Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation:
2 NaHCO3(s) <--> Na2CO3(s) + H2O(g) + CO2(g)
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 (Kp) is :
Kp = PH2O.PCO2
Kp = (3.88 atm)(3.88 atm)
Kp = 15.05 atm2 |
Pressure equilibrium constant (Kp) " 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.
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