Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation:

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

a. A sample of 1.00 x 10^2 grams of solid NaHCO3 was placed in a previously evacuated rigid 5.00-liter container and heated to 160 C. Some of the original solid remained and the total pressure in the container was 7.76 atmospheres when equilibrium was reached. Calculate the number of moles of H2O(g) present at equilibrium. (2 marks)

b. How many grams of the original solid remain in the container under the conditions described in part (a)? (2 marks)

c. Write the equilibrium expression for the equilibrium constant, Kp, and calculate its value for the reaction under the conditions in part (a). (2 marks)

d. If 1.10 x 10^2 grams of solid NaHCO3 had been placed in the 5.00-liter container and heated to 160 C, what would the total pressure have been at equilibrium? Explain. (2 marks)

As the only components exerting pressure are CO2 and Water (Because they are in gas phase), the total pressure can be splitted between the two of them so:

Pwater = Ptotal/2

Pwater = 7.76 / 2 = 3.88 atm

With this pressure, and using the ideal gas equation, we can know the moles of water:

PV = nRT

n = PV/RT using R = 0.082 L atm / K mol

n = 3.88 * 5 / 0.082 * (160+273)

n = 0.546 moles of water

b) now that we have the moles of water, we can actually know the moles that reacted originally from the sodium carbonate by stechiometry.

2NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g) MMCO2 = 84 g/mol

the moles of NaHCO3 initially:

n = 100 / 84

n = 1.19 moles

so, If 1.19 moles of NaHCO3 reacted, and only produces 0.546 moles of water and CO2, then, the remaining moles of NaHCO3 is:

remaining moles = 1.19 - 0.546 = 0.644 moles

therefore the mass remaining:

mCO2 = 0.644 * 84

mCO2 = 54.096 g

c) As it was stated before, only the gaseous components are involved in the pressure, thus, in the kp expression which is:

Kp = Pwater * PCO2

Kp = 3.88 * 3.88

Kp = 15.0544

d) As the total pressure is 7.76 atm and the fact that NaHCO3 is solid, this component is not exerting any pressure in the reaction, as seen in the Kp expression, so it won't matter that if we raise a little the quantity of the reactant, it still has some remaining.

pstnonsonjoku pstnonsonjoku · Answer 2 · 2022-02-22T11:49:44+01:00

Since NaHCO3 is a solid specie, it would not matter in the equilibrium equation.

The equation of the reaction is;

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

We know that the total pressure is 7.76 atm so;

Ptotal = Pwat + PCO2 (1) From Dalton's law for partial pressures.

But pressure can be related to number of moles as follows;

Ratio of mole = Ratio of pressure

Hence:

Pwater/PCO2 = moles water / moles CO2

Therefore:

Pwater = Ptotal/2

Pwater = 7.76 / 2 = 3.88 atm

From the ideal gas equation;

PV = nRT

n = PV/RT Where R = 0.082 L atm / K mol

n = 3.88 atm * 5 L / 0.082 L atm / K mol * (433 K)

n = 0.546 moles of water

b) Using the stoichiometry of the reaction;

2NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g) MMCO2 = 84 g/mol

Initial number of moles of NaHCO3:

n = 100 / 84

n = 1.19 moles

1.19 moles of NaHCO3 reacted, producing 0.546 moles of water and CO2 then remaining moles of NaHCO3 = 1.19 - 0.546 = 0.644 moles

So;

mass of CO2 = 0.644 * 84

= 54.096 g

c)Kp involves only the gaseous species hence:

Kp = PH2O * PCO2

Kp = 3.88 * 3.88

Kp = 15.0544

d)Since NaHCO3 is a solid specie, it would not matter in the equilibrium equation.

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