Zero Order Reaction

9 Key excerpts on "Zero Order Reaction"

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  Transport Modeling for Environmental Engineers and Scientists

  • Mark M. Clark(Author)
  • 2012(Publication Date)
  • Wiley

    (Publisher)

  These are critical tools for the environmental scientist and engineer. We point out here that kinetic studies are an important tool in understand-ing reaction mechanisms. An in-depth discussion of mechanism is outside the scope of this chapter, but the reader can follow up on this subject in some of the excellent chemistry and physical chemistry references at the end of the chapter. 9.2. FIRST-ORDER REACTIONS Consider that we find the following relationship between reactant A and product B, and that there are no other reactants or products: A -> B (9.2) Equation 9.2 represents the stoichiometry of an irreversible reaction: 1 mol of A is converted to 1 mol of B. The reaction rate is defined as the rate of change in the concentration of a reactant or product. For computational purposes, we will need to quantify the reaction rate as a differential equation. We make the assumption that the rate of change in the concentration of A and B is proportional to the concentration of A, rA = d[A] = _ k[A] ( 9 3 ) dt and r B = ^ = k[A] (9.4) dt Here r A and r B stand for the reaction rates of A and B, k is the reaction rate or kinetic constant, and [A] and [B] represent the appropriate concentrations of A and B. The concentration units could be molar, mass, or colloidal number concentration in water, or partial pressure or aerosol number concentration in air. In molar concentration units, the correct units for the r values are mol-L~ 3 -7 and the correct units for k are T~ x . From Eqs. 9.2 to 9.4, we con-clude that rA .m..m.^ <9 .5) dt dt In the present case, we have used the stoichiometry (Eq. 9.2) to guess at the kinetics law (Eqs. 9.3 and 9.4). Reactions in which the kinetics are suggested by the stoichiometry are called elementary reactions (see Section 9.6). FIRST-ORDER REACTIONS 497 Reactions 9.3 and 9.4 are called first-order reactions. First-order reactions are called linear because Eqs. 9.3 and 9.4 are linear differential equations. Integrating Eq.

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  Chemistry

  Principles and Reactions

  • William Masterton, Cecile Hurley(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Most of them take place at solid surfaces, where the rate is inde-pendent of concentration in the gas phase. A typical example is the thermal decom-position of hydrogen iodide on gold: HI( g ) 9 Au : 1 2 H 2 ( g ) 1 1 2 I 2 ( g ) When the gold surface is completely covered with HI molecules, increasing the concentration of HI( g ) has no effect on reaction rate. It can be shown using the integrated rate law that the concentration-time rela-tion for a zero-order reaction is [A] 5 [A] o 2 kt (11.4) Comparing this equation with that for a straight line, y 5 b 1 mx ( b 5 y -intercept, m 5 slope) it should be clear that a plot of [A] versus t should be a straight line with a slope of 2 k. Putting it another way, if a plot of concentration versus time is linear, the reaction must be zero-order; the rate constant k is numerically equal to the slope of that line but has the opposite sign. For a second-order reaction (a reaction whose rate depends on the second power of the reactant concentration) involving a single reactant, such as acetalde-hyde (recall Example 11.2), A 9: products rate 5 k [A] 2 it is again necessary to resort to calculus* to obtain the concentration-time relationship 1 [A] 2 1 [A] o 5 kt (11.5) In a zero-order reaction, all the reactant is consumed in a finite time, which is 2 t 1/2 . *If you are taking a course in calculus, you may be surprised to learn how useful it can be in the real world (e.g., chemistry). The general rate expressions for zero-, first-, and second-order reactions are 2 d [A]/ dt 5 k 2 d [A]/ dt 5 k [A] 2 d [A]/ dt 5 k [A] 2 Integrating these equations from 0 to t and from [A] o to [A], you should be able to derive the equa-tions for zero-, first-, and second-order reactions. Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

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  Chemistry

  • John A. Olmsted, Gregory M. Williams, Robert C. Burk(Authors)
  • 2020(Publication Date)
  • Wiley

    (Publisher)

  An example of this is the decomposition of nitrous oxide, N 2 O, into nitrogen and oxygen: 2 N 2 O(g) ⟶ 2 N 2 ( g) + O 2 ( g) If this reaction is done in the presence of a platinum surface, then the reaction takes place on that surface and not in the gas phase. The number of N 2 O molecules on the surface is limited by the area of the surface, which is constant. Thus the reaction rate is constant and does not 632 CH A P TER 1 3 Kinetics: Mechanisms and Rates of Reactions change if the partial pressure of N 2 O (g) is varied. Reactions that take place on a surface are covered in more detail in Section 13.7. The second situation occurs when there are two or more reactants, and the concentration of one is so much larger than the others that changing it does not change the rate of the reac- tion, because the rate is limited by the concentrations of the other reactants. To understand how a rate can be independent of concentration, consider the reaction a A ⟶ B, and let’s assume that the reaction has been found to be zeroth order in [A]. Thus we have rate = ( − 1 __ a)( d[A] ____ dt ) = k[A] 0 = k (because [A] 0 = 1) In other words, the rate of reaction is a constant and is not related to the concentration of A. As we did for first- and second-order reactions, we can obtain an integrated rate law for a zeroth-order reaction. From the above, we have rate = ( − 1 __ a)( d[A] _____ dt ) = k d[A] = − ak dt  d[A] = −  ak dt [A] = −akt + constant At t = 0, [A] = [A] 0 , so the constant of integration is simply [A] 0 . Thus, the final integrated rate equation is [A] = [A] 0 − akt (13.6) As we did for first- and second-order reactions, we can make a plot to determine whether a reaction is zeroth order and to determine the rate constant. In this case, the appropriate plot would be of [A] 0 − [A] versus t, in which case the plot should be linear, with a slope of −ak. The half-life for a zeroth-order reaction is found by setting [A] = ½[A] 0 and solving for t in Equation 13.6.

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  Modeling of Chemical Kinetics and Reactor Design

  • A. Kayode Coker(Author)
  • 2001(Publication Date)
  • Gulf Professional Publishing

    (Publisher)

  First order (n = 1; k = 1/sec) (–r A ) = kC A (3-19) 3. Second order (n = 2; k = m 3 /moles-sec) (–r A ) = kC 2 A (3-20) 4. Third order (n = 3; k = (m 3 /moles) 2 •sec –1 ) (–r A ) = kC 3 A (3-21) DETERMINING THE ORDER OF REACTIONS Zero Order ReactionS The rate of a chemical reaction is of a zero order if it is independent of the concentrations of the participating substances. The rate of reaction is determined by such limiting factors as: 1. In radiation chemistry, the energy, intensity, and nature of radiation. 2. In photochemistry, the intensity and wave length of light. 3. In catalyzed processes, the rates of diffusion of reactants and availability of surface sites. If the rate of the reaction is independent of the concentration of the reacting substance A, then the amount dC A by which the concentra-tion of A decreases in any given unit of time dt is constant throughout the course of the reaction. The rate equation for a constant volume batch system (i.e., constant density) can be expressed as: − ( ) = − = r dC dt k A A (3-22) The negative sign indicates that the component A is removed from the system, and k is the velocity constant with the units as moles/m 3 -sec. Assuming that at time t 1 the concentration of A is C AO , and at Reaction Rate Expression 117 time t 2 the concentration is C Af , integrating Equation 3-22 between these limits gives: − = ∫ ∫ dC k dt A C C t t AO Af 1 2 (3-23) − − ( ) = − ( ) C C k t t Af AO 2 1 k C C t t AO Af = − ( ) − ( ) 2 1 (3-24) If t 1 = 0, Equation 3-24 reduces to C Af = C AO – kt 2 (3-25) Plotting the concentration (C A ) versus time t gives a straight line, where C AO is the intercept and k is the slope. The velocity constant k may include arbitrary constants resulting from various limiting factors such as diffusion constants and a fixed intensity of absorbed light.

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  Introduction to Chemical Kinetics

  • Gordon Skinner(Author)
  • 2012(Publication Date)
  • Academic Press

    (Publisher)

  2.3 REACTION ORDER The rate of a reaction is usually found to depend on the concentra-tion of some of the reactants and is often influenced by the presence of other substances deliberately or accidentally added to the reaction mixture. At a given temperature, and perhaps within a limited range of concentrations, one can write a rate law for a reaction of the form rate = k H [AJ^M* 3 ' (2.3) i where the A» are the reactants, the X 3 are substances that are not reactants but do influence the rate, the as and f3's are coefficients that are not neces-sarily related to the stoichiometric coefficients v, and & is a rate constant. For a long time it was thought that the a 's and p's would be integers, but it is now clear that they need not be. Two kinds of reaction order are commonly defined. The overall order of a reaction is the sum of all the a 's and p's in the rate law expression. The overall order tells us how the rate responds to changes in the absolute concentration at constant relative concentration, such as one would produce by changing the pressure in a gaseous system or diluting a liquid system with an inert solvent. The order with respect to A» is a iy and similarly the order with respect to Xy is 13j. These individual orders not only tell us how sensitive the system is to changes in the concentration of each species, but may also suggest the chemical mechanism of the reaction. 2.3 REACTION ORDER For many elementary reactions, the rate law coefficients are equal to the stoichiometric coefficients. This is true for some of the elementary reactions given above, provided the concentrations are moderate. That is, for O H + CH 3 Br CH3OH + Br the rate law is rate = /c [OH-][CH 3 Br] in dilute solutions. Since the reaction is elementary, the rate depends on the number of collisions between the reactants, which in turn depends on the product of the concentrations.

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  Chemistry

  Structure and Dynamics

  • James N. Spencer, George M. Bodner, Lyman H. Rickard(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  14.12 The Integrated Form of Zero-, First-, and Second-Order Rate Laws As we have seen, the rate law for a reaction provides a basis for studying the mechanism of the reaction. It is also useful for predicting how much reactant will remain or how much product will be formed in a given amount of time. For these calculations, we use the integrated form of the rate laws. A derivation of the integrated forms of the rate laws can be found in the Special Topic section at the end of this chapter. Let’s start with the simplest of all rate laws, a reaction that is zero-order. When this equation is rearranged and both sides are integrated, we get the fol- lowing result. Integrated form of the zero-order rate law: In this equation, (X) is the concentration of X at any moment in time, (X) 0 is the initial concentration of this reagent, k is the rate constant for the reaction, and t is the time since the reaction started. (X) - (X) 0 = - kt d(X) dt = k(X) 0 = k 658 CHAPTER 14 / KINETICS E x e r c i s e 1 4 . 7 How long will it take for the concentration of a solution that was initially 2.0 M to decrease to 1.0 M if the reaction is zero-order and the rate constant, k, for the reaction is 1.5  10 2 mol/L s? Solution For a zero-order reaction, the concentration at any time (X) after the reaction has begun is related to the initial concentration (X) 0 by the following equation. Substituting the known values for the initial and final concentration of X and the rate constant for the reaction into this equation gives the amount of time it takes for the change in the concentration of X to occur. Solving for t gives the following result. t = 1.0 mol/L 1.5 * 10 - 2 mol/L # s = 67 s (1.0 mol/L) - (2.0 mol/L) = - (1.5 * 10 - 2 mol/L # s) * t (X) - (X) 0 = - kt # Let’s now turn to the rate law for a reaction that is first-order in the disap- pearance of a single reactant, X. When the equation is rearranged and both sides are integrated, we get the fol- lowing result.

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  Basic Physical Chemistry

  The Route to Understanding

  • E Brian Smith(Author)
  • 2012(Publication Date)
  • ICP

    (Publisher)

  255 256 | Basic Physical Chemistry Thus, − d [ A ] d t = k r [ A ] n [ B ] m would be a typical expression of the rate of the reaction illustrated above. n and m are often simple integers (or, sometimes, half integers) and k r is a constant at any given temperature termed the rate constant . 1 The sum of the exponents, (n + m) , is defined as the overall order of the reaction . n and m are the orders with respect to each of the reactants and a knowledge of these orders, together with a knowledge of the rate constant, k r , enables us to calculate the rate of the reaction when different concentrations of the reactants are present. A simple example is H 2 + I 2 → 2HI , for which we find d [ HI ] d t = k r [ H 2 ][ I 2 ] . This corresponds to an overall order of 2 and is first-order with respect to both H 2 and I 2 . By contrast, for H 2 + Br 2 → 2HBr, the reaction rate shows a complex dependence on H 2 and Br 2 , even though the stoichiometric equation is the same as that for the formation of HI (Section 11.10). Another example is the reaction S 2 O 2 − 8 + 2I → I 2 + 2SO 2 − 4 , which might be expected to be a third-order reaction, but experiment shows it to be second-order. Reactions are sometimes defined in terms of molecularity . This is a theoretical concept which indicates the number of molecules participating in an elementary step in a reaction. The molecularity and the order of a reaction are often different. 11.2 First-order reactions The simplest type of kinetic behaviour is that represented by first-order rate equations. An example is radioactive decay of an unstable isotope, X → Y + α or β radiation. The carbon 14 C nucleus decays in this way, with a half-life , the time for the initial concentration to be reduced to half its value, of some 5200 years. An example of a chemical reaction which occurs with first-order kinetics is the decomposition of N 2 O 5 , N 2 O 5 ( g ) → 2 NO 2 ( g ) + 1 2 O 2 ( g ).

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  Chemical and Energy Process Engineering

  • Sigurd Skogestad(Author)
  • 2008(Publication Date)
  • CRC Press

    (Publisher)

  Consider, for example, the reaction A → B with reaction rate r = kc α A Here, α is the reaction order. With α = 0, we have a zeroth order reaction and the reaction rate is independent of concentration c A . Correspondingly, a first-order reaction has α = 1, and a second-order reaction has α = 2. The reaction A + 2 B → C with r = kc A c 2 B is said to be first-order with respect to A and second-order with respect to B, and the overall reaction order is 1 + 2 = 3. Note that the order does not need to be an integer. For a reversible reaction , we also need to consider the reverse reaction. For example, for the reaction 2 A ⇋ B the reaction rate can be given by r = k 1 c 2 A − k 2 c B For gas phase reactions , partial pressure rather than concentration is usually used. For example, for the gas phase reaction A + B → C , we may write r = kp A p B where p i [bar] is the partial pressure of component i and k here has unit [mol A/m 3 , s bar 2 ]. More generally, instead of concentration or partial pressure, the activity a of the components is used in the reaction rate, that is, we write r = k ( T ) · f ( a i ). For cases with elementary reaction kinetics the stoichiometric coefficients appear directly as exponents in the rate expressions. For example, for the gas phase reaction aA + bB ⇌ cC + dD with elementary reaction kinetics we have r = k 1 p a A p b B − k 2 p c C p d D However, in general, the reaction mechanisms are not so simple. For example, for a heterogeneous catalytic reaction, the rate is often proportional to the fraction θ of active surface sites occupied ( Langmuir kinetics ). Here, θ cannot exceed 1 and the resulting reaction rates are in the form (for example, for the reaction A + B → P ) r = kθ A θ B = k b A p A · b B p B (1 + b A p A + b B p B + b C p C ) 2

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  Geochemical Kinetics

  • Youxue Zhang(Author)
  • 2021(Publication Date)
  • Princeton University Press

    (Publisher)

  A good example of a first-order (pseudo-first-order) chemical reaction is the hydration of CO 2 to form carbonic acid, Reaction 1-7f, CO 2 (aq) þ H 2 O(aq) ? H 2 CO 3 (aq). Because this is a reversible reaction, the concentration evolution is considered in Chapter 2. 1.3 KINETICS OF HOMOGENEOUS REACTIONS 21 1.3.5.3 Second-order reactions Most elementary reactions are second-order reactions. There are two types of second-order reactions: 2A ? C and A þ B ? C. The first type (special case) of second-order reactions is 2A ! C : (1-50) The reaction rate law is d x = d t ¼ k [A] 2 ¼ k ([A] 0 2 x ) 2 : (1-51) The solution can be found as follows: d x = ([A] 0 2 x ) 2 ¼ k d t : (1-51a) Then d([A] 0 2 x ) = ([A] 0 2 x ) 2 ¼ 2 k d t : (1-51b) Then 1 = ([A] 0 2 x ) 1 = [A] 0 ¼ 2 kt : (1-51c) That is 1 = [A] 1 = [A] 0 ¼ 2 kt : (1-52) Or [A] ¼ [A] 0 = (1 þ 2 k [A] 0 t ) : (1-53) The concentration of the reactant varies with time hyperbolically. The second type (general case) of second-order reactions is A þ B ! C : (1-54) The reaction rate law is d x = d t ¼ k [A][B] ¼ k ([A] 0 x )([B] 0 x ) : (1-55) If [A] 0 ¼ [B] 0 , The solution is the same as Equation 1-53. For [A] 0 = [B] 0 , the so-lution can be found as follows: d x = {([A] 0 x )([B] 0 x )} ¼ k d t : (1-56) u d x = ([A] 0 x ) u d x = ([B] 0 x ) ¼ k d t , where u ¼ 1 = ([B] 0 [A] 0 ) : u ln {([A] 0 x ) = [A] 0 } u ln {([B] 0 x ) = [B] 0 } ¼ kt : ln{([A] 0 x ) = [A] 0 } ln{([B] 0 x ) = [B] 0 } ¼ k ([B] 0 [A] 0 ) t x ¼ [A] 0 [B] 0 ( q 1) = ( q [A] 0 [B] 0 ), where q ¼ exp{ k ([B] 0 [A] 0 ) t } : (1-57) 22 1 INTRODUCTION Figure 1-1 compares the concentration evolution with time for zeroth-, first-, and the first type of second-order reactions. Table 1-2 lists the solutions for concen-tration evolution of most elementary reactions.

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