2NaHCO 3 (s) <--> Na 2 CO 3 (s) + H 2 O (g) + CO 2 (g) Solid sodium hydrogen car

Question

2NaHCO3(s) <--> Na2CO3 (s) + H2O (g) + CO2(g)

Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the reaction above.

a. A sample of 125. grams of solid NaHCO3 was placed in a previously evacuated rigid 4.00 L container and heated to 160 C. Some of the original solid remained and the total pressure in the container was 4.76 atmospheres when equilibrium was reached. Calculate the number of moles of H2O (g) present at equilibrium.

b. How many grams of the original solid remain in the container under the conditions described in (a)?

c. Calculate the value for the equilibrium constant for the reaction under the conditions in (a).

Explanation / Answer

a)

PV = nRT

n = PV/(RT)

n = (4.76*4)/(0.082*(160+273)) = 0.5362473 mol of gases

ratio is 1:1 so

mol of H2O = 1/2*0.5362473 = 0.26812365 mol of H2O

b)

mass of solid original

if 0.26812365 mol of H2O is present, then

1 mol of H2O = 2 mol of NaHCO3

0.26812365mol -> 0.5362473 mol of NaHCO3 reacted

mass = mol*MW = 0.5362473*84.007 = 45.04 g

now

dM = 125-45.04 = 79.96 g of NaHCO3 remains

c)

Keq = [H2O][CO2]

K = P-H2O * P - CO2

P-H2O =0.26812365 * 4.76 = 1.276 atm

P-CO2=0.26812365 * 4.76 = 1.276 atm

Kp =1.276 *1.276

Kp = 1.6281

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