Consider the decomposition of calcium carbonate: CaCO_3(s) CaO(s) + CO_2(g). An
ID: 1064112 • Letter: C
Question
Consider the decomposition of calcium carbonate: CaCO_3(s) CaO(s) + CO_2(g). An abundant amount of pure CaCO_3(s) is initially placed in a closed container maintained at constant temperature T = 1000K and pressure P = 1bar, where the gas phase is initially filled with N_2 (not involved in the reaction), what is the equilibrium mole fraction of CO_2? What if the pressure is maintained at P = 2bar? Assume the solids are completely immiscible and neglect gas solubility in the solids. Make and justify additional assumptions you may need. Formation reaction data for 1000K are given as follows.Explanation / Answer
Find out the free energy change of the reaction as
G0rxn = [G0f,0 (CaO) + G0f,0 (CO2)] – G0f,0 (CaCO3) = [(-531.09 kJ/mol) + (-395.81 kJ/mol(] – (-951.25 kJ/mol) = (-926.9 kJ/mol) + 951.25 kJ/mol = 24.35 kJ/mol.
Find the equilibrium constant for the reaction as
G0rxn = -RT.ln K
===> 24.35 kJ/mol = -(8.314 J/mol.K)*(1000 K)*ln K
===> 24.35 kJ/mol = -(8314 J/mol).ln K
===> 24.35 kJ/mol = -(8.314 kJ/mol).ln K
===> -2.9288 = ln K
===> K = e^(-2.9288) = 0.05346
The equilibrium constant for the reaction is 0.05346 (I have kept a few guard digits extra). Write down the equilibrium constant in terms of partial pressure of gaseous components as
K = PCO2 (since CO2 is the only gaseous species present and involved in the reaction)
===> 0.05346 = PCO2
===> PCO2 = 0.05346
The partial pressure of CO2 is 0.05346 atm. The total pressure is 1 bar = (1 bar)*(1.013 atm/1 bar) = 1.013 atm
Let x be the mole fraction of CO2 in the gaseous mixture of nitrogen and CO2.
0.05346 atm = (x).(1.013 atm) (since partial pressure = mole fraction*total pressure)
===> x = 0.05277 0.053
The mole fraction of CO2 in the gaseous mixture is 0.053 (ans).
When the total pressure is 2 bar, the mole fraction of CO2 will be 0.106. The logic is simple. Since the partial pressure of CO2 remains unchanged (temperature and hence K remains constant), the mole fraction must be doubled.