Question

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

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

1. A sample of 1.00 x 10² grams of solid NaHCO₃ was placed in a previously evacuated rigid 5.00-liter container and heated to 160 ⁰C. 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. How many grams of the original solid remain in the container under the conditions described in part (a)?
3. Write the equilibrium expression for the equilibrium constant, K_p, and calculate its value for the reaction under the conditions in part (a).
4. If 1.10 x 10² grams of solid NaHCO₃ had been placed in the 5.00-liter container and heated to 160 ⁰C, what would the total pressure have been at equilibrium? Explain.

Answer 1

Ans :-

(a). Given reaction at equilibrium is :

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

Given mass of NaHCO₃ = 1.00 x 10² g

Volume = V = 5.00 L

Temperature = T = 160 ⁰C = 433 K

Total pressure of gases mixture = P = 7.76 atm

Number of moles of gases mixture at equilibrium = n = ?

From ideal gas :

PV = nRT

n = PV/RT

n = (7.76 atm).(5.00 L) / (0.0821 L atm K^-1mol^-1).(433 K)

n = 1.09 mol

Because 2 moles of NaHCO₃ gives 1 mol of H₂O (g)

So,

Number of moles of H2O at equilibrium = nH2O = 1.09 mol / 2 = 0.545 mol

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(b). Moles of NaHCO₃ decomposed = 2 x 0.545 mol = 1.09 mol

So,

Mass of NaHCO₃ decomposed = Moles x Gram molar mass of NaHCO₃

= 1.09 mol x 84.007 g/mol

= 91.57 g

So,

Mass of Solid remains = 1.00 x 102 g - 91.57 g = 8.43 g

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(c). Expression of Pressure equilibrium constant (K_p) is :

K_p = P_H2O.P_CO2

K_p = (3.88 atm)(3.88 atm)

Kp = 15.05 atm2

Pressure equilibrium constant (K_p) " is equal to the ratio of partial pressure of products to the partial pressure of reactants raise to power stoichiometric coefficient at equilibrium stage of the reaction ".

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(d). Total pressure after the addition of 1.10 x 10² g of NaHCO₃ = 7.76 atm , because addition of solid do not affect the equilibrium stage.