Does K increase with rate?

The numerical value of k, however, does not change as the reaction progresses under a given set of conditions. The reaction rate thus depends on the rate constant for the given set of reaction conditions and the concentration of A and B raised to the powers m and n, respectively.

chem.libretexts.org - 14.3: Effect of Concentration on Reaction Rates: The Rate Law

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The factors that affect the reaction rate of a chemical reaction, which may determine whether a desired product is formed. In this section, we will show you how to quantitatively determine the reaction rate. Rate laws may be written from either of two different but related perspectives. A differential rate law expresses the reaction rate in terms of changes in the concentration of one or more reactants (Δ[R]) over a specific time interval (Δt). In contrast, an integrated rate law describes the reaction rate in terms of the initial concentration ([R] 0 ) and the measured concentration of one or more reactants ([R]) after a given amount of time (t); integrated rate laws are discussed in more detail later. The integrated rate law is derived by using calculus to integrate the differential rate law. Whether using a differential rate law or integrated rate law, always make sure that the rate law gives the proper units for the reaction rate, usually moles per liter per second (M/s). Typically, reaction rates decrease with time because reactant concentrations decrease as reactants are converted to products. Reaction rates generally increase when reactant concentrations are increased. This section examines mathematical expressions called rate laws , which describe the relationships between reactant rates and reactant concentrations. Rate laws are mathematical descriptions of experimentally verifiable data.

Reaction Orders

For a reaction with the general equation:

[aA + bB ightarrow cC + dD label{14.3.1} ]

the experimentally determined rate law usually has the following form:

[ ext{rate} = k[A]^m[B]^n label{14.3.2}]

The proportionality constant (k) is called the rate constant, and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular rate constant value under a given set of conditions, such as temperature, pressure, and solvent; varying the temperature or the solvent usually changes the value of the rate constant. The numerical value of k, however, does not change as the reaction progresses under a given set of conditions. The reaction rate thus depends on the rate constant for the given set of reaction conditions and the concentration of A and B raised to the powers m and n, respectively. The values of m and n are derived from experimental measurements of the changes in reactant concentrations over time and indicate the reaction order, the degree to which the reaction rate depends on the concentration of each reactant; m and n need not be integers. For example, Equation ( ef{14.3.2}) tells us that Equation ( ef{14.3.1}) is mth order in reactant A and nth order in reactant B. It is important to remember that n and m are not related to the stoichiometric coefficients a and b in the balanced chemical equation and must be determined experimentally. The overall reaction order is the sum of all the exponents in the rate law: m + n. Note Under a given set of conditions, the value of the rate constant does not change as the reaction progresses. Although differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, integrated rate laws are used to determine the reaction order and the value of the rate constant from experimental measurements. (Click the link for a presentation of the general forms for integrated rate laws.) To illustrate how chemists interpret a differential rate law, consider the experimentally derived rate law for the hydrolysis of t-butyl bromide in 70% aqueous acetone. This reaction produces t-butanol according to the following equation:

[(CH_3)_3CBr_{(soln)} + H_2O_{(soln)} ightarrow (CH_3)_3COH_{(soln)} + HBr_{(soln)} label{14.3.3}]

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Combining the rate expression in Equation ( ef{14.3.2}) with the definition of average reaction rate

[ extrm{rate}=-dfrac{Delta[ extrm A]}{Delta t}]

gives a general expression for the differential rate law:

[ extrm{rate}=-dfrac{Delta[ extrm A]}{Delta t}=k[ extrm A]^m[ extrm B]^n label{14.3.4}]

Inserting the identities of the reactants into Equation ( ef{14.3.4}) gives the following expression for the differential rate law for the reaction:

[ extrm{rate}=-dfrac{Delta[mathrm{(CH_3)_3CBr}]}{Delta t}=k[mathrm{(CH_3)_3CBr}]^m[mathrm{H_2O}]^n label{14.3.5}]

Experiments to determine the rate law for the hydrolysis of t-butyl bromide show that the reaction rate is directly proportional to the concentration of (CH 3 ) 3 CBr but is independent of the concentration of water. Therefore, m and n in Equation ( ef{14.3.4}) are 1 and 0, respectively, and,

[ ext{rate} = k[(CH_3)_3CBr]^1[H_2O]^0 = k[(CH_3)_3CBr] label{14.3.6}]

Because the exponent for the reactant is 1, the reaction is first order in (CH 3 ) 3 CBr. It is zeroth order in water because the exponent for [H 2 O] is 0. (Recall that anything raised to the zeroth power equals 1.) Thus, the overall reaction order is 1 + 0 = 1. The reaction orders state in practical terms that doubling the concentration of (CH 3 ) 3 CBr doubles the reaction rate of the hydrolysis reaction, halving the concentration of (CH 3 ) 3 CBr halves the reaction rate, and so on. Conversely, increasing or decreasing the concentration of water has no effect on the reaction rate. (Again, when working with rate laws, there is no simple correlation between the stoichiometry of the reaction and the rate law. The values of k, m, and n in the rate law must be determined experimentally.) Experimental data show that k has the value 5.15 × 10−4 s−1 at 25°C. The rate constant has units of reciprocal seconds (s−1) because the reaction rate is defined in units of concentration per unit time (M/s). The units of a rate constant depend on the rate law for a particular reaction. Under conditions identical to those for the t-butyl bromide reaction, the experimentally derived differential rate law for the hydrolysis of methyl bromide (CH 3 Br) is as follows:

[ extrm{rate}=-dfrac{Delta[mathrm{CH_3Br}]}{Delta t}=k'[mathrm{CH_3Br}] label{14.3.7}]

This reaction also has an overall reaction order of 1, but the rate constant in Equation ( ef{14.3.7}) is approximately 106 times smaller than that for t-butyl bromide. Thus, methyl bromide hydrolyzes about 1 million times more slowly than t-butyl bromide, and this information tells chemists how the reactions differ on a molecular level. Frequently, changes in reaction conditions also produce changes in a rate law. In fact, chemists often alter reaction conditions to study the mechanics of a reaction. For example, when t-butyl bromide is hydrolyzed in an aqueous acetone solution containing OH− ions rather than in aqueous acetone alone, the differential rate law for the hydrolysis reaction does not change. For methyl bromide, in contrast, the differential rate law becomes rate =k″[CH 3 Br][OH−], with an overall reaction order of 2. Although the two reactions proceed similarly in neutral solution, they proceed very differently in the presence of a base, providing clues as to how the reactions differ on a molecular level. Note Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements. Example (PageIndex{1}) Below are three reactions and their experimentally determined differential rate laws. For each reaction, give the units of the rate constant, give the reaction order with respect to each reactant, give the overall reaction order, and predict what happens to the reaction rate when the concentration of the first species in each chemical equation is doubled. (mathrm{2HI(g)}xrightarrow{ extrm{Pt}}mathrm{H_2(g)}+mathrm{I_2(g)}

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\ extrm{rate}=-frac{1}{2}left (frac{Delta[mathrm{HI}]}{Delta t} ight )=k[ extrm{HI}]^2) (mathrm{2N_2O(g)}xrightarrow{Delta}mathrm{2N_2(g)}+mathrm{O_2(g)}

\ extrm{rate}=-frac{1}{2}left (frac{Delta[mathrm{N_2O}]}{Delta t} ight )=k) (mathrm{cyclopropane(g)} ightarrowmathrm{propane(g)}

\ extrm{rate}=-frac{Delta[mathrm{cyclopropane}]}{Delta t}=k[mathrm{cyclopropane}]) Given: balanced chemical equations and differential rate laws Asked for: units of rate constant, reaction orders, and effect of doubling reactant concentration Strategy: Express the reaction rate as moles per liter per second [mol/(L·s), or M/s]. Then determine the units of each chemical species in the rate law. Divide the units for the reaction rate by the units for all species in the rate law to obtain the units for the rate constant. Identify the exponent of each species in the rate law to determine the reaction order with respect to that species. Add all exponents to obtain the overall reaction order. Use the mathematical relationships as expressed in the rate law to determine the effect of doubling the concentration of a single species on the reaction rate. Solution A [HI]2 will give units of (moles per liter)2. For the reaction rate to have units of moles per liter per second, the rate constant must have reciprocal units [1/(M·s)]: (k extrm M^2=dfrac{ extrm M}{ extrm s}k=dfrac{ extrm{M/s}}{ extrm M^2}=dfrac{1}{mathrm{Mcdot s}}=mathrm{M^{-1}cdot s^{-1}}) B The exponent in the rate law is 2, so the reaction is second order in HI. Because HI is the only reactant and the only species that appears in the rate law, the reaction is also second order overall. C If the concentration of HI is doubled, the reaction rate will increase from k[HI] 0 2 to k(2[HI]) 0 2 = 4k[HI] 0 2. The reaction rate will therefore quadruple. A Because no concentration term appears in the rate law, the rate constant must have M/s units for the reaction rate to have M/s units. B The rate law tells us that the reaction rate is constant and independent of the N 2 O concentration. That is, the reaction is zeroth order in N 2 O and zeroth order overall. C Because the reaction rate is independent of the N 2 O concentration, doubling the concentration will have no effect on the reaction rate. A The rate law contains only one concentration term raised to the first power. Hence the rate constant must have units of reciprocal seconds (s−1) to have units of moles per liter per second for the reaction rate: M·s−1 = M/s. B The only concentration in the rate law is that of cyclopropane, and its exponent is 1. This means that the reaction is first order in cyclopropane. Cyclopropane is the only species that appears in the rate law, so the reaction is also first order overall. C Doubling the initial cyclopropane concentration will increase the reaction rate from k[cyclopropane] 0 to 2k[cyclopropane] 0 . This doubles the reaction rate.

chem.libretexts.org - 14.3: Effect of Concentration on Reaction Rates: The Rate Law

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