Question

Cesium, a silvery-white metal used in the manufacture of vacuum tubes, is produced industrially by the reaction of CsCl with CaC₂:

2 CsCl_(l) + CaC_2(s) ----> CaCl_2(l) + 2C_(s) +  2Cs_(g)

Compare the free energy produced from this reaction at 25^oC and at 1500^oC, the temperature at which it is normally run, given these values:  ∆H^o_298K = 25.0 kJ/mol,  ∆S^o_298K = 15.0 J/(mol^.K); ∆H^o_1773K = - 41.0 kJ/mol,  ∆S^o_1773K = 1.6 J/(mol^.K).

a) If you wanted to minimize energy cost in your production facility by reducing the temperature of the reaction, what is the lowest temperature at which products are favored over reactants (assuming reaction is kinetically favorable at the lower temperature)? Assume ∆H and ∆S do not vary with temperature.

b) What is the ratio K_1773K/K_298K?

Answer 1

Given at 298 K or 25 ^oC,

∆H^o_298K = 25.0 kJ/mol = 25000 J/mol,  ∆S^o_298K = 15.0 J/(mol^.K)

Therefore, ∆G^o = ∆H^o -T∆S^o = 25000 J/mol - 298 K x 15.0 J/(mol^.K) = 20530 J/mol

Therefore the reaction is non-spontaneous at 298 K.

Given at 1773 K or 1500 ^oC,

∆H^o_1773K = -41.0 kJ/mol,  ∆S^o_1773K = 1.6 J/(mol^.K)

Therefore, ∆G^o = ∆H^o -T∆S^o = -41000 J/mol - 1773 K x 1.6 J/(mol^.K) = -43836.8 J/mol

Therefore the reaction is spontaneous at 1773 K.

a) Now for ∆G^o=0, we have

∆G^o = ∆H^o -T∆S^o

=> 0 = 25000 J/mol - T x 15 J/(mol^.K)

=> T x 15 J/(mol^.K) = 25000 J/mol

=> T = (25000 J/mol)/15 J/(mol^.K)

=> T = 1666.6 K or 1393.5 ^oC

Therefore, lowest temperature at which products are favored over reactants = 1666.6 K or 1393.5 ^oC

b) We know, ∆G^o = -2.303 RT log K

Therefore, ∆G^o_{298 K}= -2.303 RT log K_{298 K}

=> 20530 J/mol = -2.303 x 8.314 J/mol x (298 K) x log K_{298 K}

=> log K_{298 K} = -3.598

=> K_{298 K} = 2.52 x 10^-4 ------(i)

and, ∆G^o_{1773 K}= -2.303 RT log K_{1773 K}

=> -43836.8 J/mol = -2.303 x 8.314 J/mol x (1773 K) x log K_{1773 K}

=> log K_{1773 K} = 1.291

=> K_{1773 K} = 19.54 ------(ii)

Therefore (ii) / (i), we get,

( K_{1773 K} / K_{298 K}) = (19.54/2.52 x 10^-4) = 77539.6