Abstract
Productivity of a process is depending on the kinetic calculation of the reactions.
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J.O. Edstrom, J. Iron Steel Inst. 175, 289 (1953)
W.M. Mckewan, Chipman Conference on Steelmaking, ed. by J.F. Elliott (MIT Press, Cambridge, MA, 1965), p. 141
L.V. Bogdandy, H.J. Engell, The Reduction of Iron Ores (Springer, Berlin, 1971)
W. Jander, Z. Anorg, Allg Chem. 163, 1 (1927)
S.K. Dutta, A. Ghosh, Metall. Mat. Trans. B 25B, 15 (1994)
A. Ghosh, A. Chatterjee, Ironmaking and Steelmaking (Theory and Practice) 1st edn. (PHI Learning Pvt. Ltd., New Delhi, 2008)
A.E. Reif, J. Phys. Chem. 56, 773 (1952)
S. Ergun, J. Phys. Chem. 60, 480 (1956)
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Appendices
Probable Questions
-
1.
What are the assumptions made by Mckewan to develop a mathematical model? Derive Mckewan equation.
-
2.
Derive the equation for solid–solid reaction.
-
3.
What do you mean by Reif’s mechanism for gasification reaction? Derive Ergun’s rate equation for gasification reaction.
-
4.
Discuss the factors that affect kinetics of solid-state iron oxide reduction, in the stack of a BF.
Examples
Example 11.1
The initial weight of pure iron oxide (Fe2O3) powder is 295.6 mg. Iron oxide powder is reduced by hydrogen at 800 °C. In reduction experiment with time weight loss are noted as follows:
Time (s) | 36 | 60 | 120 | 180 | 240 | 300 | 420 | 600 |
---|---|---|---|---|---|---|---|---|
Wt. loss (mg) | 6.69 | 14.05 | 31.94 | 48.14 | 55.33 | 59.34 | 64.94 | 70.24 |
Calculate (i) fraction of reduction (f) w.r.t time (t), (ii) draw f versus t graph, (iii) calculate rate of reduction \( \left( {\frac{{{\text{d}}f}}{{{\text{d}}t}}} \right) \) at initial stage of reduction from the graph.
Solution
(Since iron oxide is pure, so fore and ρore are one.)
Time (s) | 36 | 60 | 120 | 180 | 240 | 300 | 420 | 600 |
---|---|---|---|---|---|---|---|---|
Wt. loss (mg) | 6.69 | 14.05 | 31.94 | 48.14 | 55.33 | 59.34 | 64.94 | 70.24 |
f | 0.075 | 0.158 | 0.36 | 0.543 | 0.624 | 0.669 | 0.732 | 0.792 |
Plot f versus t, from initial slop we get rate of reduction:
Example 11.2
The following data are obtained from an experiment of reduction of iron ore fine by the hydrogen gas at 973 K. Find out (a) initial rate of reduction from fraction of reduction (f) w.r.t time (t) graph, (b) order of reaction and rate constant.
Time (s) | 36 | 60 | 120 | 180 | 240 | 300 | 420 | 600 |
---|---|---|---|---|---|---|---|---|
Wt. loss (mg) | 4.86 | 10.06 | 24.52 | 32.40 | 34.75 | 36.32 | 38.74 | 41.30 |
Given: Total removable oxygen present in iron ore is 56.18 mg.
Solution
Fraction of reductions is calculated by Eq. (1) as follow:
Wt. loss (mg) | F | (1 − f) | Ln (1 − f) | [1/(1 − f)] |
---|---|---|---|---|
4.86 | 0.0865 | 0.9135 | −0.0905 | 1.095 |
10.06 | 0.179 | 0.821 | −0.197 | 1.218 |
24.52 | 0.436 | 0.564 | −0.573 | 1.773 |
32.4 | 0.577 | 0.423 | −0.86 | 2.364 |
34.75 | 0.6185 | 0.3815 | −0.964 | 2.621 |
36.32 | 0.646 | 0.354 | −1.038 | 2.825 |
38.74 | 0.6895 | 0.3105 | −1.17 | 3.221 |
41.3 | 0.735 | 0.265 | −1.328 | 3.774 |
Now, plot f versus t graph (Fig. 11.7), and from the initial plot we get initial rate of reduction.
For zero order reaction the equation is:
where [A] = (1 − f), so plot (1 − f) versus t (Fig. 11.8).
In Fig. 11.8 shows the line is not straight line, so reaction is not zero order.
Now, we go for first-order reaction:
So plot ln (1 − f) versus t (Fig. 11.9).
Here, also the line is not straight line, so reaction is not first order.
From general equation:
Now, we go for second-order reaction, n = 2, x = f and a = 1;
So Eq. (4) become:
Now, plot \( \left( {\frac{1}{{1 {-} {\mathbf{f}}}}} \right) \) versus t (Fig. 11.10). Now, the line is almost straight.
From the slope, we get rate constant = \( \left( {\frac{{\varvec{AB}}}{{\varvec{BC}}}} \right) = \left( {\frac{{\left( {2.8 {-} 1.0} \right)}}{{\left( {290 {-} 20} \right)}}} \right) = {\mathbf{6}}{\mathbf{.67}} \times {\mathbf{10}}^{{{\mathbf{ - 3}}}} \,{\mathbf{s}}^{{{\mathbf{ - 1}}}} \)
Hence, the reduction of iron ore fine by the hydrogen gas is second-order reaction and rate constant is 6.67 × 10−3 s−1.
Example 11.3
Initial weight of pure iron oxide (Fe2O3) fines is 296.5 mg. Oxide is reduced by hydrogen gas at 800 °C. Find out: (i) fraction of reduction with respect to time, and (ii) whether these data fit to the Mckewan equation?
The following data are obtained:
Time (s) | 36 | 60 | 120 | 180 | 240 | 300 | 420 | 600 |
---|---|---|---|---|---|---|---|---|
Wt. loss (mg) | 6.69 | 14.05 | 31.95 | 48.14 | 55.14 | 59.34 | 64.94 | 70.24 |
Solution
Total removable oxygen present in iron oxide = 296.5 × 0.3 = 88.95 mg.
Time (s) | Weight loss (mg) | Fraction of reduction (R) | (1 − R)1/3 | [1 − (1 − R)1/3] |
---|---|---|---|---|
36 | 6.69 | 0.075 | 0.974 | 0.026 |
60 | 14.05 | 0.158 | 0.944 | 0.056 |
120 | 31.95 | 0.359 | 0.862 | 0.138 |
180 | 48.14 | 0.541 | 0.771 | 0.229 |
240 | 55.14 | 0.620 | 0.724 | 0.276 |
300 | 59.34 | 0.667 | 0.693 | 0.307 |
420 | 64.94 | 0.730 | 0.646 | 0.354 |
600 | 70.24 | 0.790 | 0.594 | 0.406 |
From Eq. (11.12), by plotting [1 − (1 − R)1/3] versus time, line should be straight for Mckewan equation, below figure shows that line is not straight; hence, the given data for reduction of pure iron oxide (Fe2O3) fine does not obey the Mckewan equation.
Example 11.4
For gaseous reduction of iron ore fines, the following data are obtained for H2 reductions. Find the activation energy for H2 reduction of iron ore.
Temp (°C) | H2 reduction rate (s−1) |
---|---|
898 | 5.67 × 10−3 |
973 | 5.96 × 10−3 |
1048 | 6.66 × 10−3 |
1123 | 8.80 × 10−3 |
Solution
From Arrhenius equation: Rate, \( k = A \cdot e^{ - (E/RT)} \)
Therefore, ln \( k = \ln A - (E/RT) \)
Now, plot graph of ln k versus (1/T) and we get slope = − (E/R)
So, we calculate as follows:
Temp. (°C) | Temp. (T) (K) | 1/T | 1/T × 104 | H2 reduction rate k (s−1) | ln k |
---|---|---|---|---|---|
898 | 1171 | 0.00085 | 8.54 | 5.67 × 10−3 | −5.17 |
973 | 1246 | 0.00080 | 8.03 | 5.96 × 10−3 | −5.12 |
1048 | 1321 | 0.00076 | 7.57 | 6.66 × 10−3 | −5.01 |
1123 | 1396 | 0.00072 | 7.16 | 8.80 × 10−3 | −4.73 |
From graph, \( {\text{Slope}} = - \left( {\frac{E}{R }} \right) = - \frac{AB}{BC } = - \left[ {\frac{{\left( { - 4.88 - \left( { - 5.10} \right)} \right)}}{{\left( {7.40 - 8.12} \right) \times 10^{ - 4} }}} \right] = - 3.06 \times 10^{3} \)
Therefore, \( \left( {\frac{E}{R }} \right) = - 3.06 \times 10^{3} \)
Hence, \( E = \left( { - 3.06 \times 10^{3} } \right) \times R = \left( { - 3.06 \times 10^{3} } \right) \times 8.314 = - 25.404 \;{\text{kJ}} \)
Therefore, the activation energy for H2 reduction of iron ore is \( - 25.404 \;{\text{kJ}}. \)
Example 11.5
For gasification of carbon by carbon dioxide, the following data are given for CO–CO2 gas mixture at 950 °C and 1 atm pressure. Find out the value of k1, k2, k3.
Gas composition | r × 104 (s−1) |
---|---|
Pure CO2 | 1.16 |
45% CO | 0.16 |
30% CO | 0.20 |
Solution
Since total pressure = 1 atm.
For pure CO2, \( p_{{{\text{CO}}_{2} }} = 1\;{\text{atm}} \) and pCO = 0
Putting above value in Eq. (11.39): \( r = \frac{{ k_{1} . p_{{{\text{CO}}_{2} }} }}{{1 + k_{2} . p_{{{\text{CO}}_{2} }} + k_{3} . p_{\text{CO}} }} \)
Therefore,
For 45% CO, \( p_{\text{CO}} \) = 0.45 atm and \( p_{{{\text{CO}}_{2} }} \) = 0.55 atm
Hence, from Eq. (11.35):
Again for 30% CO, pCO = 0.3 atm and \( p_{{{\text{CO}}_{2} }} = 0.7\;{\text{atm}} \)
Hence, from Eq. (11.35):
Equation (a) is divided by Eq. (b): \( \left( {1.16/0.16} \right) = \{ \left( {\frac{{k_{1} }}{{1 + k_{2} }}} \right) /\left( {\frac{{0.55k_{1} }}{{1 + 0.55k_{2} + 0.45k_{3} }}} \right)\} \)
Or
Again Eq. (a) is divided by Eq. (c): \( \left( {1.16/0.2} \right) = \{ \left( {\frac{{k_{1} }}{{1 + k_{2} }}} \right) /\left( {\frac{{0.7k_{1} }}{{1 + 0.7k_{2} + 0.3k_{3} }}} \right)\} \)
Eq. (d) is divided by
Equation (e) is divided by
Solving (f) and (g) we get: k2 = 1.58
Putting k2 value in Eq. (f) we get: k3 = 18.71
Putting k2 and k3 values in Eq. (a) we get: k1 = 2.99 × 10−4 s−1.
Example 11.6
From an experiment on sulphur removal from molten iron to slag at 1773 K, the following data is obtained:
Time, min | 0 | 4 | 15 | 33 | 63 | 91 | 151 |
---|---|---|---|---|---|---|---|
[S] % | 0.16 | 0.14 | 0.132 | 0.091 | 0.072 | 0.033 | 0.012 |
Calculate first-order rate constant for de-sulphurization.
Solution
From first-order reaction: ln [A] = ln [A0] − kt; we can calculate of rate constant (k) by plotting a graph of ln [A] versus t (correlate with: y = mx + c)
So we calculate as per following:
Time, min | [S] % | ln ([S] %) |
---|---|---|
0 | 0.16 | −1.83258 |
4 | 0.14 | −1.96611 |
15 | 0.132 | −2.02495 |
33 | 0.091 | −2.3969 |
63 | 0.072 | −2.63109 |
91 | 0.033 | −3.41125 |
151 | 0.012 | −4.42285 |
By plotting graph, the slope of line is rate constant (–k),
So, first-order rate constant is 0.01705 min−1 for sulphur removal from molten iron to slag.
Problems
Problem 11.1
The initial weight of pure iron oxide (Fe2O3) powder is 300 mg. Iron oxide powder is reduced by hydrogen at 800 °C. In reduction experiment with time weight loss are noted as follows:
Time (s) | 35 | 60 | 120 | 180 | 240 | 300 | 420 | 600 |
---|---|---|---|---|---|---|---|---|
Wt. loss (mg) | 6.65 | 14.0 | 31.95 | 48.15 | 55.35 | 59.35 | 64.94 | 70.24 |
Calculate (i) fraction of reduction (f) w.r.t time (t), (ii) draw f versus t graph, (iii) calculate rate of reduction (df/dt) at initial stage of reduction from the graph. [Ans: 3 × 10−3 s−1].
Problem 11.2
For gaseous reduction of iron ore fines following data is obtained for reduction by CO gas. Find the activation energies for such reductions.
Temp (°C) | Rate (s−1) |
---|---|
800 | 1.04 × 10−3 |
900 | 1.20 × 10−3 |
1000 | 1.72 × 10−3 |
1100 | 2.52 × 10−3 |
[Ans: −43.76 kJ]
Problem 11.3
Find the rate of the reaction at 300 K, if activation energy of the reaction is 167.36 kJ/mol. By raising the temperature by 100 K, how much reaction rate will increase?
[Ans: 1.927 × 107]
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Dutta, S.K., Chokshi, Y.B. (2020). Kinetics. In: Basic Concepts of Iron and Steel Making. Springer, Singapore. https://doi.org/10.1007/978-981-15-2437-0_11
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