Question

At 1073 K, the equilibrium pressure of carbon dioxide above mixtures of calcium carbonate and calcium oxide is 183 torr. (a) If 5.0 g of solid calcium carbonate is placed in a 4000- mL container at 1073 K, give the final amounts of the various species. CaCO3 = Caos DOO CO2 = (b) The same as (a), except 0.50 g of CaCO3 is placed in the container. CaCO3 = DO Ca0 = CO2 =

Answer 1

Caco, (s) cao(s) + CO₂ (9) ; keq (equilibrium constant and at Equilibrium , Kp = Plo= 183 torv (latm = 760ton? = = 183 atm 760 = 0.24 atm

so, kp= 0.24 atm and we have sgm CaCO3 in 4000 m2 Contalher so mass of Ca CO₂ initially = 5gm taken molog mass of Calog = 100 gmimol so, moles of CaCO3 = for mol 100 = 0.05 mol Volume of containerlu)= 4000mL v = 42 Temperature (D = 1073K Caco (s) = CaO (s1 + CD, (3) 7=0 0.05 moles (initially at equilibrumos se ze z e

- at equilibriom if mots of CO2 A = 2 mol from PV=nRT s na part Prog = Kp = 0.24 atm R = 0.0821 atm-2 | gay-constant) mol-k n = 0.94-x 4 0:0821x1073 n= 0.0109 mols so = 0.0109 mols at Equilibrium 0.039 mous mous of mous of moly of CaCO3 = Cao = 0.0log mols cog= 0.0109 mols

so * (i mass of Caco = moly x molar mors 0.039 & 100 [3.81 pm Cao = moly x molar mars * (ii) Mars of = 0.0109 x 56 =1061 qm gr/mol molar mars of CqO= 56 * Clip) (iii) Mass of cog = 0-409 x 44 = [0.48 gm] molar mars of cog = 44 gm /mol

since keq is only function of temperature a temp. does not change in given Question so, kp remains constant er and Kp = Peoz = 0.14 atm so pressure of [Og at eam I also constant. moles of cog at pu equilibrium 0.94 x4 0.0821X 1073 = 0.0109 mous for equelibriom we should have o.0109 moles of cog at eam. so now Cowe(8) = Cad(s) + Wg (9) Q 0.5 mou Initial moly of ca Co₂ = 100 = 0.005 moles

but we have only 0.005 mous of calz so all the cacoz will conuert into cog and cao, and there is no Equilibrium becauye ay per stechiometry for equelibriom we should haue at wout 0-0109 mors of Ca CO₂ intially so that equilibriom can estabilised, so CQC,(s) 7-0 0.005 mol + - C90(s) + colg) - 0.005 mol 0.005 mol

so, finally * 0 Moley of Calog = Omol mass of CaCO3 = Ogn mols of CaO = 0.005 moly mass of cao = 0.005x50 gm = 0.98 gon 0.005 moles * 0 Moly of co= Mass of co= 0.005444 gm = 0.99 good