Question

help with this please

Ca(OH)2(aq) + 2 HCl(aq) +CaCl2(8) + 2 H2O(1) Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K ANSWER:

Answer 1

   Ca(OH)2(aq) + 2HCl(aq) -----> CaCl2(S) + 2H2O(l)

dG0rxn = (DG0f,CaCl2(S) + 2*DG0f,H2O(l))-(1*Dg0f,Ca(OH)2(aq) + 2*Dg0f,HCl(aq))

       = (-748.1 + 2*(-237.1))-(-868.1+2*(-131.2))

       = -91.8 kj

DG0rxn = - RTlnK

-91.8*10^3 = -8.314*298lnK

K = 1.235*10^16

N2(g) + O2(g) ----> 2NO2(g)

DG0rxn =(2*DG0f,NO2(g)) - (1*DG0f,N2(g) + 1*DG0f,O2(g))

       = (2*51.3)-(1*0+1*0)

       = 102.6 kj

DG0rxn = - RTlnK

102.6*10^3 = -8.314*298lnK

K = 1.035*10^-18

2CO(g) + O2(g) ----> 2CO2(g)

DG0rxn =(2*DG0f,CO2(g)) - (1*DG0f,O2(g) + 2*DG0f,CO(g))

       = (2*-386)-(1*0+2*-137.2)

       = -497.6 Kj

per 1.54 mole CO(g)

DG0rxn = -497.6*1.54/2 = -383.15 kj