Question

Consider the decomposition of Baco3(s) into Ba) and Co2(g). What is the eqilibrium partial pressure (in atm) of CO2 at 220.°c? ? atm

Answer 1

BaCO3(s)BaQ(s)+CO2(g) Enthalpy of reaction: AHo of reaction-YAH of (products)-ΣΑΗ of (reactants) AH of (BaQ) -553.5 kJ/mol AH of (CO2) 393.5 kJ/mol AH of (BaCO3)-1216.3 kJ/mol AHo (-947.0 kJ/mo) (-1216.3 kJ/mol) - +269.3 kJ/mol Entropy of reaction: ASo of reaction So of (products) So of (reactants) = [So (BAQ) + So (CO2)]-[S0 (BaCO3)] So of (BaO) = 0.0704 kJ/mol-K So of (CQ20.2136 kJ/mol-K So of (BaCO3) 0.112 kJ/mol-K ASO (+0.2840 kJ/mol-K) (+0.112 kJ/mol-K) - +0.1719 kJ/mol-K Now, we know that Where AGo Gibbs free energy AHo Enthalpy of reaction ASo- Entropy of reaction T- Temprature in Kelvin (K) -220 C 220 +213- 493 K

-(+269.3 k.J/mol (493 KX+0.1719 kJ/mol-K) l- +184.55 kJ/mol Now, we know that Where AGo Gibbs free energy R Universal gas constant -8.314 J/ol. K 0.008314 kJ/mol. K T- Temprature in Kelvin (k) Ksp Solubility product constant Sp e-AG/RT) ел [-(+184.55)(0.00831 4x493)] -2.79 x 10-20-picol