Reactions half-time

P3.05.10. HALF TIME. REACTION PITH VARIABLE DENSITY... 

Table 1. Regioselectivity and half-time reaction for 2-cyclohexenone reduction...

It was pointed out that a bimolecular reaction can be accelerated by a catalyst just from a concentration effect. As an illustrative calculation, assume that A and B react in the gas phase with 1 1 stoichiometry and according to a bimolecular rate law, with the second-order rate constant k equal to 10 1 mol" see" at 0°C. Now, assuming that an equimolar mixture of the gases is condensed to a liquid film on a catalyst surface and the rate constant in the condensed liquid solution is taken to be the same as for the gas phase reaction, calculate the ratio of half times for reaction in the gas phase and on the catalyst surface at 0°C. Assume further that the density of the liquid phase is 1000 times that of the gas phase. 

The only kinetic data reported are in a Ph.D. thesis (41). Integral order kinetics were usually not obtained for the reaction of a number of ketones with piperidine and a number of secondary amines with cyclohexanone. A few of the combinations studied (cyclopentanone plus piperidine, pyrrolidine, and 4-methylpiperidine, and N-methylpiperazine plus cyclohexanone) gave reactions which were close to first-order in each reactant. Relative rates were based on the time at which a 50% yield of water was evolved. For the cyclohexanone-piperidine system the half-time (txn) for the 3 1 ratio was 124 min and for the 1 3 ratio 121 min. It appears that an... 

FIGURE 14.4 Plot of the course of a first-order reaction. The half-time, <1/9, is the time for one-half of the starting amonnt of A to disappear. 

Hydroxy-8-azapurine was shown by rapid-reaction techniques (see Section II, E) to be anhydrous in the anion and hydrated in the neutral species. The hydration reaction has a half-time of about 0.5 second, which is too rapid for exact measurements with existing apparatus. The cation of 2-amino-8-azapurine was shown to have an anomalous value and ultraviolet spectrum, although its 6-methyl derivative is quite normal. Hydration in this case proved to be too fast to register in the rapid-reaction apparatus. 

Although it is attractive to directly convert chemical energy to electricity, PEM fuel cells face significant practical obstacles. Expensive heavy metals like platinum are typically used as catalysts to reduce energy barriers associated with the half-cell reactions. PEM fuel cells also cannot use practical hydrocarbon fuels like diesel without complicated preprocessing steps. Those significantly increase the complexity of the overall system. At this time, it appears likely that PEM fuel cells will be confined to niche applications where high cost and special fuel requirements are tolerable. 

A further factor which must also be taken into consideration from the point of view of the analytical applications of complexes and of complex-formation reactions is the rate of reaction to be analytically useful it is usually required that the reaction be rapid. An important classification of complexes is based upon the rate at which they undergo substitution reactions, and leads to the two groups of labile and inert complexes. The term labile complex is applied to those cases where nucleophilic substitution is complete within the time required for mixing the reagents. Thus, for example, when excess of aqueous ammonia is added to an aqueous solution of copper(II) sulphate, the change in colour from pale to deep blue is instantaneous the rapid replacement of water molecules by ammonia indicates that the Cu(II) ion forms kinetically labile complexes. The term inert is applied to those complexes which undergo slow substitution reactions, i.e. reactions with half-times of the order of hours or even days at room temperature. Thus the Cr(III) ion forms kinetically inert complexes, so that the replacement of water molecules coordinated to Cr(III) by other ligands is a very slow process at room temperature. 

RELATIVE RATES OF REACTION OF Ac20 WITH R CjH4FeCjHs IN CH2C12 CATALYSED BY BF3Et20 AT 0 AND 25 °C (HALF TIME AT EACH TEMPERATURE)431... 

Let us now assume that these matters have been attended to properly. At this stage we can but assume that the reaction orders were correctly identified and correct mathematical procedures followed. During the course of the work, the investigator should make the occasional quick calculation to show the values are roughly correct. (Does the rate constant yield the correct half-time ) Also, one should examine the experimental data fits to see that the data really do conform to the selected rate equation. Deviations signal an incorrect rate law or complications, such as secondary reactions. 

The half-time (or half-life) of the reaction is independent of [A]o. The reciprocal of the rate constant, t = l/k, is referred to as the lifetime or the mean reaction time. In that time [A] falls to l/e of its initial value. The pharmaceutical industry refers to the shelf life or t90, the time at which [A]/[A]o reaches 0.90. Both t and t90 are also independent of [A]o. 

The logarithmic plot is not linear, of course, since this is not a first-order reaction. Note, however, that even In [A], is linear in time to about 50 percent reaction. One cannot use these procedures to establish the kinetic order without data taken to at least two half-times, and preferably longer. 

The time required to convert a given fraction of the limiting reagent is a characteristic of the rate equation. A comparison of successive half-times, or any other convenient fractional time, reveals whether a reaction follows any simple-order rate law. Thus, the ratio of the time to reach 75 percent completion to that for 50 percent is characteristic of the reaction order. Values of this ratio for different orders are as follows ... 

Second-order kinetics, (a) Derive expressions for the half-time and lifetime of A if the rate law for its disappearance is v = fc[A]2 (b) calculate t]/i and t for the data presented in Section 2.2 (c) calculate the second half-life, t /i(2), i.e., the time elapsed between 50 percent and 75 percent completion, for the same reaction (d) compare fj/2(l) and fi/>(2), and contrast this result with that from first-order kinetics. 

Opposing reactions. Calculate half-times for equilibration in the triphenyl methyl system starting with all A or with the equilibrated mixture, for the conditions given in Table 3-3. Use algebraic equations, not the tabulated numerical values. Compare the latter with the t fi from the approximate solution given in Eq. (3-39). Compare the values of 4AT-15o and a, to assess whether Eq. (3-39) provides an adequate representation. 

Consecutive reactions. In the sequence of Eq. (4-1), prove that the half-time for the disappearance of A occurs at fmax for I when k = a 2. What is the numerical value of a ... 

The first of these reactions was carried out in 1,4-cyclohexadiene over a temperature range of 39 to 100 °C. It is fairly slow the half-times were 20 h and 3.4 min at the extremes. Reaction (7-11) is quite fast the second-order rate constant, kn, was evaluated over the range 6.4 to 47.5 °C. Values of feio and fen are presented in Table 7-1. The temperature profiles are depicted in Fig. 7-1 from their intercepts and slopes the activation parameters can be obtained. A nonlinear least-squares fit to Eq. (7-1) or... 

In this connection it seems odd that no product other than the parent compounds occurs in (CH3)2Hg and (C2H5)2Hg . Here one would be tempted to suggest that the other expectable products have been masked through exchange reactions. This, however, seems not to be the case, since the known half times for the exchange reactions are too long to be involved here. 

Under conditions where the dismutation reaction is slow the exchange between Au(III) and Au(I) has been shown to proceed at a measurable rate at 0 °C in 0.09 M HCl, an exchange half-time of about 2 min was observed. The isotopic method ( Au) and a separation method based on the precipitation of dipyridine -chloroaurate(III) was used to obtain data. An acceleration in the exchange rate was observed as the HCl concentration was increased. ... 

In perchlorate media the latter reaction was found to be rapid (100 % exchange in < 60 sec) and the former reaction slow (half times of > 200 hours at 100 °C). The isotopic method was used ( Am). 

The effective half-time for this reaction at room temperature is 9 min in comparison with 79 min without BF3-etherate.[139]... 

The migration of heavy metals into mineral lattices is very slow. Ahnstrom and Parker (2001) observed the slow migration of luCd into the residual fraction in arid soils with a weeks-to-years reaction half-time. Theoretically, the residual fraction is comprised of very refractory Cd bound in the lattices of aluminosilicate minerals. Lattice diffusion, a process necessary for isotopic exchange, can require years. 

Over zinc oxide it is clear that only a limited number of sites are capable of type I hydrogen adsorption. This adsorption on a Zn—O pair site is rapid with a half-time of less than 1 min hence, it is fast enough so that H2-D2 equilibration (half-time 8 min) can readily occur via type I adsorption. If the active sites were clustered, one might expect the reaction of ethylene with H2-D2 mixtures to yield results similar to those obtained for the corresponding reaction with butyne-2 over palladium That is, despite the clean dideutero addition of deuterium to ethylene, the eth-... 

One consequence of a first-order reaction is that it takes a constant amount of time for half the remaining substrate to be converted to product—regardless of how much of the reactant is present. It takes the same amount of time to convert 100,000 A molecules to 50,000 P molecules as it takes to convert 10 A molecules to 5 P s. A first-order reaction has a constant half-time t1/2. 

All that arithmetic is just to show you that the half-time for a first-order reaction depends only on k, not on how much A you have to start with. The whole point is that the bigger the k, the shorter the half-time, the faster the reaction. 

This process of creating ATP, known as electron transport phosphorylation, then, involves two half-cell reactions, one at the electron donation site and the other where the electrons are accepted from the transport chain. Taking aerobic sulfide oxidation as an example, the donating species H2S(aq) gives up electrons, two at a time, to a series of redox complexes. With the loss of each pair of electrons, the sulfide oxidizes first to S°, then thiosulfate, sulfite, and finally sulfate. 

In homogeneous catalysis, the quantification of catalyst activities is commonly carried out by way of TOF or half-life. From a kinetic point of view, the comparison of different catalyst systems is only reasonable if, by giving a TOF, the reaction is zero order or, by giving a half-time, it is a first-order reaction. Only in those cases is the quantification of activity independent of the substrate concentration utilized ... 

A sample of iodine-128 was produced in a Szilard-Chalmers reaction by irradiating a sample of ethyl iodide in a neutron source. The radio-iodine was extracted with sodium thiosulfate solution and then counted in a Geiger counter at various time Intervals. Use the tabulated data of t in minutes against C counts/min to find the rate equation and the half time. 

From the tabulated half time and decomposition of tetrahydrofuran (JACS 68 reaction and the Arrhenius parameters. 

Data of initial pressure or concentration and half times are given in the several parts of this problem. From problem P3.05.01, such data are related to the specific rate and reaction order by... 

Since the half time is substantially constant, the reaction order must be one. 

Given the half time and initial total pressure data with amounts of the reactants, find the overall order of the reaction,... 

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