Liquid iron containing 0.0105 wt% aluminium is equilibrated with an alumina crucible at 1727°C (2000 K). The equilibrium is represented by A1203 = 2[Al]FeiWtt+3C0]FefWtJl ; K =3162x1012 2000· ~· Calculate the residual oxygen in the melt. Given: (a) Activity coefficient of oxygen in binary Fe-0 alloy is unity. (b) Activity coefficient of aluminium in binary Fe-Al alloy is unity. (c) eA1 = -3.15. 0 (d) Atomic weights of Al and 0 are 27 and 16 respectively.
Question
Liquid iron containing 0.0105 wt% aluminium is equilibrated with an alumina crucible at 1727°C (2000 K). The equilibrium is represented by A1203 = 2[Al]FeiWtt+3C0]FefWtJl ; K =3162x1012 2000· ~· Calculate the residual oxygen in the melt. Given: (a) Activity coefficient of oxygen in binary Fe-0 alloy is unity. (b) Activity coefficient of aluminium in binary Fe-Al alloy is unity. (c) eA1 = -3.15. 0 (d) Atomic weights of Al and 0 are 27 and 16 respectively.
Solution
To calculate the residual oxygen in the melt, we need to use the equilibrium constant (K) and the given information.
The equilibrium constant K is given by:
K = [Al]^2 * [O]^3
Given that the activity coefficients of Al and O are unity, the activities are equal to the concentrations.
The weight percent of Al is given as 0.0105 wt%. To convert this to atomic percent, we need to use the atomic weights of Al and Fe.
Atomic percent of Al = (0.0105 g Al / 27 g/mol Al) / ((0.0105 g Al / 27 g/mol Al) + (99.9895 g Fe / 55.85 g/mol Fe)) * 100%
This gives an atomic percent of Al of approximately 0.00039 at%.
Substituting this into the equilibrium constant equation gives:
3162 x 10^12 = (0.00039)^2 * [O]^3
Solving for [O] gives a value of approximately 0.00018 at%.
To convert this back to weight percent, we use the atomic weights again:
Weight percent O = (0.00018 mol O / 16 g/mol O) / ((0.00018 mol O / 16 g/mol O) + (99.99982 mol Fe / 55.85 g/mol Fe)) * 100%
This gives a weight percent of O of approximately 0.000052 wt%.
So, the residual oxygen in the melt is approximately 0.000052 wt%.
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