Question

From the information in the data section, calculate the standard Gibbs energy and the equilibrium constant at 298 K and 500 K for the reaction CuO(s) + CO (g) ó Cu(s) + CO₂.  

Answer 1

Given reaction is CuO(s) + CO (g) ------------> Cu(s) + CO₂

ΔG_f^o [CO2(g)] = - 394.36 kJ/mol

ΔG_f^o [CO(g)] = - 137.17 kJ/mol

ΔG_f^o [Cu(s) ] = 0 kJ/mol

ΔG_f^o [CuO(s) ] = -129.7 kJ/mol

ΔG^orxn = ΔG_f^o(products) - ΔG_f^o(reactants)

            = ΔG_f^o [ Cu(s)] +ΔG_f^o [CO2(g)] - {ΔG_f^o [CuO(s)]+  ΔG_f^o [CO (g)] }

             = 0 + (- 394.36 kJ/mol ) - { -129.7 kJ/mol + -137.17 kJ/mol}

             = - 124.96 kJ/mol

Therefore, ΔG^orxn = - 124.96 kJ/mol

Equilibrium constant at 298 K:

ΔG^orxn = - 124.96 kJ/mol = - 124960 J/mol

ΔG^orxn = -RT InK

K = e^-[-124960/ (8.314) (298) ]

    = 8.02 x 10²¹

Therefore,

Equilibrium constant at 298 K = 8.02 x 10²¹

Equilibrium constant at 500 K:

ΔG^orxn = - 124.96 kJ/mol = - 124960 J/mol

ΔG^orxn = -RT InK

K = e^-[-124960/ (8.314) (500) ]

    = 1.13 x 10¹³

Therefore,

Equilibrium constant at 500 K = 1.13 x 10¹³