Using the data in Appendix 3, calculate the standard entropychanges for the foll

Question

Using the data in Appendix 3, calculate the standard entropychanges for the following reactions at 25 degree C. (a) Pb(s)+ O2(g) BaO(s) +CO2(g) 2 J/K PbO2(s) 1 J/K (b) BaCO3(s)

Explanation / Answer

BaCO3(s) -> BaO(s) + CO2(g) We can write the equilibrium constant for this reaction as: K = (a_BaO*a_CO2)/a_BaCO3 where a_xxx means the thermodynamic activity of xxx. But by convention, the activities of pure solids are taken to beequal to 1, so: K = a_CO2 where a_CO2 is the activity of CO2 gas at equilibrium. If we assume CO2 behaves as an ideal gas, we can write the activityin terms of the equilibrium partial pressure of CO2: a_CO2 = p_CO2/1 atm where we have chosen the standard state for CO2 as the pure gas at1 atm pressure. You should also know the fundamental relationship between thestandard free energy change of a reaction and the equilibriumconstant for that reaction: delta-G = -R*T*ln(K) where delta-G is the standard Gibbs free energy change for thereaction, T is the thermodynamic (absolute) temperature, and R isthe universal gas constant (= 8.3145 J/(K*mol) In this case, we have that: ln(p_CO2/1 atm) = -delta-G/(R*T) or p_CO2 = exp(-delta-G/(R*T)) atm To get the free energy change for the reaction, you need to look upthe standard free energies of formation of BaCO3, BaO, and CO2, andcalculate delta-G for the reaction. I happen to have a handyreference (see source) that lists the following values for the freeenergies of formation at 25 C of these compounds as: BaCO3(s): -1132.21 kJ/mol BaO(s): -548.1 kJ/mol CO2(g): -393.51 kJ/mol So the free energy change for the decomposition reaction is: (-393.51 + -548.1 - -1132.21) kJ/mol = 190.6 kJ/mol Plugging this value, along with the values for R and T, into theexpression above: p_CO2 = exp((-1.906*10^5 J/mol)/(8.3145 J/(mol*K) * 298.15 K)atm p_CO2 = 4.06 * 10^-34 atm

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