Question 13.10.2: Calculate the equilibrium oxygen content of an Fe–C–O alloy ...

Introduction to the Thermodynamics of Materials [ EXP-7693]

Calculate the equilibrium oxygen content of an Fe–C–O alloy which, at 1600°C, contains 1 wt% C and is under a pressure of 1 atm of CO.

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For $C_{(gr)}+‰ O_{2(g)}=CO_{(g)}, \Delta G^\circ =-111,700-87.65TJ$

For $C_{(gr)}=C_{(1 wt\% in Fe)}, \Delta G^\circ =22,600-42.26TJ$

For $½O_{2(g)}=O_{(1 wt\% in Fe)}, \Delta G^\circ =-111,300-6.41TJ$

Thus, for

C_{(1  wt\%)}+O_{(1  wt\%)}=CO_{(g)},   \Delta G^\circ =-23,000-38.98TJ

Therefore,

$\Delta G^\circ _{1873 K}=-96,010$ J

and

\frac{p_{CO}}{h_{C}h_{O}}=476

Thus,

h_{C}h_{O}=f_{C}[wt\%  C]f_O[wt\%  O]=2.1\times 10^{-3}p_{CO}

At 1600°C,

e^{C}_{C}=0.22

e^{O}_{O}=-0.2

e^{O}_{C}=-0.097

e^{C}_{O}=-0.13

Thus, for 1 wt% C and $p_{CO}=1$ atm,

\log [wt\%  O]-0.297[wt\%  O]=-2.768

solution of which gives [wt% O] = 0.00171. If all of the interaction parameters had been ignored, the weight percentage of O would have been calculated as 0.00210.

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