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Worksheet 7: Kinetics I

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    81872
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    Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

    Chemical kinetics is the study of the rates of chemical reactions; i.e., how fast reactants are converted into products. Experimentally determined rate law expressions show how the rate of a reaction depends upon the concentrations of the reactants and sometimes the products, too. This knowledge can be used to gain insight into the detailed molecular pathway (the mechanism) by which the reaction occurs. Understanding the mechanism allows chemists to devise ways of improving or modifying a chemical reaction for useful purposes.

    Learning Objectives

    • Understand how reaction rate is defined and measured
    • Understand the meaning of experimentally determined rate law expressions

    Success Criteria

    • Be able to define the rate of any reaction on the basis of the stoichiometry
    • Given a rate law expression, be able to predict how reaction rate changes with changes in reactant concentrations
    • Be able to determine the form of the differential rate law expression for a reaction, based on experimental kinetic data
    • Be able to calculate the rate constant value, with the proper units, from kinetic data

    Defining the Rate of a Reaction

    Rate is a change in some value over a period of time. The rate of a reaction is generally expressed in terms of the change in concentration (or sometimes pressure for a gas species) of reactants or products with time. The rate of a chemical reaction is not a constant. Rather, in most cases, it changes continuously. Generally a reaction starts with a fast rate and becomes slower as it progresses. Moreover, the rates of chemical reactions are influenced by temperature, being faster at higher temperature.

    We can define the rate of a reaction on the basis of the disappearance of any reactant or appearance of any product. But regardless of the species we choose to follow, for any point in the course of the reaction Rate should be defined so as to give the same number, regardless of the particular reactant or product on which it is based. To accomplish this, we must define Rate on the basis of the stoichiometry of the reaction. Also, we will conventionally define Rate so as to be a positive number, regardless of whether we are following the disappearance of a reactant species or the appearance of a product species over time.

    Suppose we want to define the rate of the reaction

    \[\ce{Cl_2(g) + 2NO(g) \rightarrow 2NOCl(g)} \label{eq1}\]

    over time at 300 K, starting with 1.00 mol/L \(Cl_2(g)\), 2.00 mol/L \(NO(g)\), and no \(NOCl\) in a vessel of fixed volume. (Note that the initial concentrations need not be in the stoichiometric ratio.) As the reaction proceeds, the concentrations of reactants, \(Cl_2(g)\) and \(NO(g)\), will decline, and the concentration of the product, \(NOCl(g)\), will increase, as shown in the following plot.

    If we follow the change in concentration of \(Cl_2\) over some small period of time then \(Δt = t_2 – t_1\). For example, from \(t_1 = 5 \times 10^{–5} s\) to \(t_2 = 10 \times 10^{–5} s\), the rate of change in concentration of chlorine gas will be

    \[rate = \dfrac{-\Delta [Cl_2]}{\Delta t} = \dfrac{-([Cl_2]_2-[Cl_2]_1)}{t_2-t_1}\]

    The bracket notation, such as \([Cl_2]\), is customarily taken to mean mol/L units. Since \(Cl_2\) is being consumed in the reaction, \(Δ[Cl_2]\) will be negative. We add a negative sign in front of \(Δ[Cl_2]\) so that Rate will always be defined as a positive number.

    If we choose to follow the change in \(NO\) concentration instead, we could write

    \[ \text{rate} = - \dfrac{\Delta [NO]}{\Delta t} = - \dfrac{([NO]_2-[NO]_1)}{t_2-t_1}\]

    However, based on the reaction' stoichiometry in Equation \(\ref{eq1}\), we see that two moles of \(NO\) are used for every one mole of \(Cl_2\). Therefore, the change in NO concentration would be two times as great as the change in chlorine concentration over the same period of time.

    To avoid ambiguity and to make Rate be numerically the same, regardless of which species we follow, we divide the change in \(NO\) concentration with time by two, its stoichiometry coefficient in the balanced reaction equation.

    Thus,

    \[\text{rate} = - \dfrac{1}{2} \dfrac{\Delta [NO]}{\Delta t}\]

    Likewise, if we followed the build-up of the \(NOCl\) product instead, we could define Rate on this basis as

    \[\text{rate} = + \dfrac{1}{2} \dfrac{\Delta [NOCl]}{\Delta t}\]

    Once again, we divide \(Δ[NOCl]/Δt\) by two, the stoichiometric coefficient of \(NOCl\) in the balanced equation, because \(Δ[NOCl]\) is changing at twice the rate of \(Δ[Cl_2]\) over any time period. But in this case we use a positive sign, because \(Δ[NOCl]\) is increasing, while both \(Δ[Cl_2]\) and \(Δ[NO]\) are decreasing. In general, when defining rate on the basis of a change in a product, we will use a positive sign; but when defining rate on the basis of a change in a reactant, we will use a negative sign. That way, Rate will be a positive number in either case.

    We will always want to define Rate on the basis of the stoichiometry of the reaction so that the numerical value obtained is the same, regardless of the species on which it is based.

    Therefore, for a general reaction of the form

    \[aA + bB \rightarrow cC + dD \label{gen}\]

    where \(a\), \(b\), \(c\), and \(d\) are the stoichiometric coefficients of the species \(A\), \(B\), \(C\), and \(D\), respectively, we will define

    \[ \text{rate} = - \dfrac{Δ[A]}{a Δt} = - \dfrac{Δ[B]}{b Δt} = + \dfrac{Δ[C]}{c Δt} = + \dfrac{Δ[D]}{d Δt} \]

    Any of these definitions is equivalent to the others and gives the same numerical value for a particular interval of time. Which one we choose to use is simply a matter of convenience in the case under consideration. Whichever definition we use, the numerical values for Rate will start out high at the beginning of the reaction and get smaller as the reaction continues.

    Q1

    Define the rate of the following gas phase reaction on the basis of each reactant and product, so that the measured value of Rate would be the same at any point in the reaction, regardless of which species was measured:

    \[2 N_2O_5(g) \rightarrow 4 NO_2(g) + O_2(g)\]

    Q2

    At a particular point in the course of the reaction \[2 N_2O_5(g) \rightarrow 4 NO_2(g) + O_2(g)\] the rate of disappearance of \(N_2O_5(g)\) was found to be \(1.16 \times 10^{–4} \,mol/L×s\).

    1. What is the rate of appearance of \(NO_2(g)\) at this point?
    2. What is the rate of appearance of \(O_2(g)\)?
    3. What is the Rate of the reaction at this point, regardless of the basis for its definition?

    The Differential Rate Law

    We have noted that in general the rate of a reaction slows down as time passes. This is a result of the decreased concentrations of reactants and increased concentrations of products. The dependence of the rate on concentration for a reaction can generally be stated as an experimentally determined mathematical expression, called the differential rate law (often, simply rate law). In general, the rate law has the form

    \[Rate = k[A]^m[B]^n ... \]

    where

    • \(m\) is the order with respect to \(A\)
    • \(n\) is the order with respect to \(B\)
    • \(m + n + ...\) is overall order = ω
    • \(k\) is the rate constant for the reaction at a certain temperature

    So the rate law expected for the four component reaction in Equation \(\ref{gen}\) would be

    \[Rate = k[A]^m[B]^n \]

    In general, these rate laws involve both reactions and products; if the reaction under consideration were bidirectional (instead of unidirectional like Equation \(\ref{gen}\))

    \[aA + bB \rightleftharpoons cC + dD \label{gen1}\]

    then the general rate law would be expressed as

    \[Rate = k[A]^m[B]^n [C]^o [D]^p \]

    From knowledge of the balanced equation we no NOTHING about the values of \(k\), \(m\), \(n\), \(o\) or \(p\). The explicit form of the rate law expression for a reaction ultimately depends upon the roles various species play in the mechanism of the reaction. In general, we cannot know the mechanism (more later) from looking at the overall stoichiometry of the balanced equation, and therefore there is no reliable way to predict the form of the differential rate law expression from the reaction equation.

    The explicit form of the differential rate law for a particular reaction must be experimentally determined and cannot in general be deduced from the overall stoichiometry of the reaction.

    The units of the rate constant, k, must be such that the Rate given by the differential rate law is in appropriate units of concentration per time. In general, if \(ω\) is the overall order of the rate law, the units of k will be

    \[k = (\text{conc. units})^{–(ω - 1)} (time)^{–1}\]

    Thus, if concentration is expressed in mol/L and time in seconds, a first-order reaction (\(ω = 1\)) has units of s-1 for k, and a second-order reaction (\(ω = 2\)) has units of (mol/L)–1×s–1 for k.

    Q3

    For the reaction \[CHCl_3(g) + Cl_2(g) \rightarrow CCl_4(g) + HCl(g)\]the experimentally observed rate law is \[Rate = k[CHCl_3][Cl_2]^½.\]

    1. What are the orders of the rate law with respect to each reactant, and what is the overall order?
    2. What effect on the rate of the reaction would occur if the concentration of \(CHCl_3\) were doubled while the concentration of \(Cl_2\) remained the same?
    3. What effect on the rate of the reaction would occur if the concentration of \(Cl_2\) were doubled while the concentration of \(CHCl_3\) remained the same?
    4. What effect on the rate of the reaction would occur if both reactant concentrations were doubled?
    5. If the rate of the reaction was determined with concentrations in mol/L and time in seconds, what are the units of k for this reaction?

    Determining the Rate Law

    The differential rate law for any reaction must be determined experimentally. One way of doing this is to compare the initial rate of the reaction (just after it starts) from two separate runs, with one of the reactant concentrations varied and any other reactant concentrations held the same in the two runs. This comparison allows the experimenter to see how the reaction depends on the concentration of that single reactant whose concentration was changed.

    For example, if doubling a particular reactant concentration, \([A]\), makes the reaction rate double, then the rate is directly proportional to that species’ concentration, and the rate law is first-order dependent in that species; i.e.,

    \[Rate \propto [A].\]

    If doubling a particular reactant concentration, \([B]\), makes the reaction rate quadruple, then the rate law is second-order dependent in that species (22 = 4); i.e.,

    \[Rate \propto [B]^2.\]

    Likewise, if doubling a particular reactant concentration, \([C]\), makes the reaction rate increase by a factor of eight, then the rate law is third-order in that species (23 = 8); i.e.,

    \[Rate \propto [C]^3.\]

    Sometimes, the order with respect to a species is a rational fraction (e.g., ½). For example, if quadrupling a particular reactant concentration, \([D]\), makes the reaction rate increase by a factor of only two, then the rate law is half-order dependent in that species (4½ = 2); i.e.,

    \[Rate \propto [D]^{1/2} \propto \sqrt{[D]}.\]

    In some cases, the rate of the reaction may not depend on the concentration of some particular reactant, in which case the rate dependence is zero-order in that reactant. If the rate law is zero-order dependent in any species, that species is generally omitted from the differential rate law expression, i..e,

    \[Rate \propto [D]^{0} \propto 1.\]

    Once the mathematical form of the rate law is known, the value of the rate constant, \(k\), can be calculated from the known concentrations and observed rate for a particular experiment, simply by solving the expression for \(k\). This requires knowledge of how the rate measured at any specific time (often the initial rate) changes as a function of concentration as the following examples demonstrate.

    Q4: Experimentally Determining the Rate Law

    Determine the rate law and value of \(k\) for the following reaction at 800 oC in a closed vessel. Remember how concentration is related to pressure within the Ideal Gas law.

    \[2H_2(g) + 2NO(g) \rightarrow 2H_2O(g) + N_2(g)\]

    Exp. \(P_{H_2}\) (atm) \(P_{NO}\) (atm) Initial Rate (atm/min)
    #1 0.13 0.39 0.050
    #2 0.26 0.39 0.10
    #3 0.13 0.78 0.20
    #4 0.26 0.78 0.40

    Q5: Experimentally Determining the Rate Law

    Determine the rate law for the following hypothetical reaction.

    \[H_2(g) + A_2(g) → 2HA(g)\]

    Exp. \([H_2]\) (mol/L) \([A_2]\) (mol/L) Initial Rate M/s
    #1 0.30 0.30 0.020
    #2 0.30 1.20 0.040
    #3 1.20 0.30 0.080

    Q6: Experimentally Determining the Rate Law

    Determine the differential rate law expression and value of \(k\) (with units) for the reaction

    \[A + B_2 + C → AB + BC\]

    Exp. \([A]\) \([B_2]\) \([C]\) Rate M/s
    #1 0.128 0.111 0.702 1.07 x 10-3
    #2 0.384 0.111 0.702 3.21 x 10-3
    #3 0.128 0.444 0.702 2.14 x 10-3
    #4 0.128 0.444 0.351 2.14 x 10-3