Question

5.) (Taken from Chemistry: A Molecular Approach, 1^st Ed. by Tro)

                Consider the following reaction at 298 K:

BaCO₃(s) --> BaO(s) + CO₂(g)

                (a) Show that this reaction is not spontaneous at standard conditions.

                (b) If BaCO₃ is placed in an evacuated flask, what partial pressure of CO₂ will be present when

                      the system reached chemical equilibrium?

                (c) Can the reaction be made more spontaneous by an increase or decrease in temperature? If so,

                      what temperature is required to produce a CO₂ partial pressure of 1.0 atm?

Answer 1

For the decomposition of BaCO3

(a) dGorxn = dGo(products) - dGo(reactants)

                  = (-520.4 - 394.6) - (-1134.4)

                  = 219.4 kJ/mol

As the sign for dG is +ve, the reaction would be non-sponatenous under standard condition.

(b) If BaCO3 is evacuated, CO2 partial pressure,

dGo = -RTlnK

219400 = -8.314 x 298 ln(pCO2)

[pCO2] = 3.48 x 10^-39 atm

(c) The enthalpy change for the reactin,

dHrxn = (-393.3 - 548.1) - (-1213) = 271.6 kJ/mol

The reaction is thus endothermic in nature and would be spontaneous at higher temperature.

To produce a pCO2 = 1.0 atm

T = 273 K would give a partial pressure of CO2 = 1 atm