MEASUREMENT OF INTEREST RATE AND PERIOD USING LOGARITHM EQUATIONS: A STUDY OF TIME VALUE OF MONEY

The present work is concerned with the numerical study of the elasticity consistency of the spatial rate equations using the conventional Oldroyd, Truesdell, Cotter–Rivlin, Jaumann and Green–Naghdi rates and the three novel co‐rotational ΩE‐ and Ω¯L‐based, logarithmic rates, and of the rotated material rate equation describing the relationship between the material time derivatives of the rotated Kirchhoff stress and material logarithmic strain. To this end, three integration procedures for updating stress are presented. The stress responses of several typical deformation processes are simulated. According to the numerical results we know that among the spatial rate equations only the logarithmic rate equation is consistent with elasticity under constant material parameters. Integrating the other spatial rate equations will provide path‐dependent stress response. These numerical conclusions support the arguments in H. Xiao et al. (Acta Mechanica 1999; 138:31–50). The reasons leading to elasticity inconsistency of spatial rate equations are analysed. If the material parameters are assumed to be strain‐dependent, the logarithmic rate equation loses also its elasticity‐consistent property. The numerical results prove also that the spatial logarithmic and rotated material rate equations are equivalent to each other. Copyright © 2002 John Wiley & Sons, Ltd.

Rate equations for dislocation-free and dislocation-assisted growth of gallum

Minimum entropy production and logarithmic rate equations

Quartz dissolution: Negative crystal experiments and a rate law

A fundamental study of the kinetics of zinc oxidation in the temperature range 320–415°C in atmospheres of pure oxygen and oxygen doped with gaseous impurities

Kinetic study on bonding reaction of gelatin with CdS nanopaticles by UV–visible spectroscopy

Uses and limitations of measurements of rates of isotopic exchange and incorporation in catalyzed reactions

Implementation and estimation of errors for the absolute determination of 125I activity

The surface chemistry and kinetics of tungsten chemical vapor deposition and selectivity loss

The carbonyl reagent amino-oxyacetate is frequently used in metabolic studies to inhibit individual pyridoxal phosphate enzymes. The reaction of this compound with three such enzymes, aspartate transaminase, 4-aminobutyrate transaminase and dopa (3,4-dihydroxyphenylalanine) decarboxylase, was studied to determine the extent to which the inhibition is reversible and the rates at which it takes place. Reactions were followed by observing changes in the absorption spectra of the bound coenzyme and by measuring loss of enzyme activity. The reactions with aspartate transaminase and aminobutyrate transaminase were not rapidly reversible and had second-order rate constants (21 degrees C) of 400 M-1.s.1 and 1300 M-1.s-1 respectively and all all concentrations studied showed the kinetics of a simple bimolecular reaction. The reaction with 4-aminobutyrate transaminase could not be reversed and that with aspartate transaminase could only be reversed significantly by addition of cysteinesulphinate to convert the enzyme into its pyridoxamine form. The first-order rate constant (21 degrees C) for the reverse reaction was 4 X 10(-5)s-1. Dopa decarboxylase inhibition by amino-oxyacetate was more rapid and more readily reversible, but measurements of rate and equilibrium constants were not obtained for this enzyme.

IntroductionIn these times of tight pharma R&D budgets, tightening academic budgets, and increasing merger and acquisition pressures (and following the burst of the biotechnology bubble), investments in laboratory automation demand justification beyond simple declarations that something is better or faster. Choices need to be analyzed to convince not only top management but also convince us that automation makes sense for an organization, be it an established pharmaceutical company, start-up biotech, or academic research center. Formal techniques exist to help complete this analysis and answer the key question posed by economic justification: Is this the best choice for an organization given limited funds?Laboratory automation is a technological endeavor. As such, existing economic tools and financial techniques can be used to compare technology and investments based on economic measures of effectiveness. These tools and techniques are often called economic analysis, engineering economy, or economic decision analysis. 1White J.A. Agee M.H. Case K.E. Principles of Engineering Economic Analysis.. John Wiley & Sons, New York1989Google Scholar, 2HEI Consulting Engineering Economy Glossary: http://www.hei.ca/eglossary.html.Google Scholar, 3Definition of Engineering Economy; Iowa State University class ConE241: http://www.public.iastate.edu/cone241/Engecn1.ppt.Google Scholar Traditionally used in engineering and manufacturing, these techniques have become ingrained into public sector decision making and therefore are taught widely. They run the gamut from a simple comparison of alternatives to defining the return on investment (ROI), cost-to-benefit ratios, breakeven analysis, and more. Luckily, a comparison of alternatives for laboratory operations needs only a subset of these methods. In this JALA Tutorial, we present a primer on the basics of engineering economic analysis and describe several common methods for justifying the introduction of automation into a laboratory using plenty of examples.The Present Economy StudyWhen judging alternatives, money follows one of two time frames: present economy or the time value of money. The easier of these two to understand is the present economy study, which ignores the way that the value of money changes with time. A present economy study is acceptable if one of the following criteria is met:1There is no investment of capital, only out-of-pocket costs.2After any first (capital) cost is paid, the long-term costs will be the same or proportional to the first cost, no matter which alternative is picked.3The alternatives will have essentially identical results regardless of the capital investment.These conditions allow the person comparing alternatives to view the choice as occurring only once and only right now.Example 1A lab needs to purchase extra nitrile gloves as a one-time event, and it has a choice between two suppliers. Both provide the same number of gloves per box, with the same specifications. This is a one-time purchase (i.e., not part of a standing order); there will be no costs at a future time associated with the various economic alternatives, and it is not a capital expense (meeting criterion 1 above). Therefore, only the price in the present economy is important. If one supplier’s product is $7 and the other’s is $9, then the most economically effective choice is the $7 purchase.Of course, this example is simplistic. More complexity can be added by taking into account quality issues or unquantifiable issues such as look and feel. Present economy studies come into use when various choices of equally capable equipment have different purchase prices but the same maintenance, support and consumable costs, and the same user acceptance. The purchase price at that time is the only factor needed to determine economic effectiveness.Costs TerminologyBefore looking at more realistic justifications involving the time value of money, we must define the types of costs (and income) that need attention as part of an examination of alternative solutions. These can be called the life cycle costs, which refer to all of the expenditures involved in a project or purchase throughout its entire useful life. For example:1First cost: The total initial investment required getting an item or project ready for service, usually non-recurring, including purchase price, installation, modifications to infrastructure, etc.2Operating and maintenance costs: Recurring costs necessary to keep the item in service, including service contracts and personnel salaries. These can be very large. These are also examples of future costs because they occur in the future.3Opportunity costs: The costs of missed opportunities due to investing in the project in question. For example, when manual labor is not available for other projects, it can create a missed opportunity, and there can be a cost associated with that loss.4Salvage value (income): Any money that potentially can be recovered for an item at the end of its life cycle. It equals market value less disposal costs. This can be especially difficult to calculate for laboratory automation.5Product sales (income): If a product is being produced for sale, the projected income from each unit/dose/test sold can be used to offset costs. In this situation, the costs listed here should result in a profit.Some of these costs are fixed costs, meaning that they do not change with the number of experiments, procedures, or operations performed, such as the purchase price. Otherwise, they would be variable costs, as they do scale with the amount of work performed, such as reagents consumed. All of the costs mentioned here are direct costs, meaning they are directly involved with the process in question, either as equipment, people, or material (including consumables and reagents). Indirect costs would be the costs of the building, utilities, upper management, etc. In traditional uses of economic analysis, these costs are used to help compare alternatives regarding where to locate a manufacturing plant, the infrastructure for that plant, and management organization alternatives. Very rarely does laboratory automation affect an organization to that level. Therefore, the focus of this tutorial is on the direct costs.Time Value of MoneyThe value of money changes with time. It either gains worth over time if invested properly or decreases in value due to returns that are lower than the rate of inflation. Economic analyses consider this using the interest calculation. Following is an example of the mathematical relationships between the present and future values of money.Let n be a number of years, and In the accumulated interest earned over n years. Then, the future value (Fn) is related to the present value (P) by the equationFn=P+InNow let i be the annual interest rate, which is the change in value of $1 over a 1-year period. There are two approaches to compute the value of In: simple and compound interest. With simple interest, P changes in value each year by an amount of P multiplied by i. In other words, for n years,In=PinandFn=P(1+in)Note, F is the future value of the money/investment and is almost never used in day-to-day personal life.With compound interest, i is the rate of change in the accumulated value of money. In this case, Fn–1 is the future value of the previous year. This is the kind of interest with which most people are familiar:In=iFn-1andFn=Fn-1(1+i)To judge the total effect of compound interest, we must create and summarize a table of the interest over a number of years. This relationship is shown in Table 1.Table 1Illustrating the effect of compound interestABC = A + BEnd of periodAmount owedInterest for next periodAmount owed for next period0PPip + Pi = P(1 + i)1P(1 + i)P(1 + i)iP(1 + i) + P(1 + i)i = P(1 + i)22P(1 + i)2P(1 + i)2iP(1 + i)2 + P(1 + i)2i = P(1 + i)3........n–1P(1 + i)n–1P(1 + i)n–1iP(1 + i)n–1 + P(1 + i)n–1i = P(1 + i)nnP(1 + i)n = F Open table in a new tab Single Sums of MoneyThe final equation from Table 1 can be used to find the future value of some present amount of money. A cash flow diagram can be used to describe this situation. This is a graphical representation of the flow of money into or out of an investment situation, shown on a timeline. Inflows are depicted as upward pointing arrows (income), and outflows as downward pointing arrows (cost or expense). Ideally, arrow lengths are proportional to the amount of money represented.Example 2$40,000 (the present value, P) is invested into a 5-year certificate of deposit with 3% annual interest (i). The interest is reinvested and compounded once each year for 5 years (the number of interest periods, n). What amount of money will be available at the end of five years (the future value, F)?Figure 1 illustrates the problem as a cash flow diagram. To solve this problem, use the equation F = P(1 + i)n. In standard economic analysis, this is written in a notation that is intended to be used with generated solution tables, which are provided with most economic analysis texts. For this problem, the notation is=P(F|Pi,n)which is read as “the present value times the factor to find the future value given the present value, interest rate per period and number of interest periods.” Given the information from this example, the solution for the future value F is as follows:F=$40,000(F/P3,5)=$40,000(1+0.03)=$40,000(1.159274)=$46,370.965This yields an increase of $6,370.96 on the $40,000 investment.This example illustrates a simple way to compare alternatives. Looking for the future value of multiple alternatives, even with different interest rates, allows the comparison of choices based on economic return.Series of Cash FlowsA more realistic technology investment situation involves multiple cash flows. Using present worth (also referred to as net present value), analysis means all cash flows will be converted to their present worth in order to assess the full economic impact of the project. This can be done via the following equation:P=F(1+i)-nIn standard economic analysis notation, this is written=F(P|Fi,n)which is read as “the future value times the factor to find the present value given the future value, interest rate per period, and number of interest periods.” This yields the inverse of the previous equation, reducing a future cash flow by the change in value of the money over the number of interest periods to yield the true present worth.Example 3An off-the-shelf integrated automation system is purchased today for $300,000. It saves a company $40,000 next year, $70,000 the year after, and $20,000 at the end of the third year. In year four, it requires an overhaul of its components that costs $30,000. In year five, the company sells the system for a salvage value of $50,000 (see Fig. 2). Find the present worth of the entire series given an interest rate of 4% annually, compounded once per year.Figure 2Cash flow diagram for description of example 3.View Large Image Figure ViewerDownload (PPT)Using the equationP=F(P|Fi,n)=F(1+i)-nwe sum up all of the cash flows as follows:P=F(P|Fi,n)=F(1+i)-nThis result indicates that the technological investment costs $163,587.37 in today’s money. Normally, to compare all of the possible alternatives for this task we consider costs at present worth and look for the least costly option (without taking into account other issues such as capital expenses, vendor pricing contracts, or tax deductions). Please note that different tables and calculators round differently. Some differences may appear, depending on where the factors are found or which calculator is used.Uniform Series of Cash FlowsA uniform series of cash flows is a situation in which the same amount of money flows in or out over each interest period, either to return a future value or to pay back an initial investment. It can also be used to model consistent costs such as maintenance or labor. Another name for this type of series is the amortization series, which is used to determine monthly payments on home and automobile loans. The equations to deal with this series are presented in the following example.Example 4A small building is purchased for $545,000. If it is financed for 15 years at 7% annual interest, compounded monthly, what is the monthly payment (see Fig. 3)? Use the lender’s point of view (initial amount goes out, payments come in).Figure 3Uniform series cash flow diagram for example 4.View Large Image Figure ViewerDownload (PPT)The equation for finding the present worth for a uniform series of cash flows is given asP=A[(1+i)n-1i(1+i)n]=A(P|Ai,n)The notation at the right-most part of this equation is read as “the amortized amount times the factor to find the present worth given the amortized amount, at i interest per period for n interest periods. To calculate the inverse, use the capital recovery factor for a uniform series of cash flows, which is given asP=A[(1+i)n-1i(1+i)n]=A(P|Ai,n)In our example, an annual interest rate is quoted, but actual interest is compounded monthly. This means the annual interest rate must be converted to a monthly interest rate. Note that the interest given (i) is not the annual percentage rate (APR) from typical home or automobile loan experiences. The equation to convert annual interest to interest per n annual periods is as follows. We also show the equation with substituted values from our example.iperiod=i/nperiods⇒0.07/12=0.005833perperiodClearly, the number of interest periods per year is 12 (for monthly). Substituting iperiod for i in the equations yields the following solution for example 4:A=$545,000[0.0058(1+0.0058)180(1+0.0058)180-1]A $545,000 building would cost $4,886.43 per month with a 15-year loan at 7% annual interest (compounded monthly).Other Cash Flow SeriesIn addition to the uniform series, two other cash flow series are commonly used in economic analysis, the gradient series and the geometric series. The assumption of the gradient series is that the cash flow is increasing linearly (growth). See Figure 4 for the gradient series cash flow diagram. The geometric series assumes that cash flow is increasing geometrically. Equations for the gradient series only are presented in this tutorial since they represent more realistic behavior.Figure 4Cash flow diagram for the gradient series equations.View Large Image Figure ViewerDownload (PPT)Equations for finding the present value and the amortized value for a gradient series are given as follows:P=G[1-(1+ni)(1+i)-ni2]=G(P|Gi,n)Α=G[1i-ni(A|Fi,n)]=G(A|Gi,n)Converting to an effective annual interest rateOne issue that may arise in your analyses is the need to convert from multiple compounding periods per year (m) to an effective annual interest rate (ieff) in order to accurately compare alternatives. This would be like determining the non-fee portion of the APR on a home or automobile loan. The equation for effective annual interest rate is as follows:ieff=(1+r/m)m−1where r = the nominal annual interest rate; m = the number of compounding periods per year; and i = (r/m) = the interest rate per compounding period.Example 5Find the effective annual interest for a home loan of 5% compounded monthly where the quoted annual rate (r) is 5.00% (0.05) and the number of compounding periods per year (m) is 12. From the previous equation, we can calculate the effective annual interest rate as follows:i=r/m=0.05/12=0.00417interest per periodieff=(1+r/m)m-1=(1+0.05/12)12-1=(1.00417)12-1ieff=(1.0512)-1=0.0512or5.12%As shown by the final effective annual interest rate, a 5% nominal annual interest rate compounded monthly is equivalent to an annual compounding rate of 5.12%. The APR quoted on a loan would be 5.12% plus the backed-in measurement of the closing costs and required fees.Inflation and Compounded Interest RatesInflation is a measure of the decrease in the buying power of money due to increasing costs of operations and life. This can be handled in the world of economic analysis by the following equation:i=d+j+djwhere i is the compounded interest rate, d is the base desired interest rate of return or charged, and j is the inflation rate. Following is an example showing how inflation affects interest calculations.Example 6To invest for a 5% return after inflation, where inflation is 1.5%, how much return is needed? Substituting into the previous equation, obtain the following.i=0.05+0.015+(0.05*0.015)=0.065+(0.00075)=0.06575or6.575%A return of 6.575% would be required to yield a 5% post inflation return on this investment.Profitable companies use a number of metrics to help choose the most profitable projects. One way is by designating a Minimum Attractive Rate of Return (MARR). They define this percentage to take into account the company’s cost of capital (the cost of raising funds via the sale of stock or loans), opportunity costs, and inflation while yielding a desired profit. Many profitable manufacturing-based companies use MARRs in the 15% to 20% range. If a company is profitable, then it will want to use the figure determined by its accounting and planning departments. Otherwise, inflation is the predominant factor. The average inflation from the Consumer Price Index (CPI) for the years 2000 through 2003 is 2.5%. Note that this is a historically low rate. Because I have observed that laboratory equipment prices seem to increase a little ahead of the CPI, I use 3% as the inflation rate if my accounting department does not have a number for me to use. Estimates for projects with a 5- or 10-year time span probably should use a higher inflation rate that is more in line with typical rates, such as 5%.Filling Out the Costs of Laboratory Automation and ResearchThe first step to take when performing an economic analysis of laboratory automation is to identify costs.Direct Costs of PeopleThe people who work on a project, in a lab or on a product, must be considered direct costs and must be included in any analysis. Indirect personnel, such as administrative, purchasing, safety, facilities support, top management, are not considered, as they are not changed by the choice of laboratory automation.Table 2 presents a summation of biotech salaries for a number of hotbed locations from Salary.com 4Salary.com; Wellesley, MA; June 2003.Google Scholar for U.S. jobs and WetFeet.com 5Wetfeet.com; San Francisco, CA; June 2003.Google Scholar,6SalaryExpert.com; Baker, Thomsen Associates Insurance Services, Inc.; Vancouver, WA.Google Scholar for European jobs, based on 2002 data. Even though the number may not be completely accurate, a 2080-hour work year can be used to determine hourly costs. Please note that all of these figures are based on incomes prior to this current year. The U.S. is experiencing downward pressure on salaries due to a large number of lay-offs in the pharma and biotech sectors; therefore, scientist salaries appear inflated. An individual company’s accounting or planning departments should be able to provide more accurate numbers to use in cost calculations.Table 2Yearly and hourly salaries for a number of titles and locationsMedian Salaries – Yearly and (hourly – rounded) in $U.S., based on 2002 data.Note: these figures do not include benefits.Job title and years experienceSan Diego, CASan Francisco, CABoston, MAEngland (Average)Switzerland (Average)Engineer II 2–5 years$65,549 ($31)$72,993 ($35)$68,051 ($33)Engineer III 5–8 years$78,960 ($38)$87,926 ($42)$81,974 ($39)Research Associate I 0–2 years$38,457 ($18)$42,824 ($20)$39,925 ($19)Research Associate II 2–5 years$47,394 ($23)$52,046 ($25)$48,523 ($23)Research Associate III 5+ years$57,886 ($28)$64,458 ($31)$60,095 ($29)Scientist II III all all all all Open table in a new tab and to these From with a of in the of is a to determine the realistic direct costs to a In other of the is a more accurate In the cost can be than a factor of Ideally, the personnel, or accounting departments should provide an accurate Costs of is a of equipment price based on U.S. This can be very different in or especially for through a These are considered direct off-the-shelf integrated contracts average of the purchase price per Costs of of laboratory automation do not consider related this may be acceptable for and their associated costs must be into account for large integrated If the system does not come with for personnel, the company must For the the best is the standard for and of This is available from the Scholar This studies and new It is the for the U.S. and is often used by the & It how to and possible by using and has other – for Indirect Costs of even more cost is that of as or by as U.S. of & Scholar and by other these are a common problem in laboratory equipment is especially for the cost of a is Therefore, for a that is work should focus on as as possible that are the least to to the of is that can cost a company in time throughout the organization with From studies at a number of year it can be that one out of five on a (i.e., task will have this in the an Automation & June Scholar in a at a but it is one of that companies will This cost must be included in the cost for people, especially for of can be the most part of any or These costs scale with the number of or and therefore are variable costs. As research in or is reagents can be very a example of this is the used to a The that are often are costly and can from to to these costs are the costs involved in of these consumables are often and disposal can be very Some can be more to of as than they to Even with a for and have costs associated with Even disposal has a a company’s accounting department or disposal can provide an average to account for costs associated with of and even there can be other consumables such as and these with or can be by the equipment that is purchased once and then One of the of a laboratory process is to the cost of usually by reducing the of one of the of laboratory automation is that of reducing of and are to between and data. If then current can be In companies and research this number is more difficult to than it is to One determined that its rate for a process to of the the this to the quality and quality of the reagents this to In reducing this costs also and 3% of manufacturing can meaning almost This is essentially for projects. not is the more will the following for and operations manual especially due to or of Automation but much less They do affect more when they Automation system very rarely but affect a large number of very create To cost for a process or need to be by one plus the process costs a projected per to run in consumables With an rate of how much would it cost in consumables to all of the A rate means more must be using of Therefore, the actual cost of the is as we can that rate is an of is the percentage of time that a of equipment or is capable of performing its There is an entire of engineering to especially at the is the time needed for maintenance and laboratory automation equipment do not these the best is to for To determine a current can on how a their equipment less and maintenance time. that most are only while people are can be on a for The will be used in a One research will be able to and support two The cost per run in and is plus an for and two per a per is the present worth cost per for a year useful life of the per in an interest rate of 3% compounded To solve this problem, first convert costs to an annual The costs of reagents per year is the information we use + for the plus Then, the costs of personnel per year is cash flow diagram for this problem is depicted in Figure flow diagram for example Large Image Figure ViewerDownload for equivalent present worth (P) and then by the number of per years,

Does government debt affect interest rates? Despite a substantial body of empirical analysis, the answer based on the past two decades of research is mixed. While many studies suggest, at most, a single-digit rise in the interest rate when government debt increases by 1% of gross domestic product (GDP), others estimate either much larger effects or find no effect. Comparing results across studies is complicated by differences in economic models, definitions of government debt and interest rates, econometric approaches, and sources of data. Using a standard set of data and a simple analytical framework, we reconsider and add to empirical evidence about the effect of federal government debt and interest rates. We begin by deriving analytically the effect of government debt on the real interest rate and find that an increase in government debt equivalent to 1% of GDP would be predicted to increase the real interest rate by about two to three basis points. While some existing studies estimate effects in this range, others find larger effects. In almost all cases, these larger estimates come from specifications relating federal deficits (as opposed to debt) and the level of interest rates or from specifications not controlling adequately for macroeconomic influences on interest rates that might be correlated with deficits. We present our own empirical analysis in two parts. First, we examine a variety of conventional reduced-form specifications linking interest rates and government debt and other variables. In particular, we provide estimates for three types of specifications to permit comparisons among different approaches taken in previous research; we estimate the effect of an expected, or projected, measure of federal government debt on a forward-looking measure of the real interest rate; an expected, or projected, measure of federal government debt on a current measure of the real interest rate; and a current measure of federal government debt on a current measure of the real interest rate. Most of the statistically significant estimated effects are consistent with the prediction of the simple analytical calculation. Second, we provide evidence using vector autoregression analysis. In general, these results are similar to those found in our reduced-form econometric analysis and are consistent with the analytical calculations. Taken together, the bulk of our empirical results suggests that an increase in federal government debt equivalent to 1% of GDP, all else being equal, would be expected to increase the long-term real rate of interest by about three basis points (though one specification suggests a larger impact), while some estimates are not statistically significantly different from zero. By presenting a range of results with the same data, we illustrate the dependence of estimation on specification and definition differences.

Polymerization of acrylonitrile in homogeneous medium in dimethyl formamide at 50°C and 60°C (to be inserted in all places where temperatures are mentioned) in precipitation medium in toluene at 50°C and 60°C, and in bulk at 35° to 55°C, all initiated by α,α′‐azobis‐isobutyronitrile has been studied. Transfer constants with their activation energies for polyacrylonitrile radicals and dimethyl formamide or toluene have been evaluated. By measurements of rates and degrees of polymerization, rates of initiation have been evaluated in homogeneous and precipitation media. Values of K, A′,kp/k and their respective activation energies have been evaluated. The abnormalities of exponents for initiator and monomer concentrations in the rate equation are discussed. The applicability of normal homogeneous medium kinetics is examined. Apparent values of kp/k in bulk have been evaluated at temperatures ranging from 35°C to 55°C.

The rapid growth of the organometallic chemistry of the transition metals during the last 15–20 years owes much to the development of homogeneous catalyst systems which are capable of synthesizing organic molecules under mild conditions and occasionally with remarkable selectivities. Several have been commercialized and are now used on a large scale. (1) A few have received considerable detailed study—including spectroscopic identification of the species present in solution under reaction conditions, isolation of reactive intermediates in some cases, determination of the overall rate law and measurement of rate and equilibrium constants of several individual steps, and isotopic labeling studies—so that we have a reasonably clear picture of how they operate. It is these systems that form the focus of this chapter. For more general reviews the reader is referred to some recent books,(1-3) which also discuss the special electronic properties of transition metals which are in part responsible for their catalytic behavior.†

Rapid-equilibrium rate equations for the enzymatic catalysis of A + B = P + Q over a range of pH