PSO2
1.89 × 10 4
− 3.7828) .... (A6)
PS2 1/ 2 ⋅ PO2
(T / K )
(1 / 2) S2 + (1 / 2) O2 = SO ................... (A7)
logK ( A7) = log
2935
.... (A4)
(1 / 2) S2 + O2 = SO2 ....................... (A5)
1) E. T. Turkdogan: Physical Chemistry of High Temperature Technology, Academic Press, New York, (1980), 10.
2) G. K. Sigworth and J. F. Elliott: Met. Sci., 8 (1974), 298.
3) F. D. Richardson and C. J. B. Fincham: J. Iron Steel Inst., 178 (1954),
4.
4) J.-D. Shim and S. Ban-ya: Tetsu-to-Hagané, 68 (1982), 251 (in
Japanese).
.... (A2)
H 2 + (1 / 2 ) S2 = H 2S ....................... (A3)
Acknowledgement
This work was supported by ISIJ Research Promotion
Grant 2019 and JSPS KAKENHI Grant Number 18K04798,
and these are gratefully acknowledged.
PH2O
PH2 ⋅ PO2 1/ 2
PSO
3.37 × 103
+ 0.3128) .... (A8)
PS2 1/ 2 ⋅ PO2 1/ 2
(T / K )
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ISIJ International, Vol. 61 (2021), No. 12
, where K is the equilibrium constant of the corresponding
reaction. The mole ratios of hydrogen/argon, sulfur/argon
and oxygen/argon (denoted as α, β and γ ) remain constant
during the chemical reactions.
nH
2 PH2 ° 2 PH2 + 2 PH2O + 2 PH2S
....... (A9)
=α =
nAr
PAr °
PAr
nS
PSO2 ° PH2S + 2 PS2 + PSO2 + PSO
.... (A10)
=β=
nAr
PAr °
PAr
PAr =
PAr =
x = PH2 =
Equations (A2), (A4), (A6) and (A8) to (A12) involve 8
unknowns, i.e., PAr, PH2 , PH2O , PH2S , PO2 , PS2 , PSO2 and PSO.
Hence, the equilibrium values for these 8 unknowns are
obtainable by solving these 8 equations simultaneously with
the initial values for PAr°, PH2 ° and PSO2 °. Now, PH2 , PO2 1/2 and
PS2 1/2 are abbreviated as x, y and z, respectively. By rewriting
Eqs. (A2), (A4), (A6) and (A8), PH2O , PH2S , PSO2 and PSO can
be expressed as
PH2S = K ( A3) xz ........................ (A14)
= K ( A5) y z ........................ (A15)
PSO2
(α + 4β − 2γ ) K ( A5) z + 4β K ( A1) y
+ ( −2α + 4β − 2γ ) K ( A3) K ( A5) z 2
+ {( −2α + 4β ) K ( A3) + ( 4β − 2γ ) K ( A5)
+ (α + 2β − 2γ ) K ( A1) K ( A7)} z + 4β y 2
+ {( 2α − 4γ ) K ( A1) + ( −α + 2β − 2γ ) K ( A3) K ( A7)} z 2
+ ( 2β − 2γ ) K ( A7) z y
2
Then, PAr can be given as
PAr =
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−4γ K ( A3) z + 1 z 2 = 0
...................................... (A21)
Finally, z is optimized when the partial pressures of all the
gaseous species determined from Eqs. (A13) to (A20) with
y = PO2 1/2 and z = PS2 1/2 satisfy Eq. (A12). The calculation
procedure mentioned here can be carried out with simple
spreadsheet software.
PSO = K ( A7) yz ......................... (A16)
.... (A20)
Equation (A20) indicates that, when the value for z is
assumed, y can be calculated by solving the following cubic
equation.
PAr + PH2 + PH2O + PH2S + PO2 + PS2 + PSO2 + PSO = 1 .... (A12)
2α z 2 + α K ( A5) y 2 z + α K ( A7) yz
2β + 2β K ( A1) y + ( −α + 2β ) K ( A3) z
2α y 2 + 2α K ( A5) y 2 z + α K ( A7) yz
2γ + ( −α + 2γ ) K ( A1) y + 2γ K ( A3) z
When the gas phase is present at one atmospheric pressure,
we have
PH2O = K ( A1) xy ........................ (A13)
K ( A1) xy + 2 y 2 + 2 K ( A5) y 2 z + K ( A7) yz
.... (A19)
By erasing PAr in Eqs. (A17) to (A19), we obtain
nO
2 PSO2 ° PH2O + 2 PO2 + 2 PSO2 + PSO
....(A11)
=γ =
nAr
PAr °
PAr
K ( A3) xz + 2 z 2 + K ( A5) y 2 z + K ( A7) yz
.... (A18)
2 x + 2 K ( A1) xy + 2 K ( A3) xz .......... (A17)
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...
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