NaHCO3 decomposes to sodium carbonate, water vapor, and Carbon dioxide. 100g of

Question

NaHCO3 decomposes to sodium carbonate, water vapor, and Carbon dioxide. 100g of the reactant is placed in a 5L container, and the temperature is raised to 433 K.

2NaHCO3 -> Na2CO3 +CO2 + H2O

b) The total pressure in the container at 433 K is 7.76 atm once equilibrium is reached. calculate the number of moles of water vapor present at equilibrium.

p= 7.76 atm

T=433 K

v =5L

n=?

R=.08206

n=pv/rt

n=((7.76atm)(5L))/((0.08206)(433))

n=1.09mol << not sure if that is correct

c) Write the equilibrium expression for Kp and calculate its value at 433 K

d) Use LeChatelier's principle to predict what would happen to the partial pressure of carbon dioxide in the container if the container's volume were decresed.

e) Calculate Kc for this reaction

*Please explain in detail how to do B, C, D, E. Thank you.

Explanation / Answer

2NaHCO3 --> Na2CO3 + CO2 + H2O

b. 100g NaHCO3 / 84g/mole = 1.19moles

1.19moles NaHCO3 yields 0.6moles Na2CO3, CO2, H2O

the issue is that CO2 AND H2O are gases so what you found is the total number of moles of gas, not just H2O vapor

the mole fraction of H2Ov in the gases = 0.5 (H2O and CO2 are gases at a 1:1 ratio)

0.5 x 7.76 = 3.88atm = partial pressure H2Ov

n = pv/(rt) = 3.88atm x 5L / (0.0821L-atm/mole-K x 433K) = 0.535moles

using what you did, you solved for total moles and then take this and multiply by the mole fraction of H2Ov and you get 0.545moles. same thing different method

c. Kp = [CO2][H2O] = 0.286

d. partial pressure increases with decrease in volume

e. Kc = KpRT^n.....n here = 2 since it is the sum of the superscripts for CO2 and H2O, each being 1, so n = 2

Kc = 0.286 x (0.0821L-atm/mole-K x 433K)^2 = 362

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