Kinetics

Abstract

Productivity of a process is depending on the kinetic calculation of the reactions.

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

$$ \begin{aligned} {\text{The fraction of reduction}}\,\left( f \right) & = \left[ {\frac{{\left( {\text{O}_{2} \,{\text{removed from the sample}}} \right) }}{{\left( {{\text{Total removable}}\,{\text{O}}_{2} \,{\text{present in the sample}}} \right)}}} \right] \\ & = \left[ {\frac{{\left( {\text{weight loss}} \right) }}{{\left( {{\text{Total removable}}\,{\text{O}}_{2} \,{\text{present in the sample}}} \right)}}} \right] \\ & = \left[ {\frac{{\left\{ {\left( {W_{1} - W_{2} } \right)} \right\} }}{{W_{\text{O}} }}} \right] \\ \end{aligned} $$

$$ \begin{aligned} \text{where total removable oxygen present in iron oxide} & = W_{\text{O}} = W_{1} \times f_{\text{ore}} \times \rho_{\text{ore}} \times f_{\text{O}} \\ & = 295.6 \times 1 \times 1 \times 0.3 = 88.68\,\text{mg} \\ \end{aligned} $$

(Since iron oxide is pure, so f_ore and ρ_ore 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:

$$ \left( {\frac{{{\text{d}}f}}{{{\text{d}}t}}} \right) = \left( {\frac{AB}{BC}} \right) = \left( {\frac{{\left( {55.33 - 14.05} \right)}}{{\left( {240 - 60} \right)}}} \right) = {\mathbf{0}}{\mathbf{.229}}\,{\mathbf{s}}^{ - 1} $$

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

$$ {\text{The fraction of reduction}}\,(f) = \left[ {\frac{{\left\{ {\left( {W_{1} - W_{2} } \right)} \right\} }}{{W_{\text{O}} }}} \right] = \left[ {\frac{{\left\{ {\left( {W_{1} - W_{2} } \right)} \right\} }}{56.18}} \right] $$

(1)

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.

Fig. 11.7

f versus t

$$ {\text{The}}\,{\text{initial}}\,{\text{rate}}\,{\text{of}}\,{\text{reduction}}\left( {\frac{{\varvec{df}}}{{\varvec{dt}}}} \right) = \left( {\frac{{\varvec{AB}}}{{\varvec{BC}}}} \right) = \left( {\frac{{\left( {\mathbf{0.6 {-} 0.4}} \right)}}{{\left( {\mathbf{180 {-} 50}} \right)}}} \right) = {\mathbf{1}}{\mathbf{.54}} \times {\mathbf{10}}^{{{\mathbf{ - 3}}}} \,{\mathbf{s}}^{ - 1} $$

For zero order reaction the equation is:

$$ [A] = - kt + \left[ {A_{o} } \right] $$

(2)

where [A] = (1 − f), so plot (1 − f) versus t (Fig. 11.8).

Fig. 11.8

(1 − f) versus t

In Fig. 11.8 shows the line is not straight line, so reaction is not zero order.

Now, we go for first-order reaction:

$$ \ln [A] = \ln \left[ {A_{0} } \right] - kt. $$

(3)

So plot ln (1 − f) versus t (Fig. 11.9).

Fig. 11.9

ln (1 − f) vd t

Here, also the line is not straight line, so reaction is not first order.

From general equation:

$$ \left( {\frac{{a {-} x}}{{n {-} 1}}} \right)^{1 - n} = kt + C $$

(4)

Now, we go for second-order reaction, n = 2, x = f and a = 1;

So Eq. (4) become:

$$ \left( {\frac{1}{{1 {-} f}}} \right) = kt + C $$

(5)

Now, plot $\left( {\frac{1}{{1 {-} {\mathbf{f}}}}} \right)$ versus t (Fig. 11.10). Now, the line is almost straight.

Fig. 11.10

[1/(1 − f)] versus t

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 p_CO = 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,

$$ 1.16 \, \times \, 10^{ - 4} = \left( {\frac{{k_{1} }}{{1 + k_{2} }}} \right) $$

(a)

For 45% CO, $p_{\text{CO}}$ = 0.45 atm and $p_{{{\text{CO}}_{2} }}$ = 0.55 atm

Hence, from Eq. (11.35):

$$ 0.16 \times 10^{ - 4} = \left( {\frac{{0.55k_{1} }}{{1 + 0.55k_{2} + 0.45k_{3} }}} \right) $$

(b)

Again for 30% CO, p_CO = 0.3 atm and $p_{{{\text{CO}}_{2} }} = 0.7\;{\text{atm}}$

Hence, from Eq. (11.35):

$$ 0.2 \times 10^{ - 4} = \left( {\frac{{0.7k_{1} }}{{1 + 0.7k_{2} + 0.3k_{3} }}} \right) $$

(c)

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

$$ 0.45k_{3} - 3.44k_{2} = 2.99 $$

(d)

Again Eq. (a) is divided by Eq. (c): $\left( {1.16/0.2} \right) = \{ \left( {\frac{{k_{1} }}{{1 + k_{2} }}} \right) /\left( {\frac{{0.7k_{1} }}{{1 + 0.7k_{2} + 0.3k_{3} }}} \right)\}$

$$ {\text{Or}}\;3.06 = 0.3k_{3} - 3.36k_{2} $$

(e)

Eq. (d) is divided by

$$ 0.45{:}6.64 = k_{3} - 7.64k_{2} $$

(f)

Equation (e) is divided by

$$ 0.3{:}1.02 = k_{3} - 11.2k_{2} $$

(g)

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),

$$ {\text{Slope}} \left( { - k} \right) = - \frac{AB}{BC } = - \left[ {\frac{{\left( { - 2.5 - \left( { - 4.0} \right)} \right)}}{{\left( {40 - 128} \right)}}} \right] = - 0.01705\;{\min}^{-1}$$

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]