Question 12.11.1: Compare the relative efficiencies of H2 and CO as reducing a...

The values of ΔH° and ΔS° for these two reactions cause their Ellingham lines to intersect at 1127 K, as shown in Figure 12.13, with \Delta G^{\circ }_{(ii)}   being more negative than \Delta G^{\circ }_{(i)}   at temperatures higher than 1125 K and \Delta G^{\circ }_{(i)}   being more negative than  \Delta G^{\circ }_{(ii)} at temperatures lower than 1125 K. This indicates that H_2  is the more efficient reducing agent at higher temperatures and that CO is the more efficient reducing agent at lower temperatures.

Consider the reduction of CoO by each of H_2  and CO at 1673 K and at 873 K.

For CoO(s) =Co_{(s)}+ ½O_{2(g)}   :

\Delta G^{\circ }_{(iii)} =+233,900-71.85  T  J       (iii)

Combination of  \Delta G^{\circ }_{(i)}  and  \Delta G^{\circ }_{(iii)}  gives

\Delta G^{\circ }_{(iv)}=-48,500+14.96  T            (iv)

for  CoO +Co = Co +CO_2 , and combination of \Delta G^{\circ }_{(iii)}  and  \Delta G^{\circ }_{(ii)}  gives

\Delta G^{\circ }_{(v)}=-12,500-17.05T  J            (v)

for  CoO +H_2 = Co +H_2O . The positive value of \Delta S^{\circ }_{(v)}  causes \Delta G^{\circ }_{(v)}  to become morenegative with increasing temperature and the negative value of  \Delta S^{\circ }_{(iv)}  causes  \Delta G^{\circ }_{(iv)}  to become less negative with increasing temperature. At 1673 K  \Delta G^{\circ }_{(v)}=-41,024  J,and thus,

Thus, if H_2 at 1673 K is passed through a column of CoO, which is long enough that reaction equilibrium is achieved before the gas leaves the column, the fraction of H_2 which is consumed before equilibrium is reached is 19.1/20.1 = 0.95, and thus, 1 mole of H_2 is required to reduce 0.95 moles of CoO.

At 1673 K,  \Delta G^{\circ }_{(iv)}= -23,470  J,and thus,

and thus, the fraction of CO which is consumed by the reduction reaction at 1673 K before equilibrium is reached is 5.40/6.40 = 0.844. Thus, 1 mole of CO is required to reduce 0.844 moles of CoO.

At 873 K,  \Delta G^{\circ }_{(iv)}= -27,384  J, which gives

and thus, the fraction of H_2  consumed is 43.5/44.5 = 0.978 and 1 mole of H_2 reduces 0.978 moles of CoO.
At 873, \Delta G^{\circ }_{(iv)}= -35,440   J, which gives

Thus, the fraction of CO consumed is 132/133 = 0.992, and 1 mole of CO reduces 0.992 moles of CoO. Thus, H_2  is the more efficient reducing agent at higher temperatures and CO is the more efficient at lower temperatures, and decreasing the temperature at which the reduction reaction is conducted increases the efficiencies of both reductants.