Question

Substance BaCO3(s) | BaO(s) CO2(g) BaCO3(s) = Bao(s) + CO2(g) A.HⓇ/kJ mol -1213.0 -548.0 -393.5 Sº / 1 mol K- 112.1 72.1 213.8 a. Using the information in the table above, show that the reaction is not spontaneous under standard conditions by calculating 4,Gº. {5 marks) b. If Baco, is placed in an evacuated flask, what partial pressure of CO2 will be present when the reaction reaches equilibrium? (3 marks) c. Can the reaction be made more spontaneous by an increase or decrease in temperature? If so, what temperature is required to make the reaction spontaneous under standard conditions? {3 marks) d. What temperature is required to produce a carbon dioxide partial pressure of 1.5 bar? (3 marks)

Answer 1

GS given BaCO₂ (s) Baois) & CO (9) substance DH Kimalt s / I mot kt Baco, -1213.0 112.0 Bao - 548.0 72.1 со, -393.5 213.8 Solution: W.KT 19 - AH - TAS° 6 AH - EDH poonducts ΣΔΗ reactants [-3935+ (-548.0)]- (-1213 11 941.5 + 1213 AH 271.5 kJ /mol = 271.5 x103J mol } EDs products Ens oily reactants Aso (213.8+52:1)-(112.0) 173.9 Imottat =

.. At standard conditions T-213K :D 271.5 x103 – (273 + 173.91 (271.5x10 47.48 x103) I mott 224.02 Ilmol : Ago =tve Therefore the reaction is non-spon- - taneous at standard conditions b be written as Partial pressure equation for the given reaction can KF Elco Kp = Per ko equilibrium constant Peg Partical pressure of Cog (since Bacy, and Bao au solids they are not in rolud en Kp reaction) R = constant = 8.314 J kt mot AG ==RTinka T= 273K ar in Kp -ng' RT 224.02 Imett = 29.87x102) 8.3148 mot x 273 x

Kp = antiin (-9.87X102) 0.906 . kp 0.906 Pcos Above reaction can be made spontaneous by increasing the temperature. if Reg = 1.5 bar at what temperature Tza W..I AG= – RT Inkp 224.02 IMOT -8.3145k mott & T x 100.5) since Kp = (co - 15 T 224.02 -8.314ktx 1n (15) - 66.45K T - 66.45 K