At 125 C Kp=0.254 for the reaction 2NaHCO3 (s) <-> Na2CO3 (s) + CO2 (g) + H2O (g

Question

At 125 C Kp=0.254 for the reaction 2NaHCO3 (s) <-> Na2CO3 (s) + CO2 (g) + H2O (g). At 2.50 L flask containing a substantial amount of NaHCO3 is evacuated and heated to 125 C. what mass of Na2CO3 (molar mass=106.0 g/mol) is also produced when equilibrium is established? At 125 C Kp=0.254 for the reaction 2NaHCO3 (s) <-> Na2CO3 (s) + CO2 (g) + H2O (g). At 2.50 L flask containing a substantial amount of NaHCO3 is evacuated and heated to 125 C. what mass of Na2CO3 (molar mass=106.0 g/mol) is also produced when equilibrium is established? At 125 C Kp=0.254 for the reaction 2NaHCO3 (s) <-> Na2CO3 (s) + CO2 (g) + H2O (g). At 2.50 L flask containing a substantial amount of NaHCO3 is evacuated and heated to 125 C. what mass of Na2CO3 (molar mass=106.0 g/mol) is also produced when equilibrium is established?

Explanation / Answer

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Special notes:

Typically; we use aqueous and gas phases. (Gases and Aqueous Concentrations can be related via PV = nRT, since M = n/V as well)

Solids and Liquids, i..e. (s) and (l) have activity of 1. Therefore, they must not be considered in the ratios.

then

Kp = P-CO2 * P-H2O

Kp = 0.254

V = 2.5 L ., T = 125°C

Kp = P-CO2 * P-H2O

0.254 = P-CO2 * P-H2O

P-CO2 = P-H2O, due to stoichiometry

P-CO2 ^2 = 0.254

P-CO2 = sqrt(0.254) = 0.50398 atm

now...

PV = nRT

n = PV/(RT)

n = (0.50398)(2.5)/(0.082*(125+27)) = 0.101087 mol of CO2

ratio is

1 mol of CO2 = 1 mol of Na2CO3

0.101087mol of CO2 = 0.101087 mol of Na2CO3

mass of Na2CO3 = mol*MW = 0.101087*105.9884 = 10.714 g of Na2CO3

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