Question 13.14.2: Determine the conditions under which an Fe– Cr– O melt is in...

At 1600°C, Fe– Cr melts exhibit Raoultian ideality, and the molar heat of melting of Cr, at its equilibrium melting temperature of 2173 K, is 21,000 J. Thus, for $Cr_{(s)}=Cr_{(l)}$,

$\Delta G^\circ _{m}=\Delta H^\circ _{m}-T\frac{\Delta H^\circ _{m}}{T_{m}}=21,000-9.66T$  J

and for $Cr_{(l)}=\left[Cr\right]_{(1  wt\%  in  Fe)}$,

$\Delta G=RT\ln \frac{55.85}{100\times 52.01}=-37.70T$  J

Therefore, for $Cr_{(s)}- \left[Cr\right]_{(1  wt\%  in  Fe)}$,

$\Delta G^\circ _{(iii)}=21,000-47.36T$  J              (iii)

The standard Gibbs free energy change for the reaction

$2\left[Cr\right]_{(1wt\%)} +3\left[O\right]_{(1wt\%)}=Cr_2O_{3(s)}$                 (iv)

is thus

= -829,090 + 372.13T  J

or, at 1873 K,

$\log \frac{h^2_{Cr(1  wt\%)}.h^3_{O(1  wt\%)}}{a_{Cr_2O_3}} =-3.68$                      (v)

Saturation of the melt with solid $Cr_2O_3$  occurs at ${a_{Cr_2O_3}}=1$  , and, if the interactions between Cr and O in solution are ignored, and it is assumed that oxygen obeys Henry’ s law, Equation (v) can be written as

$\log \left[wt\%  Cr\right] =-1.5\log \left[wt\%  O\right] -1.84$                      (vi)

which is the variation of [wt% Cr] with [wt% O] in liquid iron required for equilibrium with solid $Cr_2O_3$  at 1600°C. Equation (vi) is drawn as line (vi) in Figure 13.35.

For

$\Delta G^\circ _{(vii)}=-1,409,420+318.07T$  J               (vii)

and thus, for the reaction

= -1,007,140 + 436.27T  J               (viii)

or, at 1873 K,

$\log \frac{a_{Fe}.h^2_{Cr(1wt\%)}.h^4_{O{(1wt\%)}} }{a_{FeO.Cr_2O_3}} =-5.30$             (ix)

Saturation of the melt with FeO.$Cr_2O_3$  occurs at $a_{FeO.Cr_2O_3} =1$  and, with the same assumptions as before, and $a_{Fe}=X_{Fe}=1-X_{Cr}$,  the variation of [wt% Cr] with [wt% O] required for equilibrium with solid FeO.$Cr_2O_3$  at 1600°C is

$\log (1-X_{Cr})+2\log \left[wt\%  Cr\right]+4\log \left[wt\%  O\right]=-5.30$                (x)

In solutions sufficiently dilute that $X_{Fe}\sim 1$,  Equation (x) can be simplified as

$\log \left[wt\%  Cr\right]=-2\log \left[wt\%  O\right] -2.65$               (xi)

Equation (xi) is drawn as line (x) in Figure 13.35. Lines (vi) and (x) intersect at the point A , log [wt% Cr] = 0.59, log [wt% O] = – 1.62 (wt% O = 0.024, wt% Cr = 3.89), which is the composition of the melt which is simultaneously saturated with solid $Cr_2O_3$  and  FeO.$Cr_2O_3$.  From the phase rule, equilibrium in a three-component system (Fe– Cr– O) among four phases (liquid Fe– Cr– O, solid $Cr_2O_3$, solid FeO.$Cr_2O_3$,  and a gas phase) has one degree of freedom, which, in the present case, has been used by specifying the temperature to be 1873 K. Thus, the activities of Fe, Cr, and O are uniquely fixed, and hence [wt% Cr] and [wt% O] are uniquely fixed. The equilibrium oxygen pressure in the gas phase is obtained from Equation (ii) as

which, with [wt% O] = 0.024, gives $p_{O_2(eq)}=8.96\times 10^{-11}  atm$.  The positions of the lines in Figure 13.35 are such that, in melts of [wt% Cr] > 3.89, $Cr_2O_3$  is the stable phase in equilibrium with saturated melts along the line AB and, in melts in which [wt% Cr] < 3.89, FeO.$Cr_2O_3$  is the stable phase in equilibrium with saturated meltsalong the line AC . Alternatively $Cr_2O_3$  is the stable phase in equilibrium with saturated melts of [wt% O] < 0.024, and FeO.$Cr_2O_3$  is the stable phase in equilibrium with saturated melts of [wt% O] > 0.024. Consider a melt in which log [wt% Cr] = 1.5. From Figure 13.35, or Equation (vi), the oxygen content at this chromium level required for equilibrium with $Cr_2O_3$  (at the point B in Figure 13.35) is $5.93\times 10^{-3}$  wt%, or log [wt% O] = – 2.25. From Equation (v), the activity of $Cr_2O_3$  in this melt with respect to solid $Cr_2O_3$  is unity, and hence the melt is saturated with respect to solid $Cr_2O_3$.  However, from Equation (ix), in the same melt (i.e., $X_{Fe}=0.668$,  [wt% Cr] = 31.6, [wt% O] = 0.00593), the activity of FeO.$Cr_2O_3$  with respect to solid FeO.$Cr_2O_3$  is only 0.2. Thus, the melt is saturated with respect to $Cr_2O_3$  and is undersaturated with respect to FeO.$Cr_2O_3$.   Moving along the line BA from B toward $A,   a_{Cr_2O_3}=1$  , and $a_{FeO.Cr_2O_3}=1$  increases from 0.2 at B to unity at A in the doubly saturated melt. Consider a melt in which log [wt% Cr] = – 0.5. From Figure 13.35, the oxygen content required for saturation with FeO.$Cr_2O_3$  is 0.084 wt% (log [wt% O] = – 1.075 at the point C in Figure 13.35). From Equation (ix), the activity of FeO.$Cr_2O_3$  in this melt is unity. However, from Equation (v), the activity of $Cr_2O_3$  in the melt, with respect to solid $Cr_2O_3$,  is only 0.285. Thus, this melt is saturated with FeO.$Cr_2O_3$  and is undersaturated with $Cr_2O_3$  On moving along the line CA from C toward A ,$a_{FeO.Cr_2O_3}$  is unity and $a_{Cr_2O_3}$  increases from 0.285 at C to unity at A . If the various solute– solute interactions had been considered, Equation (v), with $a_{Cr_2O_3}=1$, would be written as

or

or

With

$e^{0}_{Cr}=O$         $e^{O}_{O}=-0.2$     $e^{Cr}_{O}=-0.041$     and  $e^{O}_{Cr}=-0.13$

this gives

$-0.43\left[wt\%  O\right]+0.0615\left[wt\%  Cr\right] +\log \left[wt\%Cr\right]+1.5\log \left[wt\%  O\right] =-1.84$              (xii)

which is drawn as line (xii) in Figure 13.36.

Similarly, with $a_{FeO.Cr_2O_3} =1$  , Equation (ix) would be written as

or

or

which is drawn as line (xiii) in Figure 13.36. Lines (xii) and (xiii) intersect at log [wt% Cr] = 0.615, log [wt% O] = – 1.455 ([wt% Cr] = 4.12, [wt% O] = 0.035). When the interactions among the solute were ignored, the point of intersection, A , was obtained as [wt% Cr] = 3.89, [wt% O] = 0.024.