure values obtained in Part A into this equation to find Kp.
C.) Kc is related to Kp through the ideal gas law equation: Kp = Kc(RT)^(∆n), where ∆n is the change in the number of moles of gas between products and reactants.
In this case, ∆n = (1 + 1 + 1) - (2 x 2) = 2 - 4 = -2, as two moles of NaHCO3 produce 1 mole of CO2.
Therefore, we can calculate Kc using the equation: Kc = Kp / (RT)^∆n
D.) In this scenario, the initial presence of CO2 at a partial pressure of 1 bar will affect the equilibrium.
To find the partial pressures of CO2 and H2O at equilibrium, we need to calculate the changes in the number of moles of CO2 and H2O based on the reaction stoichiometry.
Since the initial partial pressure of CO2 is given as 1 bar, we need to subtract this from the calculated partial pressure of CO2 found in Part A. This will give us the partial pressure of CO2 at equilibrium.
The partial pressure of H2O can be calculated directly by subtracting the partial pressure of CO2 calculated in Part A from the total pressure of the system at equilibrium (which is given as 7.68 bar).
I hope this explanation helps! Let me know if you have any further questions.