Reaction Kinetics PDF

REACTION KINETICS  Study of thermodynamics provides information about the chemical equilibrium of a metallurgical reaction and indicates the direction in which a reaction will proceed under the stated condition.  It does not provide any information on the rate of reaction.  There is s large negative free energy change (G =  1576 kJ) when Al is exposed to air at room temperature 298 K to form oxide Al2O3. The initial formation of the film of Al2O3 acts as a barrier to further contact between the atmosphere and the metallic Al, resulting in an extremely low overall rate of reaction.  The reaction between carbon and oxygen in liquid steel has a large thermodynamics driving force (G =  113.26 kJ) but requires the bubbling of oxygen or an inert gas, or some other form of turbulent in the liquid steel to initiate the carbon boil reaction. [C] + [O] = CO (g)  The kinetic factor which restricts the rate at which the reaction proceeds.  The thermodynamic tell about the possibility of the reaction, but not about the mechanism of change or rate of a reaction.  Studies of both the thermodynamics and reaction kinetic are required for complete understanding of any metallurgical process.  At higher temperature, rate of reaction is faster.  At higher temperature, chemical equilibrium likely to the attained with respect to some reaction even when the entire macrosystem is considered.  Knowledge of reaction rates is important for better operation and control of process analysis and design.  The subject dealing with this is known as reaction kinetics and the rate of a reaction depends on several factors  Concentration or activity of each reactant  Temperature, pressure, nature of solvent, if any  Presence of catalysts and reaction mechanism the chemical kinetic mainly deal with the effect of two factors i.e. concentration and temperature, on the rate of reaction. Types of reactions  Homogeneous reactions: Take place within one phase, such as, a gas or a liquid solution.  Heterogeneous reactions: Take place between different phases, such as gas-solid, slag-metal reactions etc. Most of the metallurgical reactions are heterogeneous. The main types of heterogeneous reactions are as following:  Gas-solid e.g. reduction of oxide with CO or H2 gas, oxidation of metal.  Gas-liquid e.g. gaseous reduction in hydrometallurgy.  Liquid-liquid e.g. metal-slag reactions.  Liquid-solid e.g. leaching and corrosion reactions.  Solid-solid e.g. reduction of ZnO with carbon. Difference between heterogeneous and homogeneous reaction HOMOGENEOUS REACTION HETEROGENEOUS REACTION 1. It takes place entirely within one phase, i.e. single 1. It involves more than one phase i.e. slag-metal phase. Reactants and products in one phase as well reaction, molten slag is in one phase and molten as reaction take place in that phase. metal in another phase, reaction take place at interface. 2. Reaction occurs in the bulk i.e. within the phase. 2. Reaction occurs not in the bulk, but at the interface. 3. Rate of chemical reaction is proportional to the 3. Rate of chemical reaction is proportional to the volume of the phase. interfacial area between the two or more phases. 4. No adsorption and desorption are involved in the 4. Heterogeneous react involves following steps: homogeneous reaction. i. Adsorption of reactants into the interfacial layer. ii. Chemical reaction between absorbed components. iii. Desorption of products into the phases. 5. Reaction between gas molecules to produce gaseous 5. Transfer of a element or iron from molten slag to products. molten metal in a refining process. 6. Reaction mechanism are not complex. 6. Reaction mechanism are complex. Rate of reaction: The rate of reaction may be define as the rate at which the concentration of a reactant decreases or the rate at which the concentration of a product increases. A + B = C ……….. (1)  d  A   d B    d C  Rate     dt     dt    dt ....................................2 [d(i)/dt] refers to the rate of change of concentration of component i with time. Where i refers to the concentration of component (i.e. either A, B or C). The rate is a (+) ve quantity, but as concentration of reactants are decreasing, so [d(i)/dt] =  ve Effect of condition on the rate of reaction Surface area Heterogeneous reactions occur at an interface of the two phases, the area of the interface will affect the rate of reaction. For solid-solid or solid-gas reactions, solid affects the rate of reaction. MOs   Cs   M s   COg .............................................3 Alternative, reduction can be with CO gas MOs   COg   M s   CO2 g .............................................4 Reaction (3) will be slow because of the small area of contact between two solids, as well as low diffusivity of solid to solid. Reaction(4) is likely to be faster because it involves a gas, that is a solid-gas reaction which benefits from the high diffusivity of gas, allowing contact of gas with hole surface of solid. Catalyst: A catalyst is a substance which can alter the rate of a reaction but remains unchanged in amount at the end.  Natural gas is reformed in presence of catalyst Ni or Al2O3 at 950C CH4(g) + H2O (g) = CO(g) + 3 H2(g)  The more familiar catalysts such as MnO2 and V2O5. 2H 2 O2 l  MnO 2 2H 2 Ol   O2 g  2SO2 g   O2 g  V  2SO3 g  2O5 MnO2 and V2O5 are positive catalysts.  Negative catalysts (inhibitors): Glycerol decreases the rate of the reaction 2H2O2(l) = 2H2O(l) + O2(g) Concentration The effect of various conditions on reaction rate can be seen drawing the figure of concentration (weight, volume etc) against the time as shown in Figure1 and Figure 2 for the following reaction. CaCO3 + 2HCl = CaCl2 + H2O + CO2 ……….. (1) Figure 1: Concentration vs time Figure 2: Concentration vs time ( temperature T2 > T1) ( at temperature T1) Figure 1 shows the volume of CO2 varies with time for the reaction (1). At higher temperature (T2), concentration (i.e. volume of CO2) increases more with time with respect to lower temperature (T1). Figure 2 shows that the concentration of hydrochloric acid varies with the time for the same reaction (1) at temperature T1. The concentration of hydrochloric acid decreases with time. ORDER OF REACTION  The order of reaction may be defined as the sum of the powers to which the concentration of the reactants must be raised to determine the rate of reaction.  The order of a reaction is the total power of the reactants.  Reactions are designated as of first order, second order, third order etc, according to the sum of the powers is equal to one, two etc.  A reaction may be classified according to its molecularity i.e. the numbers of molecules or the number molecules plus atoms, taking part in a reaction.  The order of reaction does not necessarily bear any relation to the molecularity of the reaction. NO + O3 = NO2 + O2 ……… (1) The reaction (1) shows that one molecule of nitrogen monoxide and one molecule of ozone collide, and two molecules are involved, so the molecularity is two. Case 1 2HI  H 2  I 2......................................2 Two molecules of HI take part in the reaction and therefore the reaction is bimolecular. By experiment, it is found that the reaction rate is proportional to the square of the HI concentration, so it is the second order reaction. Case 2 HI  H 2O2  HIO  H 2O........................................3  HI is present in the systems is excess. The rate of this reaction depends only on the concentration of the H2O2 because the concentration of HI is not altered significantly by the reaction and it is therefore first order reaction even though it is bimolecular.  The reaction is not completed after this first stage and a further in the reaction take place HI + HIO = H2O + I2 …………………. (4) Overall reaction: 2HI + H2O2 = 2 H2O + I2 ……….. (5)  Experiment shows that reaction still depends only on the concentration of the H2O2 and it is first order reaction. 2nd stage reaction (4) occurs much faster than the 1st stage reaction (3) , hence 1st stage reaction is the rate controlling step of the overall reaction.  Molecularity of a reaction must always have an integral value and will never be zero , but the order of a reaction, an experimentally determined, can be fractional i.e. 2.5 etc or zero. Zero order A reaction is said to be zero order, when the rate of reaction is constant. For the zero order reaction: A product …………………………. (6) The rate of reaction is constant.   d A  K............................7   dt  Where [A] is concentration of A, and K is constant. This can occur is some heterogeneous reactions or in surface catalysed reactions: N2 + 3H2 = 2NH3 ……….. (8)  The reaction (8) occurs in the presence of a solid tungsten catalyst. Under a range of gas pressures, the catalyst surface is completely covered with reactant molecules. Since almost all the reactions occur at the surface, the rate depends on the number of reactant molecules absorbed which can not increases.  Thus the reaction rate is constant (except low pressure) for a given sample of catalyst and is independent of the concentrations of the reactants.   d  A   K  dt...................................9  A  A0   Kt  A t A 0 0 A  Kt  A0 .........................................10 The slop of the line is K i.e. the rate of the reaction and intercept is [A0]. Figure 3: Concentration vs time for zero order First order For the first order reaction: A product i.e. A X + Y ……….. (9) The rate of reaction is proportional to the concentration of the reacting substances i.e. rate  [A].  d  A  rate     K  A.......................................12   dt  Where [A] is the concentration of A and K is a proportionality constant or a rate constant. The negative sign is for the decrease in the concentration of A with time as the reaction proceeds.  A d  A  d  A         K  dt t  A  A  A0 0  ln A  ln A    Kt 0  ln A  ln A   Kt.................................................14  0 If a moles of A are present initially in unit volume and x moles have reacted in unit volume in time t, [A0] = a and [A] = (a-x) lna  x   ln a  kt.............................................15 If a, x and t can be measured experimentally, a graph of ln(a-x) vs t will give a straight line for a first order reaction (figure 4).The slope of the line is  K i.e. the rate of the reaction and intercept is lna. a x ln    Kt  a  1 a  K  ln t ax 2.303 a K log.......................................16  t ax The time required for 50 percent completion of first order reaction, i.e. half- life period (t0.5), can be calculated from equation(16). Figure 4: Concentration vs time for first order x = a/2 2.303 a 2.303 log 2 t0.5  log ..........................................17  K aa K 2 SECOND ORDER: For second order reaction: A + B products  A +B = X + Y The rate of reaction is proportional to the concentration of the reacting substances, i.e. rate  [A] or [B] ,  d  A rate      K  AB .................................18  dt  where [A] and [B] are the concentrations of A and B respectively, K is a proportionally constant or a rate constant. The negative sign is for the decrease in the concentration of A and B with time as the reaction proceeds. If a moles of A and b moles of B are prevent initially in unit volume and x moles have reacted in unit volume in time t. [A] = (a  x) and [B] = (b  x)  d a  x   d b  x         K a  x b  x   dt   dt  da dx    K a  x b  x  dt dt  dx      K a  x b  x ...........................................19   dt  The initial concentration a is a constant i.e. a will not very with time t.  da   0  dt  If originally equimolar amounts of A and B are present, then a = b and equation (19) becomes:  K a  x   K a  x .....................................20 dx 2 2 dt Equation (19) is a differential form of 2nd order reaction. 1  ..................................................21 dx Now if further modified equation (19) dt  K  a  x b  x . 1 dx  dt   Integrating equation (20) t x K a  x b  x  0 0 1 1  1 1     x  dx K a  b b  x a  x  0  ba  x ......................................22 1 t K a  b   ab  x  ln  Equation (22) is the integration form of 2nd order reaction. Further equation (22):  1   b   a  x   t  ln    ln  b  x     K a  b   a     a  x   b ln    t.K.a  b   ln  ...............................23  b  x   a Figure 5: ln[(a-x)/(b-x)] vs t plot for 2nd order A plot of should be a straight line with a slope equal to K(a  b) as shown in figure. 1 x  K ...............................24 t  aa  x  The half life period, t0.5 , for a second order reaction, in which both the reactants have the same initial concentration, can be obtained from equation (24). 1 t0.5 ..............(25) Ka General nth order For the higher order reaction: A + B + ………… products Whenever the reactants are originally present in equimolar amount, then rate equation can be generalized as:  dx     K a  x ................................26 n  dt  Where n is the order of the reaction, which can be determined experimentally, and can be zero, an integer and a fraction but not n =1.  dx   n   Kdt....................................27   a  x   n  a  x  dx  K  dx  a  x 1 n     Kt  C........................................28  n  1  where C is an integration constant. A plot of (a  x)1-n vs t should be give a straight line (figure 6) and from the slop and intercept of the plot the rate Figure 6: (a-x)1-n vs t plot for general constant and integration constant (c) respectively can be determined. order Determination of order and rate constant of a reaction The concentrations of reactants or products are measured at various times during the progress of a reaction. Thus are several methods for calculating the order and rate constant as follows:  Method of integration  Half-life method  Differential method  Initial rate method Method of Integration The integrated terms of the rate equations such as for 1st order: ln a  x  ln a  Kt.............................................1 For general equation:  a  x 1 n     Kt  C..........................................2  n  1  From the curves i) ln(a  x) vs t (fig-4) and (ii) (a  x)1n vs t (figure 6) the rate constant (K) can be determined from the slope of the line. This is a trial and error method, this method is not very accurate for complex reaction. e.g; consider a 2nd order reaction. Consider a 2nd order reaction. First assumption is 1st order reaction where ln[A] is plotted against time (t). Again same fractions assuming 3rd order reaction, then 2nd order reaction, are also plotted against time (Figure 7). Figure shows only 2nd order plot gives a straight line that means reaction is 2nd order. Half-life method Figure 7: f[(A)] vs t plots (a) First order reaction: This method determines the time (t1/2) taken for half-mole of reactant to react during the reaction. The integrated forms of the 1st order reaction:  a  ln a  x  ln a  Kt...............................1 ln    Kt..........................................2  a  x      a  ln   K  t 1  ln 2  Kt 1 x = a/2 at t = t1/2 a a   2 2  2    K  ln 2 .......................................3  t 1   2  From this equation (3) it is understood that for a 1st order reaction the half-life (t1/2) is independent to the initial concentration of the reactants i.e., t1/2 is independent of a, which is not true for nth order. (b) second order  a  x 1 n  From general equation:    Kt  C............................1  n 1  x = a/2 at t = t1/2 and n = 2 (for 2nd order reaction) 1 a    K.t 1  C.........................................2 2 2 To find the value of integration constant (c), when t = 0, x = 0 and n = 2 (for 2nd order).    C......................................3 1  K.t 1     K   .........................................4 1 2 1 a a 2 a  a.t 1   2  (C) nth order reaction:  a  x 1n     Kt  C............................1  n  1  To find the value of integration constant (C); when t = 0, x =0  a 1n   C  n 1  Now x = a/2 at t = t1/2    a 1n   a        2     K.t   a 1 n   n  1     1 2  n 1       2n 1  1  K    n  1a.t 1   n 1 2    Differential method: For the first order reaction    d  A  K  A.....................................1  dt  Now generalizing for a reaction of order n involving only one reacting substance: nA product............................................2  d  A    K  An......................................3  dt  Taking two different initial concentration [A1] and [A2], then equation (3) become  d  A1   d  A2    K  A n    K  A n  dt   dt  1 2   d  A1     d  A2   log     log       log K  A1   log K  A2  n n     dt    dt   nlog A1   log A2 .................................................4  Thus n can be determined by taking two values of concentration and measuring the rate of reaction at these concentrating. Initial Rate Method During the first 5% or so of reaction, so little reactant has been used up, that is, concentration of that is virtually constant. Figure 8: Concentration vs time plot. The rate is then constant [as shown in figure 8(b)] and that rate is known as the initial rate.  d A For a single reactant: initial rate:  dt   K  An 0............................................1 0 d A Where subscript zero indicates initial conditions, log   log K  n log A0.....................................2  dt  0 A plot of log[d[A]/dT]0 vs log[A]0 gives a straight line, from slope and intercept get values of n and K, respectively. Problem: The radioactivity decay of uranium 238 is first order and the half-life is 4.51  109 years. Calculate the specific rate. In how many days will 75 percent of a given amount of uranium disappear? Solution: First order reaction rate: [d[A]/dt] = K[A] d A   Kdt A If t = 0 [A] = a and t = t [A] = a  x a  x d A    Kt  ln  a x t  a 0  a    Kt For half life a 1 x  ln   Kt 1  Kt 1  ln 2 2 2 2 2 2.303 log 2  t1  2 K 2.303 log 2 2.303 log 2 10 1 K   1.532  10 year t1 4.51  10 9 2 The specific rate of radioactive decay of uranium 238 is 1.532  1010 year1. The time for 75 percent completion, t0.75 be calculated as  a  2.303  100  ln    Kt  t  log   a  x 100  75  0.75 K 2.303  365  t 0.75  log 4 days 1.532  10 10  330.4  1010 days. The time required for 75 percent of a given amount of uranium 238 to disappear is 330.4  1010 days. Problem: A second order reaction has an initial concentration of the reactants of 0.4 moles/liter. The reaction is 30percent complete in 80 min. Calculate the rate constant and the time it would take for the reaction to be 80 percent complete? a x dx x  K a  x  dx t Solution:  a  x    Kdt  Kt  2 dt xa 2 0 aa  x  1 x 1 0.4  0.3 K    lt / mole / sec. t aa  x  80  60 0.40.4  0.4  0.3  2.232  10  4 liter / mole / sec. The time required for 80 percent completion t0.8 1 0.8  0.4 2.232  10  4   t 0.8 0.4  0.8  0.40.4 0.8  0.4  t 0.8   4.48  10 4 sec. 0.4  0.4  0.8  0.4   2.232  10 4 80 percent of the reaction will be completed in 4.48  104 sec. Reaction Rates for Homogeneous Reactions: The reversible chemical reactions are specific in nature and the rate expressions are determined experimentally. For a reaction: M + N ⇌ X + Y The rate of a reaction may be expressed as:  dC   dC N   dC X  or  dCY  ......................................1   M  or   ,    dt   dt   dt   dt  Where CM, CN are the concentration of reactants M and N, CX, CY are the concentration of products X and Y, respectively. rate  K C C N CM......................................2 Where KC is a constant known as the rate constant. This is also known as specific rate for the chemical reaction and it increases with increasing temperature as follows:   B  log K C   A   .....................................3   T  where A and B are constants. Law of Mass Action Law of mass action states that the rate of a reversible chemical reaction is proportional to the product of the active mass (means activity) of the reacting substances. For the reaction: M  N  X  Y................(1) Kf Kb n  0 The rate of reaction, according to the above law:    K C C  K C C...................................2  V  f M N b X Y where (n/V) is molar rate of reaction per unit volume, Kf and Kb are the forward and backward reaction rate constants, and Ci is molar concentration i per unit volume. At equilibrium, n = 0 i.e., the forward and the backward are equal, and the net rate is zero. K f C M C N  K b C X CY  Kf   C x CY        K.................................3  Kb   CM C N  where K is the equilibrium constant Kf   n0    C C  K b    and    K f C M C N   X Y ................................4  K  V    K  Equation (4) is the rate expression for a reversible reaction which is one occurring under conditions near equilibrium. In contract, there is another class of reactions which takes place under conditions away from equilibrium. These are known as irreversible reaction. Here the backward reaction is ignored yielding:  n0     K f C M C N..........................................5 V  The law of mass action may be applied to the overall reaction or to the elementary steps. A correct rate expression is obtained only when it is applied to the elementary reaction step which is rate controlling. Effect of Temperature on Rate of Reaction (Arrhenius Equation) According to Van’t Hoff equation :  d ln K   H    2 ...........(1)  dT   RT  Where H is change in enthalpy per mole of reaction.  K   d ln  f     K b     H ...........(2)  dT   RT 2  as K = Kf /Kb      d ln K f   d lnK b    H       RT 2 .............(3)  dT   dT     Ef  ln K f     constant ………….. (4)  RT  E  ln K b   b   constant ………….. (5)  RT  Where (Eb  Ef ) = H and I is a constant. The rate constant is: Kc = A exp( E/RT) where E is known as activation energy and A is known as the frequency factor. Scientist Arrhenius conceived the existence of an energy barrier between reactants and products as shown in Figure 9, it is termed as activation energy (E). It is well known that the rates of reactions increase with rise of temperature. This universal observation was first put as a mathematical equation by scientist Arrhenius in 1889. K = A exp( E/RT) = Ae(E/RT) …………………………… (6) where K is the rate constant A is the frequency factor or Arrhenius constant E is the activation energy of the reaction, and T is the absolute temperature Figure 9: Activation Energy Barrier Arrhenius equation can be applied successfully to many homogeneous as well as heterogeneous reactions. Arrhenius equation (6) can be used in several forms such as: ln K = ln A  (E/RT) …………………. (7) By potting: ln K vs (1/T) (Figure 10),straight line is obtained and from the slope of the, the value of (E/R) i.e. activation energy can be obtained. Differentiating equation (7) with respect to temperature.  d ln K   E    2 .................(8)  dT   RT  Figure 10: ln K vs 1/T plot The greater the activation energy of a reaction, the greater will be the increase of reaction rate with temperature. From equation (6) Arrhenius suggested that reactant molecules must be activated by acquiring an extra energy E, the activation energy of the reaction, before they could react. Reaction Rates for Heterogeneous Reactions  Most of the metallurgical reactions are heterogeneous in nature.  The over all rates of these heterogeneous reaction will be depend on three main steps which are present in the reactions:  rate of transport of reactants to the reaction site  chemical reaction rate at the reaction site  rate of transport of products away from the reaction site  These are known as kinetic steps.  The slowest amongst these, which primarily controls the rate of the overall reaction, is called the rate controlling step.  The purpose of kinetic study is to find out the slowest step in a given reaction. Chemical Reaction Control The slag- metal reaction: [Mn] + (Fe2+) = [Fe] + (Mn2+) ……………… (1) This reaction (1) between molten metal and molten slag may be produced via simultaneous anodic and cathodic reactions as follows Anodic reaction: [Mn] = (Mn2+) + 2e ……………… (2a) Cathodic reaction: (Fe2+) + 2e = [Fe] ……………… (2b) The elements are present only as ions in the slag phase, and as atom in the metal. The reaction take place only at the slag- metal interface. The overall reaction consists of several steps, known as kinetic steps. 1. Transfer of Mn to the slag-metal interface from the bulk of metal phase. 2. Transfer of Fe2+ to the interface from the bulk of slag phase. 3. Chemical reaction at the interface. 4. Transfer of Fe from the interface into bulk of the metal phase. 5. Transfer of Mn2+ from the interface into bulk of the slag phase. Figure 11: Kinetic steps of the reaction  The step 3 is a chemical reaction step and is governed by the law of chemical kinetics.  The steps 1, 2, 4, and 5 are the mass transfer steps.  Step 3 i. e. the chemical reaction step, is the slowest step and it is rate controlling step of the reaction.  The reaction (1) is chemical reaction control. 𝑆𝑢𝑚𝑚𝑎𝑟𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑅𝑎𝑡𝑒 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑓𝑜𝑟 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑂𝑟𝑑𝑒𝑟𝑠. Order Rate equation Unit of rate constant, if the units of concentration and time are Differential form Integrated form moles/ litre and sec mol/dm3 and s respectively respectively 0 𝑑𝑥 𝑥 moles/litre/sec mol/dm3/s =𝑘 𝑘= 𝑑𝑡 𝑡 1 𝑑𝑥 1/2 2 1ൗ2 1 moles1/2/litre1/2/sec mol1/2/dm3/2/s  𝑘 𝐶𝑜 − 𝑥 𝑘 = [𝐶0 − 𝐶0 − 𝑥 2 ] 2 𝑑𝑡 𝑡 1 𝑑𝑥 2.303 𝐶0 /sec /s  𝑘 𝐶𝑜 − 𝑥 𝑘= 𝑙𝑜𝑔 𝑑𝑡 𝑡 (𝐶0 − 𝑥) 3 𝑑𝑥 3/2 2 1 1 moles1/2/litre1/2/sec mol1/2/dm3/2/s  𝑘 𝐶𝑜 − 𝑥 k    2 𝑑𝑡 t  C  x  C  1 1   2 1 0 0 2 𝑑𝑥 2 1 x  liter/mole/sec dm3/mol/s  𝑘 𝐶𝑜 − 𝑥 k  𝑑𝑡 t  C C  x  0 0 3 𝑑𝑥 1  2C x  x  liter2/mole2/sec dm6/mol2/s  𝑘 𝐶𝑜 − 𝑥 2 3 k  2t  C C  x   0 𝑑𝑡 2 0 0 2 Half-Life Method If the reactant concentrations are equal, the half-life period, t0.5 for a reaction of overall order n is given by 1 t  0.5...............(1) n 1 C 0 where C0 denotes the initial concentration. Taking logarithms of both sides of Eq.(1) log t0.5 = constant  (n-1) log C0 ……………… (2) If the half-lives of a reaction are known for different initial concentrations, Eq.(2) can be used for calculating the order of reaction. The slope of the straight line obtained by plotting log t0.5 against log C0 gives the value of -(n-1) and hence that of n. Van't Hoff’s Differential Method The reaction rate (r) for an isothermal reaction can be related to the concentration of a reactant C by the following equation: r = k C n ……………………. (1) where n is the order of reaction. Taking logarithms of both sides of Eq. (1), log r = log k + n log C ……………………. (2) If the reaction rate (r) is known at various values of reactant concentration, a plot of log r against log C will result in a straight line, and the slope of the line gives the value of the order of reaction with respect to the substance whose concentration is being varied. Two different methods may be used for calculating reaction rates at different reactant concentrations. One consists of plotting reactant concentration against time. The slopes (negative) at different reactant concentrations give the values of the rates at the respective reactant concentrations. The order of reaction obtained on the basis of the above rate values is called ‘order with respect to time', nt. The other method involves plotting reactant concentration against time for different values of the initial concentration of reactant. Tangents are drawn at the beginning of the reaction (i.e. at time t = 0). The rate corresponding to a particular initial reactant concentration is given by the negative of the respective slope value. When these initial rate values at different initial reactant concentrations are used for calculating the order according to Eq.(2), the order obtained is known as 'order with respect to concentration' or 'true order', nc. In case the reaction involves more than one reactant, experiments are carried out with different initial concentrations of one reactant, while keeping the concentration of other reactants constant. The order of reaction with respect to this reactant is determined. In order to determine the total order, this order is added to the orders with respect to other reactants determined in a similar manner. Problem: For the reaction A + B = C, the following data are determined: Experiment Initial Concentration Half-life period No. A B moles/litre mol/dm3 moles/litre mol/dm3 Hour s 1 2.0 2.0 2.0 2.0 2.5 900 2 5.0 5.0 5.0 5.0 1.0 3600 Calculate the order of reaction and the specific reaction rate. Solution: Two experiments are carried out at the initial reactant concentrations (C0)1 and (C0)2 and their half-life periods are (t0.5)1 and (t0.5)2 respectively. We know that log t0.5 = constant  (n-1) log C0 ……………… (1) log (t0.5)1 = constant  (n-1) log (C0)1 ……… (1a) log (t0.5)2 = constant  (n-1) log (C0)2 …………… (1b) logt   logt  log 2.5  log1 n 1 0.5 1 0.5 2 1 2 Thus the reaction is second-order. logC 0   logC  2 0 1 log 5  log 2 The specific reaction rate for a second-order reaction when both the reactants are at the same concentration can be calculated from the following equation. 1 1 k   0.2 liter/mole/hour ct 0.5 2  2.5 Problem: The following data are obtained for the kinetics of reduction of FeO in slag by carbon in molten pig iron at a certain temperature : Concentration of FeO in slag, wt% 20.00 11.50 9.35 7.10 4.40 Time, min 0 1.0 1.5 2.0 3.0 Calculate the order of reaction w.r.t. FeO by Yan't Hoff's differential method. Also calculate the rate constant. Solution:  The concentration of FeO in the slag is plotted against time, as shown in Figure 1.  Tangents are drawn at the points corresponding to 18, 16, 10, and 6 wt% FeO. The slopes of these four tangents are measured.  The slopes of these four tangents are measured to obtain the rates of reduction of FeO corresponding to the four values of FeO concentration. Sl. Concentration of Slope of the tangent No. FeO in slag, wt% (C) 1 18  11.88 2 16  8.94 3 10  4.63 Figure 1: Variation of the Concentration of 4 6  3.13 FeO in Slag with Time. Sl. No. Concentration of FeO in log C Slope of the Rate of reduction of log (r) slag, wt% (C) tangent FeO, wt.% FeO /min (r) 1 18 1.2553  11.88 11.88 1.0749 2 16 1.2041  8.94 8.94 0.9513 3 10 1.000  4.63 4.63 0.6656 4 6 0.7782  3.13 3.13 0.4955  log (r) is plotted against log (c) 1.2 1 y = 1.1933x - 0.4674  The slope is 1.19, and the reaction can be R² = 0.966 0.8 regarded as first-orderw.r.t. FeO. log (r) 0.6  The intercept of the line at log C = 0 is equal to 0.4  0.4674 0.2 0 log k =  0.4674  k = 0.341 min 1 0 0.5 1 1.5 log (c) The specific reaction rate is equal to 0.341 min 1 Figure 2: Log-log Plot between the Rate of Reduction of FeO (r) and Concentration of FeO in Slag (C). Problem: Find the rate of the reaction at 300K , it activation energy of the reaction is 167.36 kJ/mole by raising the temperature by 100 K, how mush reaction rate will increase? Solution: Rate (K) at 300 K = A e( E/RT) = A e [167360/(8.314  300)] = A e  67.099 = A  7.232 1030 Rate (K) at 400 K = A e( E/RT) = A e [167360/(8.314  400)] = A e  50.325 = A  1.394 1022 Thus increasing temperature by 100 K, will increase the reaction rate by a factor: [Rate (K) at 400 K/ Rate (K) at 300 K] = [(A  1.394 1022 )/(A  7.232 1030 )] = 1.9727 107 Diffusion Mass transfer means the transfer of chemical species from higher concentration region to lower concentration region. Three types of mass transfer.  Diffusion:  Diffusion is a molecular movement.  In gases, molecules are always in motion and colliding with one another.  In solid, molecules are always vibrating.  Vibrations lead to jump.  It occurs when a substance diffuses through a layer of stagnant fluid and it may be due to concentration, temperature or pressure gradients.  Convection: Motion of the fluid media aids mass transfer.  Eddy diffusion: In turbulent flow, eddy like exchanges take place randomly in all direction and leads to extensive mixing. Fick’s Law of Diffusion The rate equation for diffusion is expressed by the Fick’s law of diffusion. The diffusion of component i in a binary mixture i and j, can be expressed as: n   C  Molar flux =     D  i ........................(1) i      ij A x m   C  * Mass flux =     D  ............................(2) i i      ij A x (ni/A) and (mi/A) are molar flux and mass flux of component i along x direction. Dij is a constant known as diffusion coefficient or diffusivity of components i and j (m2/s). It is a transport property of the material. It does not depend upon the units of mass. The negative sign on the right hand side of equations (1) and (2) means that the diffusive flux is along the negative concentration gradient. The value of the mass diffusion coefficient for a binary mixture of gases can be calculated by using the following equation: 1   1 1  3 2 T 2 D  0.0043    ..............(3)  pV  V  M M  ij 1 1 2 3 3 i j i j Where p = total pressure in atmosphere, T = absolute temperature, K Vi, and Vj = molecular volume of components i and j at normal boiling point (m3/kmol) Mi, Mj = molecular weight of components i and j. Problem: From the following data, calculate the diffusion for NH3 in air at 27 C temperature and one atmospheric pressure: Gas Molecular weight Molecular volume (m3/kmol) NH3 17 0.02643 Air 29 0.0306 Solution: The diffusion coefficient for binary gaseous mixture,   1 1 1  3 2 T 2 D  0.0043   ij   p V  Vi 1 3 j 1 3  2   M M  i j  T = 27 + 273 = 300 K, p = 1 atm, Vi = 0.02643 m3/kmol, Vj = 0.0306 m3/kmol Mi = 17, Mj = 29      3  1 1  300 2   1  2 Dij  0.0043     1 2  17   29  1 0.02643 3  0.03063   1     = 18.3 m3/s General Rate Equation Consider a chemical reaction aA + bB+ ………… = cC + dD+ ……………. (1) By stoichiometric considerations, 1   dC   1   dC   1  dC   1  dC    A   B  C   D ............................(2) a  dt  b  dt  c  dt  d  dt  The term C indicates concentration, and each term within ( ) brackets indicates a rate of change of concentration. Each of the i terms within ( ) brackets is an index of reaction rate, and these are interrelated.  Determination of concentration change affords a direct measure of the reaction rate for solutions.  In many cases, e.g. reaction of solids, this may not be useful. There could then be other parameters which can be measured to study the progress of a reaction, e.g. total weight of a reactant or product, volume of a product, etc.  There could be an “indirect” measure in terms of a property that is a function of changes occurring, e.g. variation in dimensions of a solid or that in mechanical, electrical or magnetic properties.  If there is no direct correlation between such variations and the amounts of reactants consumed or products formed, then a kinetic analysis based on such “indirect” indices becomes empirical. Considering any particular reactant i one may write dCi   k.C Aa.C Bb..... ………………. (3) dt where a + b + …….. n is the overall order of the reaction. By deﬁnition, the reaction is of order a w.r.t. A, or order b w.r.t. B and so on. The exponents a, b, c, etc., are usually simple integers but occasionally they may be fractions or even negative depending on the complexity of the reaction. For the nth order reaction of a single component dC   kC.................................(4) n dt where C is concentration and k is rate constant. For n  1; one obtains by integration 1  1 1    kt......................(5) n  1  C n 1 C  0 n 1 Here C0 is concentration at zero time and C is concentration at time t. For n = 0, C0  C = kt ……………………. (6) n = 1, ln(C0/C) = kt ……………………. (7) It is interesting to note that while for a zero-order reaction, the time t0.5 for half the reactant to get consumed depends upon the initial concentration, and for the ﬁrst-order reaction this is not so. 0.69315 t0.5 ...........................(8) k Thus half-life is independent of initial concentration. Radioactive decay and decomposition of several compounds (e.g. N2O5, acetone, etc.) follow ﬁrst-order reactions, and the initial number of radioactive particles of concentration of compound decreases exponentially. There are many examples of ﬁrst-order growth process; for example, population growth, bank deposit earning compound interests, etc. For radioactive decay, say N0 number of particles becomes N number of particles in time t. ln(N0/N) = kt and, for N = N0/2 at t = t0.5, ln(2N0/N0) = kt0.5 ………………………..(9) Hence, ln(N0/N) = t ln2/t0.5 ………………………………. (10) Thermodynamic and kinetic orders of reactions. The order deﬁned above refers to kinetics of a process and must be distinguished from thermodynamic order of a reaction. For a thermodynamically ﬁrst-order reaction, there is a discontinuous change in the slope of the plot of free energy against temperature at the transition temperature: the ﬁrst derivative of free energy with temperature is discontinuous at the transition temperature. This means that the transformation involves a ﬁnite latent heat of transformation. Most structural transformations in physical metallurgy, for example, are thermodynamically ﬁrst order. In thermodynamically second (or higher)-order transformation, discontinuity at the transition temperature is noted for the second (or higher) order of derivative of free energy with temperature. Examples of second-order transitions are some ordering transformations. For example, in -brass, the order steadily falls to zero as the temperature rises. Thermodynamically second-order transition involves discontinuous change in speciﬁc heat at the transition temperature. Clearly, the thermodynamic order of a transformation has a clear-cut physical basis and must be an integer. In contrast, the kinetic order of a reaction is an empirical parameter and need not be an integer. yrs Problem: A wooden antique has 25.6% as much C14 as a recently grown piece of wood. If the amount of C14 in the atmosphere was the same, when the old wood died as it is now what is the age of the antique? Half-life for C14 is 5600 yrs. Solution: ln(N0/N) = t ln2/t0.5 and N = 0.256N0  t0.5 ln N0   t  ln 2 N  5600 1  0.693  ln 0.  256  11,010 Problem: Amount of a product from a reaction increases 20% per minute. What is the order of the reaction and what is the rate constant? Solution: If the product amount is P then dP/P/dt = 20/100 dP/dt = 0.2P. The reaction is ﬁrst order and the rate constant is 0.2 min1. Problem: The energy consumption of a country during 1990 was twice that consumed during 1980. Assuming that the country consumed a total amount of energy up to 1980, equal to E units, how much did it consume during the period 1980–1990? Make suitable assumptions. Solution: If we assume a natural growth in energy consumption, then we can assume a ﬁrst- order model for the increase. dE E k dt where k is the growth rate. Thus, (dE/dt) = kE, that is, the rate of energy consumption is proportional to energy consumption. Hence, the doubling time for both must be the same. Therefore, energy consumption during 1980–1990 must equal that consumed up to 1980, i.e. E units. Problem: During a reaction, the amount of product formed increases at a ﬁxed rate of 10% per minute. If after 10 minutes the amount is 10 g, then what is the amount formed after 1 h? Solution: Let the weight formed in time t be W. dW W  0.1  dW  0.1dt  ln W  0.1t dt W We have ln 10 log10 10   ln W log W 60 60  log W  log10  6 10  W  10 6 Weight after an hour there is W =106 g (if the reagent supply is unlimited). Degree of Reaction It is often more useful to express the kinetic equation in terms of a dimensionless parameter , called the degree of reaction or conversion. At time t C0  C ..........................(11) C0  C where C denotes the concentration of this reactant after the inﬁnite time of reaction. C If C = 0, then   1  C  C0 1   ..........................(12) C0 dC   kC n.................................(4) dt The differential form of the rate equation (equation 4) then becomes d k.C0n.1   n where k is another constant—the rate constant when   k.1   .............(13) reaction rate is expressed in terms of . ' n dt C0 k can be written for k. The term (1  ) denotes fraction of unreacted material which, together with the temperature-dependent rate constant, governs the reaction rate. In the simplest case of a zero-order reaction, n equals zero and the rate has a constant value k. d  k.1   0  k dt The integral form of rate equation is  = kt ………………. (14) For ﬁrst-order reactions (n =1), d  k 1   ..................(15) dt  ln (1  ) = kt ……………………… (16) For higher orders, the integrated form of the rate equation is obtained by substituting Equation 12 in Equation 5. C   1   C  C 1   ..........................(12) 0 C 0 1  1 1   n 1  n 1   kt......................(5) n  1  C C0  In general, the integral form of the rate equation is written as 1  1 1   n 1   n 1   kt n  1  Co 1   n 1 C0   1 n  1C n 1 1   1 n   kt o g () = kt ………………………… (17) where g () is an appropriate function of . Differentiating with respect to time, d d k g  . ' k  '  k. f  ...................(18) dt dt g   f () in the differential form of rate equation equals 1/g(). One advantage of using dimensionless  is that the dimension of k is always t1 irrespective of the order of reaction. Evaluation of Kinetic Parameters E and A The kinetic parameters E and A can be obtained using either a conventional integral approach or two differential approaches. Integral Approach  If the integrated form of the kinetic law, i.e. Equation 17 (g () = kt), is known, then one can plot, for different isothermal experiments, values of g() against time  Each of these plots will be linear, the slope being the value of k at that temperature. k = A exp ( E/RT)  The plot of ln k versus reciprocal temperature would be a straight line.  The slope of such a line gives  E/R and intercepts ln A.  Such an approach requires prior knowledge of the form of the function g(). There is also an implicit assumption that the activation energy E does not change during the course of the reaction. Differential Approaches In the differential approach, it is not necessary to assume any kinetic law at all. One can evaluate the activation energy E without knowing the forms of g() or f(). d k  '  k. f    A. exp E / RT . f  ................(19) dt g    d   ln A  ln f    E / RT................(20) Considering a ﬁxed value of ; ln    dt  Since f () has a ﬁxed value for a ﬁxed , it follows that a plot of the left-hand side against reciprocal temperature would be a straight line, the slope of which should yield the value of E/R. Advantage It is not necessary to know the form of either g() or f(). Also, one does not assume E to be independent of , and, therefore, E can be calculated at different levels of . Disadvantage One has to know d/dt values. One can obtain these from  t plots but calculation of slopes often involves uncertainties. Second differential approach d k  '  k. f    A. exp E / RT . f  ................(19) dt g   For a ﬁxed value of  1  d t    dt....................(21) A. exp E / RT  0 f   0 t  const.  exp E  ..................(22) A  RT  This shows that, provided f() remains unchanged, the time required for a ﬁxed value of  is directly proportional to exp(E/RT). Thus, a plot of ln t against reciprocal temperature should be a straight line with slope E/R. Advantages:  It is not necessary to know the form of either g() or f(). Also, one does not assume E to be independent of , and, therefore, E can be calculated at different levels of .  Using t values which are more reliably determined than d/dt values. However, the differential approaches cannot evaluate the pre-exponential constant A and, therefore, the rate constant k. Metallurgical Systems and Approaches in Kinetic Analysis Factors Determining Rate o Metallurgical transformations and reactions are mostly heterogeneous in nature, i.e. they involve more than one phase and there are distinct phase boundaries. Table 1: Types of heterogeneous non-catalysed reactions of metallurgical importance Interface Type Examples Solid–gas S 1 + G S2 Physical S 1 S2 + G Adsorption S1 + G1 S2 + G2 Chemical Oxidation of metals Decomposition of carbonates, or sulphates Oxidation of sulphide or gaseous reduction of oxides Solid–liquid SL Physical S + L1 ⇌ L2 Melting S + L1 L2 Dissolution–crystallization S + L1 S2 + L2 Chemical Leaching Cementation Table 1: Types of heterogeneous non-catalysed reactions of metallurgical importance Interface Type Examples Solid–solid S1 S2 Physical S1 + S 2 S3 + G Sintering, phase transformation S1 + S 2 S3 + S 4 Chemical Reduction of oxides by carbon Reduction of oxides or halides by metals Liquid - gas L⇌G Physical L 1 + G1 L 2 + G2 Distillation–condensation L1 + G L2 Absorption Chemical Steelmaking by pneumatic process Absorption of gases in water Liquid -Liquid L 1 ⇌ L2 Solvent extraction Slag–metal reaction Liquid metal–liquid metal extraction o In any given system, the rate of reaction at a given time depends basically on three factors, namely the nature of the system, the time of reaction and the temperature. Rate = f (nature of system, t, T) ……………………. (23) o Rate is measured using a suitable parameter which changes with reaction. At times, it may be difﬁcult to have a definite index of rate. o Uncertainties in kinetic studies can also arise out of poor incomplete characterization of the “nature” of the system. The word implies not only the system under test but also the environment. o It encompasses chemical composition, the presence of impurities which are inert and those which have a catalytic effect, the distribution of impurities, and physical factors such as the size and shape of particles. o Reproducible kinetic studies, therefore, require strict characterization. Factors which lead to complications for real systems involving solid–gas reactions, for example, may be summarized as follows:  The presence of impurities in the solid and the gas composition.  Attempts at interpreting packed bed reaction rates using experimental data on single particles of ﬁxed geometry (sphere, cube, etc.).  Multiparticle nature of systems with or without uniform particle size and shape.  Reagent starvation and heat transfer problems.  Variations in the pore structure of solid during reaction.  Non-isothermal condition, etc. To simplify kinetic studies on a ﬁxed system, it is often advantageous to ﬁx the temperature and study the variation of rate with time. For studying the effect of temperature, such isothermal experiments are repeated at several temperatures. Approaches Towards Rate Laws  One of the main aims of a kinetic study is frequently to establish a rate law which can be used in the prediction of reaction rates under a given set of circumstances.  There can be basically three approaches, namely 1. An empirical approach, 2. A semi-empirical approach, 3. A mechanistic approach.  When a system is rather complex and the transformation phenomena are not well understood, it is not possible to establish a theoretical basis for mathematical formulations.  In such cases, often it is advisable to employ an empirical approach with aims at establishing mathematical relationships to explain the trend in experimental results.  Where some theoretical understanding is available, a semi-empirical approach may be possible. A rational mechanistic approach is possible only when the system is well-deﬁned, and there is evidence to suggest a well-deﬁned reaction mechanism. The Empirical and Semi-empirical Approaches Densification of powder compacts  A good example of an empirical approach is provided by the studies on densification kinetics.  It should be noted that sintering and densification do not mean the same thing.  Sintering is seemingly a simple process which transforms green powder agglomerates into dense products which are strong.  The essential features are densification and bonding of particles leading to product strength. The driving force for sintering is provided by the tendency of the agglomerate to reduce the total surface area of the powder particles and therefore, total surface free energy.  Densification merely implies a reduction in porosity. It is often not well understood that sintering does not necessarily imply densification or vice versa.  One can have one without the other. In most cases, however, sintering leads to densification.  The subject of sintering may be approached, for simplicity, from a point of view that any green compact, at best is a two –phase material-porosity and the solid material, each with own morphology i.e. shape, size, distribution and amount.  Densification, a common feature of sintering, is the process of elimination of pores. For actual sintering, it is not possible to have a better, thoroughly rational approach unless one is restricted to the study of bounding between two spherical particles or rods of cylindrical geometry. For an actual powder mass, d  k. f  .................................1 dt where  represents the fraction of the original porosity destroyed. v p  vs  can be expressed as follows .............(2) vp where vS and the vP denote the volume of pores in the sintered compact and volume of pores in the original green compact. If V0 volume of the green compact, VS volume of the sintered compact, Vth volume of the compact when it attains the theoretical density, th and p porosity of the green compact expressed as a fraction. Vs  Vs   Vs  V 1 1    1    v p  vs Vo  Vs Vo  V0   Vo  Vo       vp Vo  Vth 1  th V   p p 1   o  Vo  th  where p is the porosity of the green pellet, 0 is the density of the green compact and V is the volume change. Densification studies are usually carried out using a cylinder compact. For such a compact, there is a change in both diameter and length as sintering progresses. While one can measure continuous change in length easily using a dilatometer, continuous measurement of changes in diameter is not possible. It will, therefore, be useful if densification could be studied only in terms of length changes only. This is possible under some assumptions. Let L0 length of the green cylindrical compact, D0 diameter of the green cylindrical compact, L length after densification for time t, D diameter after densification for time t, L maximum change in length after densification for infinite time, D maximum change in diameter after densification for infinite time. D D D We assume that at all values of t,   D .L..................................3 L L L  L  L0  L D  D0  D VS .LD 2 4 2    D  VS . L0  L  D0  D    L0  L  D0  2 L  4 4  L  2     LO  L  DO  D.L  2 V0  L0 D02 VS 4  L   L  D L    1  1 . ..............4  4 VO   LO  L DO .LO DO2 4 L V V If D  L  L.D0  1 S ..................................5 LO VO VO This will be especially true for long cylinders. Again if D ≈ L and then D0  L0 L 1 VO  VS  1 V 3 VS  L  3L If L

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