Question

In the following reaction, assume that the reaction enthalpy is independent of temperature, and the data given below are at 25°C. Calculate the standard reaction Gibbs energy and reaction enthalpy (kJ/mol) and the equilibrium constant at 298K and 450K. (a) ∆rG°(298K)=___________ (b)K(298K)=___________ (c)∆rH°(298K)=___________ (d)∆rG°(450K)=___________ (e)K(450K)=_____________

PbO(s) + CO(g) = Pb(s) + CO2(g).

∆fG°(KJmol-1) -188.93 -137.17 0 -394.36

∆fH°(KJmol-1) -218.99 -110.53 0 -393.51

Answer 1

       PbO(s) + CO(g) <----> Pb(s) + CO2(g)

a) at 298 k , DG0rxn = Products - reactants

           = (1*0 + 1*-394.36) - (1*-188.93 + 1*-137.17)

            = -68.26 kj

b) at 298 k , DG0= - RTlnK

                -68.26*10^3 = -8.314*298lnk

            k = equilibrium constant = 9.23*10^11

   c) at 298 k ,  

   DH0rxn = ( 1*0 + 1*-393.51)-(1*-218.99 +1*-110.53)

          = -63.99 kj

d) DG0 = DH0-TDS0

-68.26 = (-63.99)-(298*x)

x = DS0 = 0.0143 kj/mol.k

DG0 = DH0-TDS0

    = (-63.99)-(450*0.0143)

    = -70.425 Kj/mol

e) at 450 k , DG0= - RTlnK

                -70.425*10^3 = -8.314*450lnk

            k = equilibrium constant = 1.5*10^8