When Solid NaHCO3 was placed in a rigid container witha volume of 2.5L and heated to 160 degrees Celsius, the equilibrium was reached.

Question

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!!

Answer 1

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.